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Chapter 3

Physics: Mechanics and Properties of Matter

TU BSc CSIT entrance past exam questions

Past questions and answers

316 questions set from this chapter, 1 of them more than once; 3 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 13 exams
  • 2075 · 1 mark

Light of wavelength 550 nm falls normally on a slit of width 22×10−722\times10^{-7}m, the angular position of second minima from central maxima will be (a) 14.5∘14.5^\circ (b) 30∘30^\circ (c) 42∘42^\circ (d) 62∘62^\circ

Similar questions: Light of wavelength 550 nm falls normally on a slit of... (2077) · Light of wavelength 550 nm falls normally on a slit of... (2079)

Answer

(b) 30∘30^\circ

For the nn-th minimum of a single slit, asin⁡θ=nλa\sin\theta=n\lambda.

sin⁡θ=2×550×10−922×10−7=0.5⇒θ=30∘\sin\theta=\frac{2\times550\times10^{-9}}{22\times10^{-7}}=0.5\Rightarrow\theta=30^\circ

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  • Most repeated · 3 of 13 exams
  • 2077 · 1 mark

Light of wavelength 550 nm falls normally on a slit of width 22×10−522\times10^{-5} m, the angular position of second minima from central maxima will be (a) 14.5∘14.5^\circ (b) 30∘30^\circ (c) 42∘42^\circ (d) 62∘62^\circ

Similar questions: Light of wavelength 550 nm falls normally on a slit of... (2075) · Light of wavelength 550 nm falls normally on a slit of... (2079)

Answer

(b) 30∘30^\circ

For the second minimum, asin⁡θ=2λa\sin\theta=2\lambda. Taking the slit width as a=2.2×10−6a=2.2\times10^{-6} m (the only value that gives one of the listed angles):

sin⁡θ=2×550×10−92.2×10−6=0.5⇒θ=30∘\sin\theta=\frac{2\times550\times10^{-9}}{2.2\times10^{-6}}=0.5\Rightarrow\theta=30^\circ

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 13 exams
  • 2079 · 1 mark

Light of wavelength 550 nm falls normally on a slit of width 22×10−722\times10^{-7}m, the angular position of second minima form central maxima will be (a) 14.5∘14.5^\circ (b) 42∘42^\circ (c) 30∘30^\circ (d) 62∘62^\circ

Similar questions: Light of wavelength 550 nm falls normally on a slit of... (2075) · Light of wavelength 550 nm falls normally on a slit of... (2077)

Answer

(c) 30∘30^\circ

For minima, asin⁡θ=nλa\sin\theta=n\lambda. For n=2n=2:

sin⁡θ=2×550×10−922×10−7=11002200=0.5\sin\theta=\frac{2\times550\times10^{-9}}{22\times10^{-7}}=\frac{1100}{2200}=0.5

So θ=30∘\theta=30^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2078 · 1 mark
  • 2079 · 1 mark

The mean free path of a gas varies with pressure (p) as (a) P (b) P−1P^{-1} (c) P2P^2 (d) P−2P^{-2}

Answer

(b) P−1P^{-1}

Mean free path λ=kT2πd2P∝1P\lambda=\frac{kT}{\sqrt2\pi d^2P}\propto\frac1P.

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  • 2074 · 1 mark

The self-inductance of coil is 10mH. If a current of 2A is flown. What is the magnetic flux through the coil? (a) 0.02 Wb (b) 0.01Wb (c) 0.05Wb (d) 0.10Wb

Similar questions: The self-inductance of a coil is 5 mH. If current of 2A is... (2078)

Answer

(a) 0.02 Wb

Flux linkage Φ=LI=10×10−3×2=0.02\Phi=LI=10\times10^{-3}\times2=0.02 Wb.

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  • 2075 · 1 mark

If length of wire is doubled keeping the diameter constant, it's Young's modulus will (a) Increase (b) decreases (c) remain same (d) depend upon nature of matter

Similar questions: If length of wire is doubled keeping the diameter constant,... (2077)

Answer

(c) remain same

Young's modulus is a property of the material only. It does not depend on the length or the diameter of the wire.

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  • 2075 · 1 mark

A sound wave has frequency 500 HZ and velocity 360m/s. What is the distance between 2 particles having phase difference 60∘60^\circ (a) 0.7 cm (b) 1.2 cm (c) 70 cm (d) 12 cm

Similar questions: A sound wave has frequency 500 Hz and velocity 360 m/s.... (2077)

Answer

(d) 12 cm

λ=vf=360500=0.72\lambda=\frac vf=\frac{360}{500}=0.72 m.

Δx=λ2πΔϕ=0.726=0.12 m=12 cm\Delta x=\frac{\lambda}{2\pi}\Delta\phi=\frac{0.72}{6}=0.12\ \text{m}=12\ \text{cm}

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  • 2075 · 1 mark

n-equal capacitors are first connected in series and then in parallel. The ratio of maximum to minimum capacitance. (a) n2n^2 (b) 1/n21/n^2 (c) nn (d) 1/n1/n

Similar questions: n equal capacitors are first connected in series and then... (2077)

Answer

(a) n2n^2

Series: Cmin=C/nC_{min}=C/n. Parallel: Cmax=nCC_{max}=nC.

CmaxCmin=nCC/n=n2\frac{C_{max}}{C_{min}}=\frac{nC}{C/n}=n^2

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  • 2075 · 1 mark

Whenever the magnetic flux linked with a coil changes, there is emf induced in the circuit the emf last (a) for a short time (b) for a long time (c) forever (d) so long as the change in flux takes place

Similar questions: Whenever the magnetic flux linked with a coil changes,... (2081)

Answer

(d) so long as the change in flux takes place

By Faraday's law e=−dΦdte=-\frac{d\Phi}{dt}, so the emf exists only while the flux is changing.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2075 · 1 mark

The kinetic energy carried by emitted electron depends on (a) Frequency (b) intensity (c) work function (d) momentum

Similar questions: The kinetic energy carried by emitted electron depends on (2081)

Answer

(a) Frequency

From Einstein's photoelectric equation KEmax=hf−ϕKE_{max}=hf-\phi, the kinetic energy depends on the frequency of the incident light (and the work function of the metal), not on its intensity.

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  • 2075 · 1 mark

A nucleus reactor has a power of 16KW if the energy per fission is 200 MeV then number of fission per second is (a) 5×10175\times10^{17} (b) 5×10165\times10^{16} (c) 5×10155\times10^{15} (d) 5×10145\times10^{14}

Similar questions: A nucleus reactor has a power of 10 kw if the energy per... (2081)

Answer

(d) 5×10145\times10^{14}

Energy per fission =200×106×1.6×10−19=3.2×10−11=200\times10^6\times1.6\times10^{-19}=3.2\times10^{-11} J.

n=16×1033.2×10−11=5×1014 s−1n=\frac{16\times10^3}{3.2\times10^{-11}}=5\times10^{14}\ \text{s}^{-1}

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  • 2077 · 1 mark

The horizontal range is 434\sqrt3 times the maximum height in a projectile motion. The angle of projectile is (a) 15∘15^\circ (b) 45∘45^\circ (c) 30∘30^\circ (d) 60∘60^\circ

Similar questions: The horizontal range is 4 3 times the maximum height in a... (2081)

Answer

(c) 30∘30^\circ

RH=u2sin⁡2θ/gu2sin⁡2θ/2g=4cot⁡θ=43⇒tan⁡θ=13\frac{R}{H}=\frac{u^2\sin 2\theta/g}{u^2\sin^2\theta/2g}=4\cot\theta=4\sqrt3\Rightarrow\tan\theta=\frac1{\sqrt3}

So θ=30∘\theta=30^\circ.

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  • 2077 · 1 mark

If length of wire is doubled keeping the diameter constant, it's Young's modulus will (a) increases (b) decreases (c) remain same (d) depend upon nature of matter

Similar questions: If length of wire is doubled keeping the diameter constant,... (2075)

Answer

(c) remain same

Young's modulus is a property of the material. It does not depend on the length or the diameter of the wire.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

The work done to blow a soap bubble of radius 'R' is W, then work done to increase the radius from R to 3R is (a) 2W (b) 8W (c) 4W (d) 9W

Similar questions: The work done to blow a soap bubble of radius 'R' is W,... (2081)

Answer

(b) 8W

A soap bubble has two surfaces, so the work done is W=2T×4πR2=8πTR2W=2T\times4\pi R^2=8\pi TR^2.

Increasing the radius from RR to 3R3R:

W′=8πT(9R2−R2)=8×(8πTR2)=8WW'=8\pi T(9R^2-R^2)=8\times(8\pi TR^2)=8W

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  • 2077 · 1 mark

Latent heat of a substance is zero at (a) boiling (b) critical temperature (c) melting point (d) freezing point

Similar questions: Latent heat of substance is zero at (2081)

Answer

(b) critical temperature

At the critical temperature the liquid and vapour phases become indistinguishable, so the latent heat of vaporisation is zero.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

A sound wave has frequency 500 Hz and velocity 360 m/s. What is the distance between 2 particles having phase difference 60∘60^\circ? (a) 0.7 cm (b) 70 cm (c) 3.2 cm (d) 12 cm

Similar questions: A sound wave has frequency 500 HZ and velocity 360m/s. What... (2075)

Answer

(d) 12 cm

λ=vf=360500=0.72 m\lambda=\frac vf=\frac{360}{500}=0.72\text{ m}

Δx=λ2πΔϕ=0.72360∘×60∘=0.12 m=12 cm\Delta x=\frac{\lambda}{2\pi}\Delta\phi=\frac{0.72}{360^\circ}\times60^\circ=0.12\text{ m}=12\text{ cm}

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  • 2077 · 1 mark

n equal capacitors are first connected in series and then in parallel. The ratio of maximum to minimum capacitance is (a) n2n^2 (b) 1n2\frac{1}{n^2} (c) nn (d) 1n\frac1n

Similar questions: n-equal capacitors are first connected in series and then... (2075)

Answer

(a) n2n^2

Series: Cmin=C/nC_{min}=C/n. Parallel: Cmax=nCC_{max}=nC.

CmaxCmin=nCC/n=n2\frac{C_{max}}{C_{min}}=\frac{nC}{C/n}=n^2

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  • 2077 · 1 mark

A magnetic needle kept in a high-uniform magnitude field. It experiences (a) a torque but not a force (b) a force and a torque (c) neither a force nor a torque (d) a force but not a torque

Similar questions: A magnetic needle kept in a non-uniform magnetic field. It... (2079)

Answer

(a) a torque but not a force

In a uniform field the forces on the two poles are equal and opposite, so the net force is zero. They form a couple that gives a torque τ=MBsin⁡θ\tau=MB\sin\theta.

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  • 2077 · 1 mark

In LCR circuit, the inductive reactance at resonance frequency is 100 Ω\Omega and resistance is 5 Ohm, the quality factor of the circuit is (a) 3000 (b) 500 (c) 20 (d) 95

Similar questions: In LCR circuit, the inductive reactance at resonance... (2081)

Answer

(c) 20

Q=XLR=1005=20Q=\frac{X_L}{R}=\frac{100}{5}=20

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  • 2077 · 1 mark

The half-life of a radioactive sample is 10 years. Its mean life is (a) 12.43 years (b) 15.43 years (c) 14.43 years (d) same as half life

Similar questions: The half-life of radioactive sample is 10 years. It means... (2081)

Answer

(c) 14.43 years

τ=T1/2ln⁡2=1.443×10=14.43 years\tau=\frac{T_{1/2}}{\ln2}=1.443\times10=14.43\text{ years}

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  • 2077 · 1 mark

NPN transistors are most preferred than that of PNP transistor. It is because of (a) low cost (b) capable of handling low power (c) low dissipation of energy (d) high mobility of electrons that holes

Similar questions: NPN transistors are most preferred than that of PNP... (2081)

Answer

(d) high mobility of electrons than holes

In an NPN transistor the majority carriers are electrons, whose mobility is higher than that of holes. This gives faster response and better performance.

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  • 2078 · 1 mark

The self-inductance of a coil is 5 mH. If current of 2A is flown, what will be the magnetic flux through the coil? (a) 1.000 weber (b) 0.100 weber (c) 0.010 weber (d) 0.001 weber

Similar questions: The self-inductance of coil is 10mH. If a current of 2A is... (2074)

Answer

(c) 0.010 weber

Φ=LI=5×10−3×2=0.010 Wb\Phi=LI=5\times10^{-3}\times2=0.010\text{ Wb}.

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  • 2079 · 1 mark

A vector is multiplied by -2 then (a) direction reverses and unit changes (b) direction reverses and magnitude is doubled (c) direction remains unchanged but unit changes (d) neither direction reverses nor unit changes but magnitude is doubled.

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Answer

(b) direction reverses and magnitude is doubled

Multiplying a vector by a negative scalar reverses its direction. Multiplying by −2-2 makes the magnitude 22 times as large. The unit does not change.

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  • 2079 · 1 mark

A magnetic needle kept in a non-uniform magnetic field. It experiences (a) Torque but not a force (b) Neither a force nor a torque (c) Force and a torque (d) A force but not a torque

Similar questions: A magnetic needle kept in a high-uniform magnitude field.... (2077)

Answer

(c) Force and a torque

In a non-uniform field the two poles experience unequal forces, so there is a net force. The forces are not along the same line, so there is also a torque.

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  • 2081 · 1 mark

A vector is multiplied by 2, then (a) direction reverses and unit change. (b) direction reverses and magnitude is doubled. (c) direction remains unchanged but unit changes. (d) neither direction reverses nor unit changes but magnitude is doubled

Similar questions: A vector is multiplied by -2 then (2079)

Answer

(d) neither direction reverses nor unit changes but magnitude is doubled

Multiplying a vector by a positive scalar changes only its magnitude (doubled); direction and unit stay the same.

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  • 2081 · 1 mark

The horizontal range is 434\sqrt3 times the maximum height in a projectile motion. The angle of projection is (a) 15∘15^\circ (b) 30∘30^\circ (c) 45∘45^\circ (d) 60∘60^\circ

Similar questions: The horizontal range is 4 3 times the maximum height in a... (2077)

Answer

(b) 30∘30^\circ

RH=2u2sin⁡θcos⁡θ/gu2sin⁡2θ/2g=4tan⁡θ=43\frac{R}{H}=\frac{2u^2\sin\theta\cos\theta/g}{u^2\sin^2\theta/2g}=\frac{4}{\tan\theta}=4\sqrt3, so tan⁡θ=13\tan\theta=\frac1{\sqrt3} and θ=30∘\theta=30^\circ.

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  • 2081 · 1 mark

The work done to blow a soap bubble of radius 'R' is W, then work done to increase radius from R to 3R is (a) 2W2W (b) 4W4W (c) 8W8W (d) 9W9W

Similar questions: The work done to blow a soap bubble of radius 'R' is W,... (2077)

Answer

(c) 8W8W

A bubble has two surfaces, so W=T⋅2⋅4πR2=8πR2TW=T\cdot2\cdot4\pi R^2=8\pi R^2T. From RR to 3R3R: 8πT(9R2−R2)=8W8\pi T(9R^2-R^2)=8W.

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  • 2081 · 1 mark

Latent heat of substance is zero at (a) boiling point (b) melting point (c) critical temperature (d) freezing point

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Answer

(c) critical temperature

At the critical temperature the liquid and vapour phases become indistinguishable, so there is no latent heat.

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  • 2081 · 1 mark

Whenever the magnetic flux linked with a coil changes, there is emf induced in a circuit. The emf lasts (a) for a short time (b) forever (c) for a long time (d) so long as the change in flux takes place

Similar questions: Whenever the magnetic flux linked with a coil changes,... (2075)

Answer

(d) so long as the change in flux takes place

By Faraday's law e=−dϕdte=-\frac{d\phi}{dt}, so emf exists only while the flux is changing.

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  • 2081 · 1 mark

In LCR circuit, the inductive reactance at resonance frequency is 100Ω100\Omega and resistance is 5 Ω5\ \Omega the quality factor of circuit is. (a) 5000 (b) 20 (c) 500 (d) 95

Similar questions: In LCR circuit, the inductive reactance at resonance... (2077)

Answer

(b) 20

At resonance XL=XC=100 ΩX_L=X_C=100\,\Omega, so Q=XLR=1005=20Q=\frac{X_L}{R}=\frac{100}{5}=20.

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  • 2081 · 1 mark

The kinetic energy carried by emitted electron depends on (a) frequency (b) work carried (c) intensity (d) momentum

Similar questions: The kinetic energy carried by emitted electron depends on (2075)

Answer

(a) frequency

Kmax=hν−ϕK_{max}=h\nu-\phi, so the kinetic energy depends on the frequency of light (not its intensity).

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  • 2081 · 1 mark

A nucleus reactor has a power of 10 kw if the energy per fission is 200 Mev then the number of fission per second is (a) 5×10175\times10^{17} (b) 5×10155\times10^{15} (c) 5×10165\times10^{16} (d) 5×10145\times10^{14}

Similar questions: A nucleus reactor has a power of 16KW if the energy per... (2075)

Answer

(d) 5×10145\times10^{14}

Number per second =PE=104200×106×1.6×10−19≈3.1×1014=\frac{P}{E}=\frac{10^4}{200\times10^6\times1.6\times10^{-19}}\approx3.1\times10^{14}. The nearest option is 5×10145\times10^{14} (obtained by taking 1 eV≈10−191\,eV\approx10^{-19} J).

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  • 2081 · 1 mark

The half-life of radioactive sample is 10 years. It means life is (a) 12.43 years (b) 14.43 years (c) 16.43 years (d) Same as half life

Similar questions: The half-life of a radioactive sample is 10 years. Its mean... (2077)

Answer

(b) 14.43 years

Mean life τ=T1/20.693=1.443×10=14.43\tau=\frac{T_{1/2}}{0.693}=1.443\times10=14.43 years.

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  • 2081 · 1 mark

NPN transistors are most preferred than that of PNP transistor. It is because of (a) Low Cost (b) Low dissipation of energy (c) capable of handling low power (d) Electrons have high mobility than holes

Similar questions: NPN transistors are most preferred than that of PNP... (2077)

Answer

(d) Electrons have high mobility than holes

In an NPN transistor the majority carriers are electrons, which are more mobile than holes, giving faster response.

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  • 2069 · Entrance Examination 2069 · 1 mark

The duration in which the displacement becomes half the amplitude in the case of a simple harmonic motion if its period is 12 second, is (a) 6 sec (b) 3 sec (c) 2 sec (d) 1 sec

Answer

(d) 1 sec

For SHM starting from the mean position, x=Asin⁡(ωt)x=A\sin(\omega t) When displacement becomes half the amplitude, x=A2⇒sin⁡(ωt)=12x=\frac{A}{2}\Rightarrow \sin(\omega t)=\frac12 ωt=π6\omega t=\frac{\pi}{6} Since ω=2πT\omega=\frac{2\pi}{T}, t=T12=1212=1 sect=\frac{T}{12}=\frac{12}{12}=1\text{ sec}

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  • 2069 · Entrance Examination 2069 · 1 mark

A train of length 600 m is travelling with a velocity of 64 km/hr. The time taken by the train to cross clearly a bridge of 1 km will be (a) 1.5 sec (b) 1.5 min (c) 15 sec (d) 15 min

Answer

(b) 1.5 min

Total distance to cross the bridge = 600 m + 1000 m = 1600 m.
Speed = 64 km/hr = 64×10003600=1609\frac{64 \times 1000}{3600} = \frac{160}{9} m/s.
Time = 1600160/9=90\frac{1600}{160/9} = 90 s = 1.5 min.

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  • 2069 · Entrance Examination 2069 · 1 mark

Which one of the following pairs of physical quantities has the same dimension? (a) Work and power (b) Energy and power (c) Work and energy (d) Angular momentum and linear momentum

Answer

(c) Work and energy

Work and energy both have the same dimensional formula, ([ML^2T^{-2}]). Power has ([ML^2T^{-3}]), while angular momentum and linear momentum have different dimensions.

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  • 2069 · Entrance Examination 2069 · 1 mark

A body of mass 1 kg is rotated in a horizontal circle of radius 50 cm. If F and v represent the centripetal force and the constant speed respectively, which one of the following expressions may be correct? (a) v=Fv = \sqrt{F} (b) v=2Fv = \sqrt{2F} (c) v=12Fv = \frac{1}{2}\sqrt{F} (d) V=F2V = \sqrt{\frac{F}{2}}

Answer

(d) V=F2V = \sqrt{\frac{F}{2}}

Centripetal force is
F=mv2rF=\frac{mv^2}{r} Given m=1 kgm=1\,\text{kg} and r=50 cm=0.5 mr=50\,\text{cm}=0.5\,\text{m}, F=v20.5=2v2F=\frac{v^2}{0.5}=2v^2 so v=F2v=\sqrt{\frac{F}{2}}

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  • 2069 · Entrance Examination 2069 · 1 mark

The root mean square (rms) speed of hydrogen molecule at STP is 2 km per second. The rms speed of oxygen molecule at 810C81^0 C is (a) 2 kms−12\ km s^{-1} (b) 1 kms−11\ km s^{-1} (c) 0.5 kms−10.5\ km s^{-1} (d) 4 kms−14\ km s^{-1}

Answer

(c) 0.5 km s−10.5\ \text{km s}^{-1}

For rms speed,
vrms=3RTM∝TMv_{\text{rms}}=\sqrt{\frac{3RT}{M}}\propto \sqrt{\frac{T}{M}}

For hydrogen at STP:
TH=273 K, MH=2 g/mol, vH=2 km/sT_H=273\ \text{K},\ M_H=2\ \text{g/mol},\ v_H=2\ \text{km/s}

For oxygen at 81∘C=354 K81^\circ\text{C}=354\ \text{K}:
MO=32 g/molM_O=32\ \text{g/mol}

vO=vHTOTH⋅MHMOv_O=v_H\sqrt{\frac{T_O}{T_H}\cdot\frac{M_H}{M_O}} =2354273⋅232=2\sqrt{\frac{354}{273}\cdot\frac{2}{32}} =20.0810≈0.57 km/s≈0.5 km/s=2\sqrt{0.0810}\approx 0.57\ \text{km/s}\approx 0.5\ \text{km/s}

So the correct option is (c).

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  • 2069 · Entrance Examination 2069 · 1 mark

Bolometer measures the (a) coefficient of thermal conductivity (b) thermo emf (c) Specific heat (d) heat radiation

Answer

(d) heat radiation

A bolometer measures the energy of incident heat (infrared) radiation — the radiation is absorbed by a thin blackened conductor, and the resulting rise in its resistance gives the intensity of the radiation.

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  • 2069 · Entrance Examination 2069 · 1 mark

A temperature of 140F14^0 F is equivalent to (a) 296 K (b) 263 K (c) 250 K (d) 1.50C1.5^0 C

Answer

(b) 263 K

Using K=59(F−32)+273K = \frac{5}{9}(F - 32) + 273: K=59(14−32)+273=59(−18)+273=−10+273=263 KK = \frac{5}{9}(14 - 32) + 273 = \frac{5}{9}(-18) + 273 = -10 + 273 = 263\ \text{K}

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  • 2069 · Entrance Examination 2069 · 1 mark

A myopic eye cannot see object beyond 1.5m. What power of lens will be required to correct this defect of vision? (a) 1.5 D (b) 0.66 D (c) -1.5 D (d) -0.66 D

Answer

(d) -0.66 D

For a myopic eye, the far point is 1.5 m. A concave lens must form the image of a distant object (at infinity) at the eye’s far point, so its focal length is
f=−1.5 mf = -1.5\ \text{m} Power of the lens: P=1f=1−1.5=−0.66 DP = \frac{1}{f} = \frac{1}{-1.5} = -0.66\ \text{D}

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  • 2069 · Entrance Examination 2069 · 1 mark

The critical angle of glass water interface is (a) sin⁡−1(89)\sin^{-1}(\frac{8}{9}) (b) sin⁡−1(45)\sin^{-1}(\frac{4}{5}) (c) sin⁡−1(34)\sin^{-1}(\frac{3}{4}) (d) sin⁡−1(2/3)\sin^{-1}(2/3)

Answer

(a) sin⁡−1(89)\sin^{-1}\left(\frac{8}{9}\right)

For glass to water, sin⁡C=nwaternglass=4/33/2=89\sin C=\dfrac{n_{\text{water}}}{n_{\text{glass}}}=\dfrac{4/3}{3/2}=\dfrac{8}{9}, so C=sin⁡−1(89)C=\sin^{-1}\left(\frac{8}{9}\right).

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  • 2069 · Entrance Examination 2069 · 1 mark

A diffraction grating has 5000 lines per cm. The maximum order visible with the radiation of wavelength 600nm in diffraction experiment is (a) 2 (b) 3 (c) 4 (d) 0

Answer

(b) 3

For a grating, (d\sin\theta = m\lambda), so the maximum order satisfies (m \le \dfrac{d}{\lambda}).
Here, (d=\dfrac{1}{5000}\text{ cm}=2\times10^{-6}\text{ m}=2000\text{ nm}), and (\lambda=600\text{ nm}).
Thus, (m \le \dfrac{2000}{600}=3.33), so the maximum visible order is (3).

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  • 2069 · Entrance Examination 2069 · 1 mark

Light travels through a glass plate of thickness t and refractive index n. If C is the velocity of light in vacuum, the time taken by light to travel this thickness is (a) ntnt (b) ntc\frac{n t}{c} (c) tc\frac{t}{c} (d) tn\frac{t}{n}

Answer

(b) ntc\frac{n t}{c}

In glass, speed of light is v=cnv=\frac{c}{n}, so time =tv=tc/n=ntc=\frac{t}{v}=\frac{t}{c/n}=\frac{nt}{c}.

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  • 2069 · Entrance Examination 2069 · 1 mark

Two charges of 2 μC2\ \mu C and 4 μC4\ \mu C are 1m apart. The electric field intensity is zero at a point between the charges. This point is approximately at a distance. (a) 50 cm from 2 μC2\ \mu C charge (b) 59 cm from 2 μC2\ \mu C charge (c) 41 cm from 2 μC2\ \mu C charge (d) 66 cm from 2 μC2\ \mu C charge

Answer

(c) 41 cm from 2 μC2\ \mu C charge

Let the zero-field point be at distance xx from the 2 μC2\ \mu C charge. Then k(2)x2=k(4)(1−x)2\frac{k(2)}{x^2}=\frac{k(4)}{(1-x)^2} (1−x)2=2x2⇒1−x=2 x⇒x=11+2≈0.414 m=41 cm(1-x)^2=2x^2 \Rightarrow 1-x=\sqrt2\,x \Rightarrow x=\frac{1}{1+\sqrt2}\approx 0.414\ \text{m}=41\ \text{cm}

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  • 2069 · Entrance Examination 2069 · 1 mark

An open organ pipe can produce (a) only odd harmonics (b) only even harmonics (c) both odd and even harmonics (d) depends upon the length of pipe

Answer

(c) both odd and even harmonics

An open organ pipe has antinodes at both open ends, so its fundamental frequency is f=v/2Lf = v/2L and all integer multiples — f,2f,3f,…f, 2f, 3f, \dots — are present. Hence both odd and even harmonics occur. (A closed pipe, with a node at the closed end, gives only odd harmonics.)

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  • 2069 · Entrance Examination 2069 · 1 mark

Watt per ampere is the unit of (a) electric potential (b) electric field intensity (c) electric power (d) electric energy

Answer

(a) electric potential

Watt per ampere = WA=J/sA=JA⋅s=V\frac{\text{W}}{\text{A}} = \frac{\text{J/s}}{\text{A}} = \frac{\text{J}}{\text{A}\cdot\text{s}} = \text{V}, and volt is the unit of electric potential.

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  • 2069 · Entrance Examination 2069 · 1 mark

Two wires A and B of equal radii are drawn from the same material. Their lengths are 1m and 2m respectively. The resistivity of A is (a) less than that of B (b) equal to that of B (c) more than that of B (d) depends upon the amount of current flowing in wires

Answer

(b) equal to that of B

Resistivity is a property of the material itself and does not depend on the length, radius, or shape of the wire. Since both wires are made of the same material, their resistivities are equal.

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  • 2069 · Entrance Examination 2069 · 1 mark

When a capacitor is added in series in an LCR circuit, the impedance of the circuit (a) increase (b) decrease (c) remains unchanged (d) increases or decreases depending upon the frequency of a.c

Answer

(d) increases or decreases depending upon the frequency of a.c.

Adding another capacitor in series decreases the equivalent capacitance, so the capacitive reactance XC=1ωCX_C=\frac{1}{\omega C} increases. Since impedance is
Z=R2+(XL−XC)2,Z=\sqrt{R^2+(X_L-X_C)^2},
the value of ZZ may increase or decrease depending on whether the original circuit is inductive or capacitive, i.e. on the frequency of the a.c.

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  • 2069 · Entrance Examination 2069 · 1 mark

A cell of emf X if connected across a resistance R. The potential difference across the wire is measured as Y. The internal resistance of the cell should be (a) X−(Y/R)X - (Y/R) (b) (X−Y)R(X-Y)R (c) (X−Y)R/X(X-Y)R/X (d) (X−Y)R/Y(X-Y)R/Y

Answer

(d) (X−Y)RY\dfrac{(X-Y)R}{Y}

The current is I=YRI=\dfrac{Y}{R}. Since terminal voltage Y=X−IrY=X-Ir, r=X−YI=X−YY/R=(X−Y)RYr=\frac{X-Y}{I}=\frac{X-Y}{Y/R}=\frac{(X-Y)R}{Y}

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  • 2069 · Entrance Examination 2069 · 1 mark

The ratio of wavelength of the first line of Lyman series to the first line of Balmer series is (a) 5:27 (b) 20:27 (c) 27:20 (d) 1:4

Answer

(a) 5:27

For Lyman’s first line, n1=1, n2=2n_1=1,\ n_2=2: 1λL=R(1−14)=3R4\frac{1}{\lambda_L}=R\left(1-\frac{1}{4}\right)=\frac{3R}{4} For Balmer’s first line, n1=2, n2=3n_1=2,\ n_2=3: 1λB=R(14−19)=5R36\frac{1}{\lambda_B}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36} So, λLλB=4/3R36/5R=527\frac{\lambda_L}{\lambda_B}=\frac{4/3R}{36/5R}=\frac{5}{27} Hence the ratio is 5:27.

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  • 2069 · Entrance Examination 2069 · 1 mark

If work function of a metal is 3.3 eV, the threshold frequency in photoelectric effect will be (a) 2×10152 \times 10^{15} Hz (b) 8×10148 \times 10^{14} Hz (c) 4×10164 \times 10^{16} Hz (d) 6×10176 \times 10^{17} Hz

Answer

(b) 8×1014 Hz8 \times 10^{14}\ \text{Hz}

Threshold frequency is given by
f0=ϕh=3.3×1.602×10−196.626×10−34≈8×1014 Hzf_0=\frac{\phi}{h}=\frac{3.3\times1.602\times10^{-19}}{6.626\times10^{-34}}\approx 8\times10^{14}\ \text{Hz}

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  • 2069 · Entrance Examination 2069 · 1 mark

A radioactive material of half life 5 days has its initial mass of 1 kg. The amount decayed after 20 days will be. (a) 125 g (b) 250 g (c) 550 g (d) 937.5 g

Answer

(d) 937.5 g

Number of half-lives in 20 days =205=4= \frac{20}{5}=4.
Remaining mass =1000(12)4=62.5 g= 1000\left(\frac12\right)^4 = 62.5\ \text{g}.
Amount decayed =1000−62.5=937.5 g= 1000 - 62.5 = 937.5\ \text{g}.

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  • 2069 · Entrance Examination 2069 · 1 mark

A charged particle is undergoing a circular motion in a uniform magnetic field. The time period is independent of (a) speed (b) mass (c) charge (d) magnetic induction A

Answer

(a) speed

The time period of a charged particle moving in a uniform magnetic field is
T=2πmqBT=\frac{2\pi m}{qB} which depends on mass mm, charge qq, and magnetic induction BB, but not on the speed of the particle.

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  • 2069 · Entrance Examination 2069 · 1 mark

Velocity of sound in a gas is 330 ms−1330\ ms^{-1} at STP. Its value at 300k will nearly be (a) 250 ms−1250\ ms^{-1} (b) 300 ms−1300\ ms^{-1} (c) 360 ms−1360\ ms^{-1} (d) 450 ms−1450\ ms^{-1}

Answer

(c) 360 ms−1360\ ms^{-1}

Velocity of sound in a gas is proportional to T\sqrt{T}.
v300=330300273≈346 m/sv_{300}=330\sqrt{\frac{300}{273}}\approx 346\ \text{m/s} This is nearest to 360 m/s360\ \text{m/s}.

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  • 2069 · Entrance Examination 2069 · 1 mark

Time of flight T of a projectile is related to the horizontal range R by an equation gT2=2RgT^2 = 2R. the angle of projection (in degrees) is (a) 90090^0 (b) 60060^0 (c) 45045^0 (d) 30030^0

Answer

(c) 45∘45^\circ

For a projectile, T=2usin⁡θg,R=u2sin⁡2θg=2u2sin⁡θcos⁡θgT=\frac{2u\sin\theta}{g},\qquad R=\frac{u^2\sin2\theta}{g}=\frac{2u^2\sin\theta\cos\theta}{g} Given: gT2=2RgT^2=2R g(2usin⁡θg)2=2(2u2sin⁡θcos⁡θg)g\left(\frac{2u\sin\theta}{g}\right)^2=2\left(\frac{2u^2\sin\theta\cos\theta}{g}\right) 4u2sin⁡2θ=4u2sin⁡θcos⁡θ4u^2\sin^2\theta=4u^2\sin\theta\cos\theta sin⁡θ=cos⁡θ⇒tan⁡θ=1⇒θ=45∘\sin\theta=\cos\theta \Rightarrow \tan\theta=1 \Rightarrow \theta=45^\circ

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  • 2069 · Entrance Examination 2069 · 1 mark

An electron in a field of 1 NC−11\ NC^{-1} will have an approximate acceleration. (a) 1.7×1011ms−21.7 \times 10^{11} ms^{-2} (b) 1×106ms−21 \times 10^6 ms^{-2} (c) 1×1012ms−21 \times 10^{12} ms^{-2} (d) 1×107ms−21 \times 10^7 ms^{-2}

Answer

(a) 1.7×1011 ms−21.7 \times 10^{11}\ \text{ms}^{-2}

Acceleration a=Fm=eEm=(1.6×10−19)(1)9.1×10−31≈1.76×1011 ms−2a=\dfrac{F}{m}=\dfrac{eE}{m}=\dfrac{(1.6\times10^{-19})(1)}{9.1\times10^{-31}}\approx 1.76\times10^{11}\ \text{ms}^{-2}.

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  • 2069 · Entrance Examination 2069 · 1 mark

A bullet gets heated if stopped by a target. If its velocity is doubled the rise in temperature of the bullet will be (a) same (b) 2 times (c) 4 times (d) 16 times

Answer

(c) 4 times

When the bullet is stopped, its kinetic energy 12mv2\frac{1}{2}mv^2 is converted into heat. Since the rise in temperature is proportional to v2v^2, doubling the velocity makes the rise in temperature 22=42^2 = 4 times.

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  • 2070 · 1 mark

The dimensional formula for potential difference is (a) [ML2T−3I−1][ML^2T^{-3}I^{-1}] (b) [ML2T−3I][ML^2T^{-3}I] (c) [ML2T3I−1][ML^2T^3I^{-1}] (d) [ML2T−2I−2][ML^2T^{-2}I^{-2}]

Answer

(a) [ML2T−3I−1][ML^2T^{-3}I^{-1}]

V=Wq=ML2T−2IT=ML2T−3I−1V=\frac{W}{q}=\frac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}.

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  • 2070 · 1 mark

A stone is thrown vertically upward with a speed uu from the top of a tower reaches the ground with a speed of 3u3u. The height of the tower is (a) 3u2g\frac{3u^2}{g} (b) 4u2g\frac{4u^2}{g} (c) 6u2g\frac{6u^2}{g} (d) 9u2g\frac{9u^2}{g}

Answer

(b) 4u2g\frac{4u^2}{g}

The stone returns to the tower top with speed uu downward, and the net displacement to the ground is hh downward. Taking downward as positive, (3u)2=u2+2gh(3u)^2=u^2+2gh, so h=8u22g=4u2gh=\frac{8u^2}{2g}=\frac{4u^2}{g}.

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  • 2070 · 1 mark

The acceleration of a particle in SHM is (a) Always zero (b) always constant (c) Maximum at extreme position (d) Maximum at the equilibrium position

Answer

(c) Maximum at extreme position

In SHM, a=−ω2xa=-\omega^2x, so acceleration is proportional to displacement. It is maximum at the extreme positions and zero at equilibrium.

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  • 2070 · 1 mark

An iron ball is heated. The percentage increase will be largest in (a) volume (b) density (c) diameter (d) surface area

Answer

(a) volume

For a small temperature rise, the percentage increase is αΔT\alpha\Delta T for length or diameter, 2αΔT2\alpha\Delta T for area and 3αΔT3\alpha\Delta T for volume. Density decreases. So volume increases the most.

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  • 2070 · 1 mark

Two wires A and B are of the same material. Their lengths are in the ratio of 1:2 and diameters are in the ratio of 2:1. If they are pulled by the same force, their increase in length will be in the ratio. (a) 2 : 1 (b) 1 : 4 (c) 1 : 8 (d) 8 : 1

Answer

(c) 1 : 8

Δl=FlAY∝ld2\Delta l=\frac{Fl}{AY}\propto\frac{l}{d^2}. So ΔlAΔlB=1/222/12=1/42=18\frac{\Delta l_A}{\Delta l_B}=\frac{1/2^2}{2/1^2}=\frac{1/4}{2}=\frac18.

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  • 2070 · 1 mark

Water is flowing through a tube of non-uniform cross section. If the radius of the tube at the entrance and the exit is 3:2, then the ratio of velocity of liquid entering and leaving the tube is: (a) 8 : 27 (b) 4 : 9 (c) 9 : 4 (d) 1 : 1

Answer

(b) 4 : 9

By the equation of continuity, A1v1=A2v2A_1v_1=A_2v_2, so v1v2=r22r12=2232=49\frac{v_1}{v_2}=\frac{r_2^2}{r_1^2}=\frac{2^2}{3^2}=\frac49.

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  • 2070 · 1 mark

The thermodynamic process in which the pressure of the system remains constant is called (a) isothermal (b) adiabatic (c) isochoric (d) isobaric

Answer

(d) isobaric

A process at constant pressure is called isobaric.

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  • 2070 · 1 mark

50 g of ice at −6∘-6^\circC is dropped into water at 0∘0^\circC. How many grams of water freeze? Specific heat capacity of ice =2000 J kg−1C−1=2000\,\text{J kg}^{-1}\text{C}^{-1}. (a) 1.785 g (b) 4.25 g (c) 3.16 g (d) 9.33 g

Answer

(a) 1.785 g

Heat needed to warm the ice to 0∘0^\circC: 0.05×2000×6=6000.05\times2000\times6=600 J. This comes from freezing water of mass mm: m×3.36×105=600m\times3.36\times10^5=600, so m=1.786×10−3m=1.786\times10^{-3} kg ≈1.785\approx1.785 g.

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  • 2070 · 1 mark

The image of an object placed at the focus of a concave mirror is at (a) f (b) Infinity (c) 2f (d) f/2

Answer

(b) Infinity

Rays from a point at the focus become parallel after reflection, so the image forms at infinity.

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  • 2070 · 1 mark

The temperature of source and sink of a Carnot engine are 400K and 300K respectively. What is its efficiency? (a) 100% (b) 75% (c) 33.3% (d) 25%

Answer

(d) 25%

η=1−T2T1=1−300400=0.25=25%\eta=1-\frac{T_2}{T_1}=1-\frac{300}{400}=0.25=25\%.

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  • 2070 · 1 mark

The refractive index of air with respect to glass is 2/3. The refractive index of diamond with respect to air is 12/5. Then the refractive index of glass with respect to diamond will be (a) 5/8 (b) 8/9 (c) 5/18 (d) 18/5

Answer

(a) 5/8

Given gμa=23{}_g\mu_a=\frac23, so aμg=32{}_a\mu_g=\frac32, and aμd=125{}_a\mu_d=\frac{12}{5}. Then dμg=aμgaμd=3/212/5=1524=58{}_d\mu_g=\frac{{}_a\mu_g}{{}_a\mu_d}=\frac{3/2}{12/5}=\frac{15}{24}=\frac58.

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  • 2070 · 1 mark

The focal length of a double convex lens for which radius of curvature of each surface is R will be (μ=1.5\mu=1.5) (a) R/2 (b) R (c) 2R (d) 4R

Answer

(b) R

1f=(μ−1)(1R+1R)=0.5×2R=1R\frac1f=(\mu-1)\left(\frac1R+\frac1R\right)=0.5\times\frac2R=\frac1R, so f=Rf=R.

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  • 2070 · 1 mark

Light of wave length 7200 Å in air has a wave length in glass equal to (Given refractive index of glass with respect to air = 1.5). (a) 7200 Å (b) 4800 Å (c) 1080 Å (d) 10800 Å

Answer

(b) 4800 Å

λg=λaμ=72001.5=4800\lambda_g=\frac{\lambda_a}{\mu}=\frac{7200}{1.5}=4800 Å.

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  • 2070 · 1 mark

Three capacitors of capacitance 3μ\muF, 9μ\muF and 18μ\muF are connected first in series and then in parallel. The ratio of equivalent capacitance in two cases (Cs/CpC_s/C_p) will be (a) 1:15 (b) 15:1 (c) 1:1 (d) 1:3

Answer

(a) 1 : 15

Series: 1Cs=13+19+118=918\frac1{C_s}=\frac13+\frac19+\frac1{18}=\frac{9}{18}, so Cs=2 μC_s=2\ \muF. Parallel: Cp=3+9+18=30 μC_p=3+9+18=30\ \muF. Hence Cs/Cp=2/30=1:15C_s/C_p=2/30=1:15.

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  • 2070 · 1 mark

Three equal resistors each of resistance r are connected to form a triangle. The equivalent resistance across any two corners of the triangle is (a) 2r (b) r/3 (c) 2r/3 (d) 3r

Answer

(c) 2r/3

Between two corners, one resistor rr is in parallel with the other two in series (2r2r): r⋅2rr+2r=2r3\frac{r\cdot2r}{r+2r}=\frac{2r}{3}.

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  • 2070 · 1 mark

The resistances of two lamps are in the ratio of 1:2. Their wattage will be in the ratio of (a) 1:2 (b) 2:1 (c) 4:1 (d) 1:4

Answer

(b) 2:1

For the same supply voltage, P=V2/RP=V^2/R, so P∝1/RP\propto 1/R. With R1:R2=1:2R_1:R_2=1:2, P1:P2=2:1P_1:P_2=2:1.

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  • 2070 · 1 mark

The energy stored in a 50mH inductor carrying a current of 4A is (a) 0.1 J (b) 0.4 J (c) 0.3 J (d) 0.01 J

Answer

(b) 0.4 J

U=12LI2=12(50×10−3)(4)2=0.4 JU=\tfrac12 LI^2=\tfrac12(50\times10^{-3})(4)^2=0.4\ \text{J}

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  • 2070 · 1 mark

The alternating current in an LCR circuit is maximum when (a) XL=0X_L=0 (b) XC=0X_C=0 (c) XL=XCX_L=X_C (d) XL2+XC2=0\sqrt{X_L^2+X_C^2}=0

Answer

(c) XL=XCX_L=X_C

Impedance Z=R2+(XL−XC)2Z=\sqrt{R^2+(X_L-X_C)^2} is minimum (equal to RR) at resonance, when XL=XCX_L=X_C, so the current I=V/ZI=V/Z is maximum.

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  • 2070 · 1 mark

A particle of mass 10−3110^{-31} kg is moving with a speed of 10510^5 m/s. The de Broglie wavelength of the particle is (a) 6.63×10−86.63\times10^{-8} m (b) 6.63 Å (c) 66.3 Å (d) 6.63×10−76.63\times10^{-7} m

Answer

(a) 6.63×10−86.63\times10^{-8} m

λ=hmv=6.63×10−3410−31×105=6.63×10−8 m\lambda=\frac{h}{mv}=\frac{6.63\times10^{-34}}{10^{-31}\times10^{5}}=6.63\times10^{-8}\ \text{m}

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  • 2070 · 1 mark

Two waves of the same wavelength and amplitude interfere to give a minimum when their phase difference is (a) π\pi (b) π/2\pi/2 (c) 3π/23\pi/2 (d) 0

Answer

(a) π\pi

Destructive interference for equal amplitudes occurs when the phase difference is an odd multiple of π\pi (path difference λ/2\lambda/2), giving minimum (zero) resultant amplitude.

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  • 2070 · 1 mark

Hard X-rays as compared to soft X-rays have (a) Higher intensity (b) lower frequency (c) Higher frequency (d) higher speed

Answer

(c) Higher frequency

Hard X-rays are produced at higher accelerating voltage, so they have higher frequency (shorter wavelength) and greater penetrating power. All X-rays travel at the speed of light.

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  • 2070 · 1 mark

The mass of α\alpha-particle is (a) Equal to the mass of four protons (b) Equal to the mass of four neutrons (c) Equal to the masses of 2 protons and 2 neutrons (d) Less than the sum of masses of 2 protons and 2 neutrons

Answer

(d) Less than the sum of masses of 2 protons and 2 neutrons

An α\alpha-particle is a helium nucleus (2 protons + 2 neutrons). Its mass is less than the sum of its nucleons because of the mass defect, which is converted into binding energy.

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  • 2070 · 1 mark

A common emitter transistor amplifier has a current gain of 50. If the load resistance is 4 KΩ\Omega and the input resistance is 500 Ω\Omega. The voltage gain of the amplifier is (a) 400 (b) 6.2 (c) 500 (d) 300

Answer

(a) 400

Av=βRLRin=50×4000500=400A_v=\beta\frac{R_L}{R_{in}}=50\times\frac{4000}{500}=400

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  • 2070 · 1 mark

A tube closed at one end has a length 1 meter. The air column in the pipe can resonate for sound of frequency (V = 320 ms−1^{-1}) (a) 166 Hz (b) 249 Hz (c) 575 Hz (d) 80 Hz

Answer

(d) 80 Hz

A pipe closed at one end resonates at f=(2n+1)v4Lf=\dfrac{(2n+1)v}{4L}. Here v4L=3204=80\dfrac{v}{4L}=\dfrac{320}{4}=80 Hz, so the fundamental is 80 Hz, with overtones at 240 Hz, 400 Hz and so on. Only 80 Hz is among the options.

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  • 2070 · 1 mark

A displacement wave is represented by y=0.25sin⁡(500t−0.0025x)y=0.25\sin(500t-0.0025x), where y, t and x are in cm, second and meter respectively. The wave length of wave is (a) 20π20\pi m (b) 40π40\pi m (c) 60π60\pi m (d) 80π80\pi m

Answer

(d) 80π80\pi m

Comparing with y=Asin⁡(ωt−kx)y=A\sin(\omega t-kx) gives λ=2π/k\lambda=2\pi/k. The option list corresponds to k=0.025 m−1k=0.025\ \text{m}^{-1}, giving λ=2π0.025=80π m\lambda=\frac{2\pi}{0.025}=80\pi\ \text{m} With k=0.0025k=0.0025 exactly as printed the value would be 800π800\pi m, so 80π80\pi m is the intended option.

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  • 2071 · 1 mark

The dimensions of (h/e) are same as that of (a) Magnetic field (b) electric field (c) Magnetic flux (d) electric flux

Answer

(c) Magnetic flux

[h]=J s[h]=\text{J s} and [e]=C[e]=\text{C}, so h/e=J s/C=V s=Wbh/e=\text{J s/C}=\text{V s}=\text{Wb}, the unit of magnetic flux.

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  • 2071 · 1 mark

The number of significant figures in 0.0030×10110.0030\times10^{11} m are (a) 5 (b) 4 (c) 11 (d) 2

Answer

(d) 2

0.0030×10110.0030\times10^{11}: leading zeros are not significant; the trailing zero after the decimal point is, so the significant figures are 3 and 0, i.e. 2.

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  • 2071 · 1 mark

μs\mu_s and μk\mu_k are coefficient of static and kinetic frictions between two bodies. Then (a) μs≥μk\mu_s\geq\mu_k (b) μs<μk\mu_s<\mu_k (c) μs=0\mu_s=0 and μk>0\mu_k>0 (d) μs<0\mu_s<0 and μk<0\mu_k<0

Answer

(a) μs≥μk\mu_s\geq\mu_k

Limiting static friction is greater than kinetic friction, so the coefficient of static friction is at least that of kinetic friction.

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  • 2071 · 1 mark

When a torque acts on a body, the change in angular momentum of the body is (a) Perpendicular to LL (b) parallel to LL (c) Always zero (d) 180∘L180^\circ L

Answer

(a) Perpendicular to LL

Since τ⃗=dL⃗/dt\vec\tau=d\vec L/dt, the change in L⃗\vec L is along the torque. When the torque is perpendicular to L⃗\vec L (as in a gyroscope or precessing top), ΔL⃗\Delta\vec L is perpendicular to L⃗\vec L: its direction changes while its magnitude stays the same.

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  • 2071 · 1 mark

The Poisson's ratio of material can't be (a) 0.10 (b) 0.20 (c) 0.40 (d) 0.65

Answer

(d) 0.65

For an isotropic material Poisson's ratio lies between −1-1 and 0.50.5; practical values are 0 to 0.5, so 0.65 is not possible.

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  • 2071 · 1 mark

The depth of water at which air-bubble of radius 0.4 mm may remain in equilibrium (surface tension of water =72×10−3=72\times10^{-3} N/m, g=9.8g=9.8 m/s2^2) is given by (a) 7.348 cm (b) 0.918 cm (c) 3.674 cm (d) 1.837 cm

Answer

(c) 3.674 cm

An air bubble in water has a single surface, so the excess pressure is 2T/r2T/r. For equilibrium ρgh=2Tr\rho g h=\frac{2T}{r}: h=2Trρg=2×72×10−30.4×10−3×1000×9.8=0.03674 m=3.674 cm.h=\frac{2T}{r\rho g}=\frac{2\times72\times10^{-3}}{0.4\times10^{-3}\times1000\times9.8}=0.03674\text{ m}=3.674\text{ cm}.

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  • 2071 · 1 mark

When a cylinder is heated, its length increases by 2%, the area of its base will increase by (a) 0.5% (b) 1% (c) 2% (d) 4%

Answer

(d) 4%

The base is a circle of area A=πr2A=\pi r^2, and the radius also increases by 2% (linear expansion). So ΔA/A=2×2%=4%\Delta A/A=2\times2\%=4\%.

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  • 2071 · 1 mark

In a pressure cooker food is cooked usually at a pressure (a) One atmosphere (b) half atmosphere (c) Two atmosphere (d) 50 atmosphere

Answer

(c) Two atmosphere

A pressure cooker raises the pressure (about 2 atm) so water boils at about 120∘120^\circC, cooking food faster.

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  • 2071 · 1 mark

The mean free path of a gas varies with absolute temperature (T) as (a) T4T^4 (b) 1/T41/T^4 (c) 1/T1/T (d) TT

Answer

(d) TT

λ=kT2 πd2p\lambda=\dfrac{kT}{\sqrt2\,\pi d^2 p}, so at constant pressure the mean free path is proportional to TT.

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  • 2071 · 1 mark

The efficiency of Carnot engine working between steam point and ice point is (a) 16.8% (b) 26.81% (c) 36.8% (d) 46.8%

Answer

(b) 26.81%

η=1−T2T1=1−273373=100373=0.2681=26.81%.\eta=1-\frac{T_2}{T_1}=1-\frac{273}{373}=\frac{100}{373}=0.2681=26.81\%.

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  • 2071 · 1 mark

The bending of sound waves through the corners of the obstacle is called (a) Refraction (b) interference (c) Beats (d) diffraction

Answer

(d) Diffraction

Bending of waves around the edges of an obstacle or aperture is diffraction.

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  • 2071 · 1 mark

When a tuning fork is vibrated its prongs have a phase difference of (a) 0 (b) π/4\pi/4 (c) π\pi (d) 2π2\pi

Answer

(c) π\pi

The two prongs of a tuning fork move in opposite directions (towards and away from each other), so they are in antiphase, with a phase difference of π\pi.

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  • 2071 · 1 mark

The image of an object placed at the focus of convex mirror is at a distance (a) ∞\infty (b) f (c) 2f (d) f/2

Answer

(d) f/2f/2

For a convex mirror ff is positive and u=−fu=-f: 1v=1f−1u=1f+1f=2f\frac1v=\frac1f-\frac1u=\frac1f+\frac1f=\frac2f, so v=f/2v=f/2 (a virtual image behind the mirror).

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  • 2071 · 1 mark

If sky is seen from moon's surface, it will appear (a) Black (b) Blue (c) White (d) Red

Answer

(a) Black

The sky is blue on Earth because of scattering by the atmosphere. The Moon has no atmosphere, so there is no scattering and the sky looks black.

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  • 2071 · 1 mark

A double convex air bubble in a glass sheet would behave as a (a) Convergent lens (b) concave mirror (c) Divergent lens (d) plane mirror

Answer

(c) Divergent lens

An air bubble is a region of lower refractive index than the surrounding glass. A double convex shape of a rarer medium in a denser medium acts like a concave (diverging) lens.

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  • 2071 · 1 mark

When two prisms are combined it gives (a) Reflection only (b) lateral Shift (c) Dispersion and deviation both (d) Dispersion or deviation

Answer

(d) Dispersion or deviation

Two prisms combined (of different materials, in opposite orientation) can give deviation without dispersion (achromatic combination) or dispersion without deviation (direct vision combination).

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  • 2071 · 1 mark

Which property of electrons is used in an electron microscope? (a) Negative charge (b) wave nature (c) Spin (d) small size

Answer

(b) Wave nature

Fast electrons have a very short de Broglie wavelength λ=h/p\lambda=h/p, which gives a much higher resolving power than visible light.

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  • 2071 · 1 mark

A system of three charges may be in equilibrium. If they are placed (a) On a straight line (b) On corners of equilateral triangle (c) On three corners of a rectangle (d) On the corners of a cube

Answer

(a) On a straight line

Three charges can be in equilibrium when collinear, with the middle charge opposite in sign to the other two and placed so that the forces on each balance. This is not possible for the other arrangements.

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  • 2071 · 1 mark

The radius of Earth is 6400 km. The capacitance is (a) 6400 MF (b) 6400 F (c) 1 F (d) 711 μ\muF

Answer

(d) 711 μ\muF

C=4πε0R=6.4×1069×109=7.11×10−4 F=711 μF.C=4\pi\varepsilon_0R=\frac{6.4\times10^{6}}{9\times10^{9}}=7.11\times10^{-4}\text{ F}=711\ \mu\text{F}.

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  • 2071 · 1 mark

Two wires of same material have lengths L and 2L and cross section areas 4A and A respectively. The ratio of their specific resistances would be (a) 1:1 (b) 1:8 (c) 8:1 (d) 1:2

Answer

(a) 1:1

Specific resistance (resistivity) depends only on the material, not on length or area, so the ratio is 1:1.

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  • 2071 · 1 mark

The magnetic field induction inside a long solenoid is (a) zero (b) 2μ0ni2\mu_0ni (c) μ0ni\mu_0ni (d) μni\mu ni

Answer

(c) μ0ni\mu_0ni

By Ampere's law, the field inside a long solenoid is uniform: B=μ0niB=\mu_0 n i, where nn is the number of turns per unit length.

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  • 2071 · 1 mark

Electrons in the atoms are held due to (a) Nuclear forces (b) Coulomb forces (c) Van der Waals force (d) Gravitational forces

Answer

(b) Coulomb forces

Electrons are bound to the nucleus by the electrostatic (Coulomb) attraction between negative electrons and the positive nucleus.

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  • 2071 · 1 mark

Which series of H atom lies in visible region? (a) Lyman (b) Bracket and Paschen (c) Balmer (d) Paschen

Answer

(c) Balmer

The Balmer series (transitions to n=2n=2) lies in the visible region. Lyman is ultraviolet; Paschen and Brackett are infrared.

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  • 2071 · 1 mark

The half-life of a substance is 1600 years. The mean life is (a) 1600 yrs (b) 3200 yrs (c) 800 yrs (d) 2309 yrs

Answer

(d) 2309 yrs

τ=T1/2ln⁡2=16000.693≈2309 years.\tau=\frac{T_{1/2}}{\ln 2}=\frac{1600}{0.693}\approx2309\text{ years}.

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  • 2071 · 1 mark

Rydberg constant is (a) Same for all elements (b) Different for different elements (c) A universal constant (d) Is different for lighter elements but same for heavier elements.

Answer

(c) A universal constant

R=me48ε02h3cR=\dfrac{me^4}{8\varepsilon_0^2h^3c} is built from fundamental constants and has the same value (1.097×107 m−11.097\times10^7\ \text{m}^{-1}) for hydrogen-like spectra, independent of the element.

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  • 2072 · 1 mark

A force F⃗=(2i^+3j^−4k^)\vec F=(2\hat i+3\hat j-4\hat k) Newton acts on a body. In 5 seconds the force displaces the body through a displacement s⃗=(3i^+2j^−2k^)\vec s=(3\hat i+2\hat j-2\hat k) meter. The power (in watt) of the force is (a) 4 (b) 3 (c) 5 (d) 6

Answer

(a) 4

Work W=F⃗⋅s⃗=(2)(3)+(3)(2)+(−4)(−2)=6+6+8=20W=\vec F\cdot\vec s=(2)(3)+(3)(2)+(-4)(-2)=6+6+8=20 J. Power =W/t=20/5=4=W/t=20/5=4 W.

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  • 2072 · 1 mark

Two simple harmonic waves are represented respectively by y1=A1sin⁡wty_1=A_1\sin wt and y2=A2cos⁡wty_2=A_2\cos wt. The phase difference between the two waves is (a) 3π/23\pi/2 (b) π\pi (c) π/2\pi/2 (d) π/4\pi/4

Answer

(c) π/2\pi/2

y2=A2cos⁡wt=A2sin⁡(wt+π/2)y_2=A_2\cos wt=A_2\sin(wt+\pi/2), so the phase difference from y1y_1 is π/2\pi/2.

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  • 2072 · 1 mark

A river is flowing west to east at a speed of 5 m/min. A man on the south bank of the river, capable of swimming at 10m/min in still water, wants to swim across the river by the shortest path. He should swim in the direction (a) due North (b) 30∘30^\circ East of North (c) 60∘60^\circ East of North (d) 30∘30^\circ West of North

Answer

(d) 30∘30^\circ West of North

To cross by the shortest path (straight across) the swimmer must cancel the river current: 10sin⁡θ=5⇒θ=30∘10\sin\theta=5\Rightarrow\theta=30^\circ, directed upstream. The river flows east, so he must aim 30∘30^\circ west of north.

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  • 2072 · 1 mark

Dimension of surface energy is (a) [M1L0T−2][M^1L^0T^{-2}] (b) [M1L2T−2][M^1L^2T^{-2}] (c) [M0L1T−2][M^0L^1T^{-2}] (d) [M2L0T−2][M^2L^0T^{-2}]

Answer

(a) [M1L0T−2][M^1L^0T^{-2}]

Surface energy is energy per unit area: ML2T−2L2=ML0T−2\dfrac{ML^2T^{-2}}{L^2}=ML^0T^{-2}.

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  • 2072 · 1 mark

The terminal velocity of a small sized spherical body of radius r falling vertically in a viscous liquid is given as one of the following proportionality with radius. (a) r−2r^{-2} (b) r−1r^{-1} (c) rr (d) r2r^2

Answer

(d) r2r^2

At terminal velocity v=2r2(ρ−σ)g9ηv=\dfrac{2r^2(\rho-\sigma)g}{9\eta}, so v∝r2v\propto r^2.

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  • 2072 · 1 mark

A constant torque acting on a uniform circular wheel changes its angular momentum from L0L_0 to 4L04L_0 in 4 seconds. The magnitude of this torque is (a) (34)L0\left(\frac34\right)L_0 (b) L0L_0 (c) 4L04L_0 (d) 12L012L_0

Answer

(a) (34)L0\left(\dfrac34\right)L_0

τ=ΔLt=4L0−L04=3L04\tau=\dfrac{\Delta L}{t}=\dfrac{4L_0-L_0}{4}=\dfrac{3L_0}{4}.

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  • 2072 · 1 mark

At what temperature the rms velocity of a hydrogen molecule be equal to that of oxygen at 47∘C47^\circ C? (a) −273∘-273^\circ C (b) 273 K (c) 20 K (d) 20∘20^\circC

Answer

(c) 20 K

vrms=3RT/Mv_{rms}=\sqrt{3RT/M}. Equal speeds give TH2=TO32\dfrac{T_H}{2}=\dfrac{T_O}{32}, with TO=320T_O=320 K. So TH=320×232=20T_H=\dfrac{320\times2}{32}=20 K.

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  • 2072 · 1 mark

Dimension of the change in entropy is (a) [M1L2T−2K−1][M^1L^2T^{-2}K^{-1}] (b) [M1L1T2K−2][M^1L^1T^2K^{-2}] (c) [M1L2T−2][M^1L^2T^{-2}] (d) [M0L2T−2K−1][M^0L^2T^{-2}K^{-1}]

Answer

(a) [M1L2T−2K−1][M^1L^2T^{-2}K^{-1}]

ΔS=QT\Delta S=\dfrac{Q}{T}, so the dimension is ML2T−2K\dfrac{ML^2T^{-2}}{K}.

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  • 2072 · 1 mark

A square plate of side 2 units is heated through 10∘10^\circ C. If the linear expansivity of the material is x, increase in the surface area of the plate is (a) 480 x (b) 80 x (c) 240 x (d) 240 x2240\ x^2

Answer

(b) 80x80x

Areal expansivity is 2x2x. Original area =2×2=4=2\times2=4, so ΔA=A⋅2x⋅Δθ=4×2x×10=80x\Delta A=A\cdot2x\cdot\Delta\theta=4\times2x\times10=80x.

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  • 2072 · 1 mark

A black body radiates heat energy at the rate of 2×1052\times10^5 Js−1^{-1} at a temperature of 127∘127^\circC. The temperature of the black body at which the rate of heat radiation is 32×10532\times10^5 Js−1^{-1}m−2^{-2} is (a) 327∘327^\circ C (b) 927∘927^\circC (c) 527∘527^\circ C (d) 727∘727^\circ C

Answer

(c) 527∘527^\circC

By Stefan's law E∝T4E\propto T^4 (the rates are compared per unit area, so the ratio is 1616). E2E1=16=(T2T1)4⇒T2=2T1=2×400=800\dfrac{E_2}{E_1}=16=\left(\dfrac{T_2}{T_1}\right)^4\Rightarrow T_2=2T_1=2\times400=800 K =527∘=527^\circC.

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  • 2072 · 1 mark

A ray of light in air enters a glass slab. Its (a) wave length decreases (b) wave length increases (c) speed increases (d) frequency decreases

Answer

(a) wavelength decreases

On entering glass the frequency stays the same, the speed falls (v=c/nv=c/n), and so the wavelength λ=v/f\lambda=v/f decreases.

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  • 2072 · 1 mark

By using white light in double slit interference experiment the central fringe is white. What color of fringe can be seen nearest to the central fringe? (a) blue (b) green (c) red (d) violet

Answer

(d) violet

Fringe width β=λD/d\beta=\lambda D/d is smallest for the shortest wavelength, so violet lies nearest to the central white fringe.

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  • 2072 · 1 mark

A convex lens is immersed in water. Compared to its power in air, the power in water will (a) Increase for red and decrease for blue (b) Decrease for all colors (c) be same for all colors (d) Increase for all colors

Answer

(b) Decrease for all colors

Power P=(μ−1)(1R1−1R2)P=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right) for a lens in air; in water the factor becomes (μgμw−1)\left(\frac{\mu_g}{\mu_w}-1\right), which is smaller for every colour. So the power decreases for all colours.

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  • 2072 · 1 mark

Two vehicles are approaching each other with a speed of 10 ms−1^{-1}. If A emits a note of 1000 vibrations, what will be the frequency of the note as heard by a person sitting in B if the velocity of sound is 330 ms−1^{-1}? (a) 950 Hz (b) 1000 Hz (c) 1062.5 Hz (d) 1215.5 Hz

Answer

(c) 1062.5 Hz

Each vehicle moves towards the other at 1010 m/s. f′=f v+vov−vs=1000×330+10330−10=1000×340320=1062.5f'=f\,\dfrac{v+v_o}{v-v_s}=1000\times\dfrac{330+10}{330-10}=1000\times\dfrac{340}{320}=1062.5 Hz.

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  • 2072 · 1 mark

An open pipe and a closed pipe resonate to the same tuning fork. The ratio of their lengths is (a) 1:1 (b) 2:1 (c) 1:2 (d) 4:1

Answer

(b) 2:1

For the same frequency (same λ\lambda): open pipe Lo=λ/2L_o=\lambda/2, closed pipe Lc=λ/4L_c=\lambda/4 (fundamental). So Lo:Lc=2:1L_o:L_c=2:1.

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  • 2072 · 1 mark

The radius of a copper wire carrying a current is doubled, the drift velocity of electrons (a) Remains same (b) increases four times (c) Decreases four times (d) Increases first and then decreases

Answer

(c) Decreases four times

I=neAvdI=neAv_d, so vd=IneA∝1r2v_d=\dfrac{I}{neA}\propto\dfrac1{r^2}. Doubling the radius makes the area four times as large, so vdv_d becomes one quarter.

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  • 2072 · 1 mark

A 25 W, 120 V lamp and 100 W, 120 V lamp are connected in series across a 120 V line. Which lamp will burn more brightly? (a) 25 W, 120 V (b) 100W, 120 V (c) Both will have same brightness. (d) Alternatively both may be bright, one at a time.

Answer

(a) 25 W, 120 V lamp

In series the same current flows through both lamps, and power dissipated is P=I2RP=I^2R. Since R=V2/PR=V^2/P, the 25 W lamp has the larger resistance (576 Ω\Omega against 144 Ω\Omega), so it dissipates more power and glows brighter.

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  • 2072 · 1 mark

A Spherical condenser of 1 Farad will have dimension of (a) a small metallic ball of radius 0.04 m (b) a big metallic ball of radius 100m (c) radius of earth whose radius is 6.4×1066.4\times10^6 m (d) a sphere having radius of more than that of earth

Answer

(d) a sphere having radius of more than that of earth

C=4πε0RC=4\pi\varepsilon_0 R, so R=C4πε0=1×9×109 mR=\dfrac{C}{4\pi\varepsilon_0}=1\times 9\times10^{9}\ \text{m}, which is far greater than the earth's radius (6.4×1066.4\times10^6 m).

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  • 2072 · 1 mark

The ratio of induced emf in a coil having 50 turns and rotating with 50 Hz to that of induced emf in a coil having 100 turns and rotating with 150 Hz in the same magnetic field is (a) 1:2 (b) 2:3 (c) 1:6 (d) 3:2

Answer

(c) 1:6

εmax=NBAω∝Nf\varepsilon_{max}=NBA\omega\propto Nf for the same field and area, so the ratio is 50×50100×150=250015000=1:6\dfrac{50\times50}{100\times150}=\dfrac{2500}{15000}=1:6.

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  • 2072 · 1 mark

A galvanometer (0-100) mA of resistance 1 Ω\Omega is to be converted to an ammeter in the range 0-1 A. We have to use a shunt (a) 1 Ω\Omega (b) 0.1 Ω\Omega (c) 0.111 Ω\Omega (d) 0.055 Ω\Omega

Answer

(c) 0.111 Ω\Omega

S=IgGI−Ig=0.1×11−0.1=0.111 ΩS=\frac{I_gG}{I-I_g}=\frac{0.1\times1}{1-0.1}=0.111\ \Omega

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  • 2072 · 1 mark

The ratio of specific charge of a proton to that of an alpha particle is (a) 1:4 (b) 1:2 (c) 4:1 (d) 2:1

Answer

(d) 2:1

Specific charge q/mq/m: proton =e/mp=e/m_p; alpha particle =2e/4mp=e/2mp=2e/4m_p=e/2m_p. Ratio =2:1=2:1.

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  • 2072 · 1 mark

The angular momentum of electron in the hydrogen atom in the ground state is (a) h/2π\pi (b) h/4π\pi (c) h (d) 2h

Answer

(a) h/2π\pi

Bohr's quantisation gives L=nh2πL=n\dfrac{h}{2\pi}, and for the ground state n=1n=1, so L=h/2πL=h/2\pi.

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  • 2072 · 1 mark

The approximate radius of aluminum nucleus (Al) is (a) 1.5 Fermi (b) 2.5 Fermi (c) 4.5 Fermi (d) 13.5 Fermi

Answer

(c) 4.5 Fermi

R=R0A1/3R=R_0A^{1/3} with R0≈1.2R_0\approx1.2 to 1.51.5 fm and A=27A=27, so A1/3=3A^{1/3}=3 and R≈3.6R\approx3.6 to 4.54.5 fm. The closest option is 4.5 fm.

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  • 2072 · 1 mark

The kinetic energy of a particle is doubled, de Broglie wavelength becomes (a) 2 times (b) 4 times (c) 2 times (d) 1/21/\sqrt2 times

Answer

(d) 1/21/\sqrt2 times

λ=h2mK∝K−1/2\lambda=\dfrac{h}{\sqrt{2mK}}\propto K^{-1/2}. Doubling KK makes λ\lambda become 1/21/\sqrt2 times.

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  • 2072 · 1 mark

Half-life and mean life of a radioactive element are related. Choose correct answer from the following: (a) half-life is less than the mean life (b) half-life is more than the mean life (c) mean life and the half-life are the same (d) mean life is exactly two times the half life

Answer

(a) half-life is less than the mean life

T1/2=ln⁡2λ=0.693 τT_{1/2}=\dfrac{\ln2}{\lambda}=0.693\,\tau, where τ=1/λ\tau=1/\lambda is the mean life. So the half-life is shorter.

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  • 2073 · 1 mark

In the gas equation (p+aV2)(v−b)=RT\left(p+\frac{a}{V^2}\right)(v-b)=RT, the dimensions of 'a' are (a) ML2T−2ML^2T^{-2} (b) ML5T−2ML^5T^{-2} (c) L6L^6 (d) L−6L^{-6}

Answer

(b) ML5T−2ML^5T^{-2}

a/V2a/V^2 has the dimensions of pressure, so [a]=[P][V2]=ML−1T−2⋅L6=ML5T−2[a]=[P][V^2]=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}.

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  • 2073 · 1 mark

The number of significant figures in 1400 are (a) 0 (b) 1 (c) 2 (d) 4

Answer

(c) 2

In 1400 the trailing zeros are not counted as significant (no decimal point), so only 1 and 4 are significant.

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  • 2073 · 1 mark

A man walks 8m forwards East and 6m towards North. The magnitude of displacement is (a) 14m (b) 2m (c) 8m (d) 82+62\sqrt{8^2+6^2} m

Answer

(d) 82+62\sqrt{8^2+6^2} m

Displacement is the straight-line distance between start and end: 82+62=10\sqrt{8^2+6^2}=10 m.

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  • 2073 · 1 mark

If initial velocity of a projectile is doubled, the maximum range will be (a) 2 times (b) 4 times (c) 8 times (d) 16 times

Answer

(b) 4 times

Range R=u2sin⁡2θg∝u2R=\frac{u^2\sin2\theta}{g}\propto u^2, so doubling uu makes the range 4 times.

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  • 2073 · 1 mark

Which of the following friction is maximum? (a) rolling (b) kinetic (c) rolling + kinetic (d) static

Answer

(d) static

Static friction can rise up to its limiting value, which is greater than kinetic friction, which in turn exceeds rolling friction.

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  • 2073 · 1 mark

If the speed of rotation of earth increases, the weight of the body on earth's surface becomes (a) smaller (b) zero (c) greater (d) unaffected

Answer

(a) smaller

At the surface the effective weight is mg−mω2Rcos⁡2λmg-m\omega^2R\cos^2\lambda; increasing the angular speed ω\omega reduces it.

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  • 2073 · 1 mark

A weightless rubber balloon has 100 g of water in it. Its weight in water will be (a) 100g (b) 50g (c) > 100g (d) zero

Answer

(d) zero

The balloon with water displaces its own volume of water, so the buoyant force equals the weight of the water inside; the apparent weight is zero.

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  • 2073 · 1 mark

The temperature of a substance increase by 270∘270^\circC. This increase is equal to (a) 300K (b) 0K (c) 270K (d) 273K

Answer

(c) 270K

A change of 1∘1^\circC equals a change of 1 K, so an increase of 270∘270^\circC is an increase of 270 K.

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  • 2073 · 1 mark

Which of the following gases possess maximum rms speed at a given temperature? (a) H2H_2 (b) N2N_2 (c) O2O_2 (d) He

Answer

(a) H2H_2

vrms=3RT/Mv_{rms}=\sqrt{3RT/M} is inversely proportional to M\sqrt M; H2H_2 has the smallest molar mass.

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  • 2073 · 1 mark

The black body emits (a) line spectrum (b) band spectrum (c) continuous spectrum (d) mixed spectrum

Answer

(c) continuous spectrum

A black body emits radiation of all wavelengths, giving a continuous spectrum that depends only on temperature.

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  • 2073 · 1 mark

The field of view is maximum for (a) plane mirror (b) concave mirror (c) convex mirror (d) cylindrical mirror

Answer

(c) convex mirror

A convex mirror diverges rays and forms a diminished erect image of a wide region, giving the largest field of view (used as a rear-view mirror).

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  • 2073 · 1 mark

Sky seen from earth appears blue because of (a) scattering of lights (b) dispersion of lights (c) reflection (d) polarization

Answer

(a) scattering of lights

By Rayleigh scattering, intensity ∝1/λ4\propto1/\lambda^4, so blue light is scattered much more than red by air molecules.

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  • 2073 · 1 mark

A person cannot see objects clearly beyond 50 cm. The power of the lens to correct the vision is (a) +5 D (b) -5 D (c) -0.5 D (d) -2 D

Answer

(d) −2-2 D

The lens must form the image of distant objects at the far point: f=−0.5f=-0.5 m, so P=1/f=−2P=1/f=-2 D (concave lens).

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  • 2073 · 1 mark

An isolated conducting sphere is given a positive charge. Its mass (a) Increases (b) decreases (c) remains constant (d) may increase or decrease depending on material

Answer

(b) decreases

A body becomes positively charged by losing electrons, so its mass decreases slightly.

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  • 2073 · 1 mark

Three capacitors each of capacitance 3μ\muF are connected in series. The net capacitance is (a) 1 μ\muF (b) (1/3) μ\muF (c) 3 μ\muF (d) 9 μ\muF

Answer

(a) 1 μ\muF

In series 1C=13+13+13=1\frac1C=\frac13+\frac13+\frac13=1, so C=1 μC=1\,\muF.

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  • 2073 · 1 mark

Kirchhoff's voltage law is based on the principle of conservation of (a) charge (b) energy (c) momentum (d) mass of charges

Answer

(b) energy

The algebraic sum of potential changes around a closed loop is zero, which expresses conservation of energy.

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  • 2073 · 1 mark

A charged particle enters in a strong magnetic field. Then its kinetic energy (a) increases (b) decreases (c) remains constant (d) increases and becomes constant

Answer

(c) remains constant

The magnetic force is perpendicular to velocity, so it does no work; the speed and kinetic energy stay constant.

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  • 2073 · 1 mark

The self-inductance of a straight wire is (a) zero (b) ∞\infty (c) negative (d) positive

Answer

(d) positive

Self-inductance L=Φ/IL=\Phi/I is always positive, since induced emf opposes the change of current (Lenz's law); it is very small for a straight wire but not zero.

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  • 2073 · 1 mark

The photoelectric is based on the law of conservation of (a) energy (b) mass (c) momentum (d) angular momentum of photons

Answer

(a) energy

Einstein's photoelectric equation hν=ϕ+Kmaxh\nu=\phi+K_{max} is a statement of conservation of energy.

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  • 2073 · 1 mark

When a radioactive element emits a β\beta-particle the mass number of the atom (a) increase by one (b) decreases by one (c) remains the same (d) decreases by four

Answer

(c) remains the same

In beta decay a neutron turns into a proton, so the atomic number rises by one while the mass number is unchanged.

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  • 2073 · 1 mark

On increasing the reverse bias to a large value in a p-n junction diode, the current (a) increases slowly (b) remains fixed (c) decreases slowly (d) suddenly increases

Answer

(d) suddenly increases

At a large reverse voltage breakdown (Zener or avalanche) occurs and the current increases suddenly.

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  • 2073 · 1 mark

The equation of sound wave is y=0.0015sin⁡(62.4x+316t)y=0.0015\sin(62.4x+316t). The wave length of sound wave is (a) 0.2 unit (b) 0.1 unit (c) 3.3 unit (d) not known from the given equation

Answer

(b) 0.1 unit

Comparing with y=Asin⁡(kx+ωt)y=A\sin(kx+\omega t), k=62.4k=62.4, so λ=2πk=2π62.4≈0.1\lambda=\frac{2\pi}{k}=\frac{2\pi}{62.4}\approx0.1 unit.

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  • 2073 · 1 mark

A wave is reflected from a rigid support. The change of phase on reflection will be (a) 0 (b) π/2\pi/2 (c) π/4\pi/4 (d) π\pi

Answer

(d) π\pi

Reflection from a rigid (fixed) support inverts the wave, a phase change of π\pi.

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  • 2073 · 1 mark

A car at rest is playing 'HORN'. If it start moving towards you, the pitch heard would be (a) higher (b) less (c) same (d) nil

Answer

(a) higher

By the Doppler effect, a source approaching the listener raises the apparent frequency, so the pitch is higher.

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  • 2073 · 1 mark

In which region of electromagnetic spectrum does the Lyman series of H atom lie? (a) Ultraviolet (b) visible (c) infrared (d) microwave

Answer

(a) Ultraviolet

The Lyman series (transitions to n=1n=1) has wavelengths of about 91 to 122 nm, in the ultraviolet.

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  • 2074 · 1 mark

The order of magnitude of radius of an atom is (a) 10−710^{-7} m (b) 10−1010^{-10} m (c) 10−1210^{-12} m (d) 10−1510^{-15} m

Answer

(b) 10−1010^{-10} m

An atomic radius is about 1 angstrom, i.e. 10−1010^{-10} m.

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  • 2074 · 1 mark

A projectile goes farthest away from the earth, when the angle of projection is (a) 0∘0^\circ (b) 45∘45^\circ (c) 90∘90^\circ (d) 120∘120^\circ

Answer

(c) 90∘90^\circ

For a projectile, the maximum height H=u2sin⁡2θ2gH=\frac{u^2\sin^2\theta}{2g} is greatest at θ=90∘\theta=90^\circ, so it goes farthest from the earth's surface.

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  • 2074 · 1 mark

A simple pendulum when set into vibration comes to rest after sometime because of (a) friction of air (b) tension of air (c) gravity (d) its mass

Answer

(a) friction of air

Air resistance (and friction at the support) dissipates the pendulum's energy, so the oscillation is damped.

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  • 2074 · 1 mark

Power clings to the skin because of (a) compression (b) cohesion (c) adhesion (d) capillarity

Answer

(c) adhesion

Powder sticks to skin because of the attractive force between unlike molecules (powder and skin), which is adhesion.

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  • 2074 · 1 mark

Viscosity is the inherent property of liquids and gases and is more closely related to (a) inertia (b) Shearing strain (c) transfer of momentum (d) surface tension

Answer

(c) transfer of momentum

Viscosity arises from transfer of momentum between adjacent layers of a fluid moving at different velocities.

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  • 2074 · 1 mark

A bridge of steel has length 700m, the temperature varies from 243K to 131K. The change in length of the bridge for above seasonal variation in temperature is (α\alpha steel =11×10−6=11\times10^{-6}/K) (a) 15.4cm (b) 15.4 cm (c) 15.8cm (d) 86.24 cm

Answer

(d) 86.24 cm

ΔT=243−131=112\Delta T=243-131=112 K, so ΔL=LαΔT=700×11×10−6×112=0.8624\Delta L=L\alpha\Delta T=700\times11\times10^{-6}\times112=0.8624 m =86.24=86.24 cm.

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  • 2074 · 1 mark

The internal energy of a perfect gas does not change during (a) adiabatic process (b) isothermal process (c) isobaric process (d) isochoric process

Answer

(b) isothermal process

Internal energy of a perfect gas depends only on temperature, and temperature is constant in an isothermal process.

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  • 2074 · 1 mark

The efficiency of steam engine is of the order of (a) 10% (b) 20% (c) 60% (d) 100%

Answer

(a) 10%

Steam engines are very inefficient, typically converting only about 10% of the heat into work.

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  • 2074 · 1 mark

A gas does maximum work when it expands (a) isothermally (b) isobarically (c) adiabatically (d) isochorically

Answer

(b) isobarically

Work is the area under the P-V curve. For expansion between the same volumes, the constant (highest) pressure of the isobaric process gives the largest area.

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  • 2074 · 1 mark

When two wave overlap the result is large disturbance means (a) The interface is constructive. (b) The interface is destructive. (c) The interference depends on the type of wave. (d) All of the above

Answer

(a) The interference is constructive.

When waves overlap to give a larger disturbance, the displacements add, which is constructive interference.

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  • 2074 · 1 mark

While standing still, an observer measure the frequency of a tuning fork to be 524 Hz. What frequency does he measure as he approaches the fork at 5m/s? (Speed of sound in air = 340m/s) (a) 500Hz (b) 420Hz (c) 520Hz (d) 532Hz

Answer

(d) 532Hz

f′=f(v+vov)=524×345340≈531.7≈532f'=f\left(\frac{v+v_o}{v}\right)=524\times\frac{345}{340}\approx531.7\approx532 Hz.

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  • 2074 · 1 mark

In longitudinal wave the displacement of the particles of the medium is (a) parallel to the wave velocity (b) anti- parallel to the wave velocity (c) perpendicular to the wave velocity (d) parallel or anti parallel to the wave velocity

Answer

(d) parallel or anti parallel to the wave velocity

In a longitudinal wave, particles vibrate along the direction of propagation, back and forth.

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  • 2074 · 1 mark

The illuminance at a distance of 2m from a 100 candela lamp is (a) 10 lux (b) 15 lux (c) 25 lux (d) 100 lux

Answer

(c) 25 lux

E=Ir2=10022=25E=\frac{I}{r^2}=\frac{100}{2^2}=25 lux.

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  • 2074 · 1 mark

When light beam is incident on a prism, it causes (a) Lateral shift (b) dispersion only (c) deviation only (d) dispersion and deviation both

Answer

(d) dispersion and deviation both

A prism bends the beam (deviation) and, because the refractive index varies with wavelength, splits white light into colours (dispersion).

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  • 2074 · 1 mark

If critical angle for a material placed in air is 30∘30^\circ, the refraction index will be (a) 2 (b) 0.5 (c) 1.5 (d) 2.5

Answer

(a) 2

n=1sin⁡C=1sin⁡30∘=2n=\frac{1}{\sin C}=\frac{1}{\sin30^\circ}=2.

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  • 2074 · 1 mark

A body has 1C of negative charge. How many excess electrons it has as compared to its neutral state? (a) 6.25×10186.25\times10^{18} (b) 6.25×10−196.25\times10^{-19} (c) 1.6×10−191.6\times10^{-19} (d) 1.6×10181.6\times10^{18}

Answer

(a) 6.25×10186.25\times10^{18}

n=Qe=11.6×10−19=6.25×1018n=\frac{Q}{e}=\frac{1}{1.6\times10^{-19}}=6.25\times10^{18}.

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  • 2074 · 1 mark

On heating a liquid, the refraction index generally (a) does not change (b) decrease (c) increase (d) may increase or decrease depending on the rate of health

Answer

(b) decrease

On heating, the density of a liquid falls, so its refractive index generally decreases.

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  • 2074 · 1 mark

The torque on electric dipole of moment vector (p) when placed in an electric field vector (E) is proportional to (a) P/E (b) E/P (c) P.E (d) PEtan⁡θPE\tan\theta

Answer

(c) P⋅EP\cdot E (the product of pp and EE)

The torque is τ⃗=p⃗×E⃗\vec\tau=\vec p\times\vec E, with magnitude τ=pEsin⁡θ\tau=pE\sin\theta, so for a given angle it is proportional to the product of pp and EE.

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  • 2074 · 1 mark

Wheatstone bridge is used to measure (a) emf (b) potential (c) current (d) resistance

Answer

(d) resistance

A Wheatstone bridge compares an unknown resistance with known ones; at balance P/Q=R/SP/Q=R/S, which gives the unknown resistance.

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  • 2074 · 1 mark

Magnetism of magnet (a) is Zero at absolution zero temperature (b) increase with temperature (c) is maximum at curie temperature (d) becomes practically zero at curie temperature

Answer

(d) becomes practically zero at curie temperature

Above the Curie temperature thermal agitation destroys the alignment of magnetic domains, so a ferromagnet becomes paramagnetic and its magnetism practically vanishes.

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  • 2074 · 1 mark

The positive rays are (a) electromagnetic waves (b) positrons (c) ions (d) protons

Answer

(c) ions

Positive rays (canal rays) are streams of positively charged gas ions.

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  • 2074 · 1 mark

Wavelength of a photon is 4000 Å. Its energy will be (a) 3KeV (b) 4000 eV (c) 3.1 J (d) 3.09 eV

Answer

(d) 3.09 eV

E=hcλ=12400 eV A˚4000 A˚=3.1 eV≈3.09E=\dfrac{hc}{\lambda}=\dfrac{12400\ \text{eV Å}}{4000\ \text{Å}}=3.1\ \text{eV}\approx3.09 eV.

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  • 2074 · 1 mark

According to Bohr's model of the H atom, the radius of the stationary orbit characterized by the principle quantum number n is proportional to (a) 1/n1/n (b) n2n^2 (c) nn (d) 1n2\frac{1}{n^2}

Answer

(b) n2n^2

In Bohr's model rn=n2h2ε0πme2∝n2r_n=\dfrac{n^2h^2\varepsilon_0}{\pi m e^2}\propto n^2.

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  • 2074 · 1 mark

In some substance, charge can flow at ordinary temperature but not at very low temperature. These are (a) semiconductors (b) conductors (c) insulators (d) dielectrics

Answer

(a) semiconductors

Semiconductors conduct at ordinary temperature because thermally excited carriers exist, but at very low temperature they behave like insulators.

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  • 2075 · 1 mark

The rate of tension and surface tension has dimension of (a) [M0L0T0][M^0L^0T^0] (b) [L][L] (c) [L−1][L^{-1}] (d) [MT−2][MT^{-2}]

Answer

(d) [MT−2][MT^{-2}]

Surface tension is force per unit length: MLT−2L=MT−2\frac{MLT^{-2}}{L}=MT^{-2}.

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  • 2075 · 1 mark

A force of 5N acts on a body of weight 10N the acceleration of body is (a) 50 m/s2^2 (b) 2 m/s2^2 (c) 5 m/s2^2 (d) 0.5 m/s2^2

Answer

(c) 5 m/s2^2

Weight 10 N gives mass m=10/10=1m=10/10=1 kg (taking g=10g=10 m/s2^2).

a=Fm=51=5 m/s2a=\frac{F}{m}=\frac{5}{1}=5\ \text{m/s}^2

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  • 2075 · 1 mark

The depth from surface of earth at which value of acceleration due to gravity become half of on the earth is (a) R (b) 2\sqrt2 R (c) R/2 (d) R/4

Answer

(c) R/2

At depth dd, g′=g(1−dR)g'=g\left(1-\frac dR\right). Setting g′=g/2g'=g/2 gives 1−dR=121-\frac dR=\frac12, so d=R/2d=R/2.

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  • 2075 · 1 mark

Spokes in a bicycle wheel is used to (a) increase mass (b) give better mass (c) increase moment of inertia (d) drive easily

Answer

(c) increase moment of inertia

Spokes carry the mass of the wheel out to the rim, which increases the moment of inertia about the axle, so the wheel runs more steadily.

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  • 2075 · 1 mark

A rain drop of radius 0.3mm has terminal velocity 1m/s in air. The viscosity of air is 18×10−518\times10^{-5} poise, then viscous force is (a) 1.01×10−71.01\times10^{-7} N (b) 1.01×10−81.01\times10^{-8} N (c) 1.6×10−91.6\times10^{-9} N (d) 1.6×10−81.6\times10^{-8} N

Answer

(a) 1.01×10−71.01\times10^{-7} N

Viscous force (Stokes' law) is F=6πηrvF=6\pi\eta rv, with η=18×10−5\eta=18\times10^{-5} poise =18×10−6=18\times10^{-6} Pa s.

F=6π(18×10−6)(0.3×10−3)(1)≈1.01×10−7 NF=6\pi(18\times10^{-6})(0.3\times10^{-3})(1)\approx1.01\times10^{-7}\ \text{N}

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  • 2075 · 1 mark

The material of least linear expansivity is (a) aluminum (b) invar (c) steel (d) nickel

Answer

(b) invar

Invar (a nickel-iron alloy) has a very small linear expansivity, about 1×10−6 /∘1\times10^{-6}\,/^\circC, so it is used in precision instruments and clocks.

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  • 2075 · 1 mark

A glass vessel contains air at 27∘27^\circC the temperature to which it must be heated so that one-fourth of air is expelled from it at constant pressure is (a) 47∘47^\circC (b) 77∘77^\circC (c) 102∘102^\circC (d) 127∘127^\circC

Answer

(d) 127∘127^\circC

The vessel has fixed volume and pressure, so nTnT is constant. If one-fourth of the air is expelled, n2=34n1n_2=\frac34n_1.

T2=43×300=400 K=127∘CT_2=\frac{4}{3}\times300=400\ \text{K}=127^\circ\text{C}

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  • 2075 · 1 mark

The efficiency of heat engine will be 100 % if the temperature of sink is (a) 0K (b) 0∘0^\circC (c) equal to source (d) none of above

Answer

(a) 0K

η=1−T2T1\eta=1-\frac{T_2}{T_1}, which equals 1 (100%) only when the sink temperature T2=0T_2=0 K.

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  • 2075 · 1 mark

A cylinder of 0.6 m long and 6 mm radius is rated at 1.5 KW. The temperature of filament if Stefan's constant is 6×10−86\times10^{-8} W m−2^{-2}K−4^{-4} is (a) 920K (b) 1125 K (c) 1025 K (d) 1320 K

Answer

(c) 1025 K

Power radiated P=σAT4P=\sigma AT^4, with curved area A=2πrl=2π(0.006)(0.6)=0.0226A=2\pi rl=2\pi(0.006)(0.6)=0.0226 m2^2.

T4=15006×10−8×0.0226≈1.1×1012⇒T≈1025 KT^4=\frac{1500}{6\times10^{-8}\times0.0226}\approx1.1\times10^{12}\Rightarrow T\approx1025\ \text{K}

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  • 2075 · 1 mark

Two lens of power 5D and -7.5D are placed in contact. The equivalent focal length of combination will be (a) -40cm (b) +40cm (c) -25cm (d) +25 cm

Answer

(a) -40cm

P=5−7.5=−2.5P=5-7.5=-2.5 D, so f=1P=−0.4f=\frac{1}{P}=-0.4 m =−40=-40 cm.

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  • 2075 · 1 mark

The length of microscope is 14cm and magnifying power is 25 for relaxed eye the focal length of eyepiece is 5cm than distance of object from objective is (a) 2.4 cm (b) 2.1 cm (c) 1.5 cm (d) 1.8cm

Answer

(d) 1.8cm

For a relaxed eye (final image at infinity) M=v0u0⋅DfeM=\frac{v_0}{u_0}\cdot\frac{D}{f_e} and the tube length is L=v0+feL=v_0+f_e.

v0=14−5=9v_0=14-5=9 cm. 25=9u0⋅255⇒u0=95=1.825=\frac{9}{u_0}\cdot\frac{25}{5}\Rightarrow u_0=\frac{9}{5}=1.8 cm.

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  • 2075 · 1 mark

In bringing an electron towards another electron, electrostatic potential energy of the system (a) decreased (b) increases (c) remains same (d) zero

Answer

(b) increases

Two like charges repel, so work must be done against the repulsion to bring them closer. This work is stored as electrostatic potential energy, which increases (U=kq1q2rU=k\frac{q_1q_2}{r} with rr decreasing).

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  • 2075 · 1 mark

A wire of resistance 12 Ω\Omega is bent to form a circle. The effective resistance between the two points on any diameter of circle is (a) 12 Ω\Omega (b) 9 Ω\Omega (c) 6 Ω\Omega (d) 3 Ω\Omega

Answer

(d) 3 Ω\Omega

A diameter divides the circle into two semicircles of 6 Ω\Omega each, connected in parallel.

R=6×66+6=3 ΩR=\frac{6\times6}{6+6}=3\ \Omega

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  • 2075 · 1 mark

The resistance of an ideal ammeter is (a) 0 (b) low (c) high (d) infinite

Answer

(a) 0

An ammeter is connected in series and must not change the current, so an ideal ammeter has zero resistance.

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  • 2075 · 1 mark

A 10 ev electron is circulating in a plane at right angles to a uniform field of magnetic induction 10−410^{-4}T, the orbital radius of electron is (a) 18 cm (b) 16 cm (c) 11 cm (d) 9 cm

Answer

(c) 11 cm

KE=10KE=10 eV =1.6×10−18=1.6\times10^{-18} J, so v=2KEm=3.2×10−189.1×10−31≈1.87×106v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{3.2\times10^{-18}}{9.1\times10^{-31}}}\approx1.87\times10^6 m/s.

r=mveB=9.1×10−31×1.87×1061.6×10−19×10−4≈0.107 m≈11 cmr=\frac{mv}{eB}=\frac{9.1\times10^{-31}\times1.87\times10^6}{1.6\times10^{-19}\times10^{-4}}\approx0.107\ \text{m}\approx11\ \text{cm}

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  • 2075 · 1 mark

The property of retentivity of material is useful in construction of (a) permanent magnet (b) electromagnet (c) transformer (d) none of above

Answer

(a) permanent magnet

Retentivity is the ability of a material to retain magnetism after the magnetising field is removed. High retentivity (with high coercivity) is needed for permanent magnets.

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  • 2075 · 1 mark

The color of the positive column in a gas discharge tube depends on (a) type of glass (b) glass used for the tube (c) the gas in the tube (d) pressure inside the tube

Answer

(c) the gas in the tube

The colour of the glow in the positive column comes from the excitation of the gas atoms, so it is characteristic of the gas (for example neon gives red-orange).

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  • 2075 · 1 mark

To obtain p-type silicon semiconductor, pure silicon may doped with (a) aluminum (b) oxygen (c) phosphorus (d) copper

Answer

(a) aluminum

p-type silicon is made by doping with a trivalent impurity (acceptor) such as aluminium, which creates holes. Phosphorus is pentavalent and gives n-type.

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  • 2076 · 1 mark

The displacement of a particle moving along x-axis is given by x=a+bt+ct2x=a+bt+ct^2. The acceleration of the particle is (a) b (b) c (c) b+d (d) 2c

Answer

(d) 2c v=dxdt=b+2ctv=\frac{dx}{dt}=b+2ct, so a=dvdt=2ca=\frac{dv}{dt}=2c.

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  • 2076 · 1 mark

Two masses 1g and 4g are moving with equal K-E. The ratio of their momentum is (a) 1:6 (b) 1:2 (c) 1:4 (d) 1:21:\sqrt2

Answer

(b) 1:2 p=2mKp=\sqrt{2mK}, so for equal KK, p1p2=m1m2=14=12\frac{p_1}{p_2}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac14}=\frac12.

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  • 2076 · 1 mark

The force of gravitational attraction on a body is (a) Minimum at the equator (b) Minimum at the poles (c) Minimum midway between poles and equator (d) Same at all points on the earth's surface

Answer

(a) Minimum at the equator The earth bulges at the equator, so gg is smaller there. The earth's rotation also reduces the effective weight at the equator.

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  • 2076 · 1 mark

The surface energy of a drop of water of radius 'r' is proportional to (a) r3r^3 (b) r2r^2 (c) rr (d) 1/r1/r

Answer

(b) r2r^2 Surface energy = surface tension ×\times area =T⋅4πr2∝r2=T\cdot4\pi r^2\propto r^2.

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  • 2076 · 1 mark

A body of volume 36cc floats with 3/4 of its volume submerged in water. The density of body is (a) 0.25 g/cc (b) 0.75 g/cc (c) 0.9 g/cc (d) 0.1 g/cc

Answer

(b) 0.75 g/cc For floating, ρbodyV=ρwater⋅34V\rho_{body}V=\rho_{water}\cdot\frac34V, so ρbody=0.75\rho_{body}=0.75 g/cc.

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  • 2076 · 1 mark

A perfect gas (γ=1.5\gamma=1.5) is compressed to one fourth of the original volume. If initial pressure of the gas is 1 atmosphere and the compression be adiabatic, then the final pressure will be (a) 2 atm (b) 4 atm (c) 8 atm (d) 16 atm

Answer

(c) 8 atm P1V1γ=P2V2γP_1V_1^\gamma=P_2V_2^\gamma gives P2=P1(V1V2)1.5=41.5=8P_2=P_1\left(\frac{V_1}{V_2}\right)^{1.5}=4^{1.5}=8 atm.

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  • 2076 · 1 mark

A black body is at a temperature 727∘727^\circC. It emits energy at the rate which is proportional to (a) 727 (b) (727)4(727)^4 (c) 1000 (d) (1000)4(1000)^4

Answer

(d) (1000)4(1000)^4 By Stefan's law, E∝T4E\propto T^4 with TT in kelvin: 727+273=1000727+273=1000 K.

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  • 2076 · 1 mark

8 g of water at 30∘30^\circC is poured on a large block of ice at 0∘0^\circC. The mass of ice that melts is (a) 30 g (b) 50 g (c) 150g (d) 1600 g

Answer

(a) 30 g The option 30 g corresponds to 80 g of water, since 80×1×30=m×8080\times1\times30=m\times80 gives m=30m=30 g. For exactly 8 g the result would be 3 g, which is not an option.

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  • 2076 · 1 mark

Liquids boils when its vapor pressure is (a) Equal to total pressure on surface (b) Greater than total pressure on surface (c) Less than total pressure on surface (d) Equal to atmospheric pressure

Answer

(a) Equal to total pressure on surface A liquid boils when its saturated vapour pressure equals the external pressure on its surface.

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  • 2076 · 1 mark

The focal length of a convex lens is (a) Maximum for red color (b) Maximum for violet color (c) Maximum for yellow color (d) Same for all colors

Answer

(a) Maximum for red color Red has the least refractive index, and 1f=(μ−1)(1R1−1R2)\frac1f=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right), so ff is greatest for red and least for violet.

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  • 2076 · 1 mark

The refracting angle of a prism of refractive index 2\sqrt2 is 60∘60^\circ. The minimum deviation from it is (a) 30∘30^\circ (b) 45∘45^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(a) 30∘30^\circ μ=sin⁡A+D2sin⁡A2\mu=\frac{\sin\frac{A+D}{2}}{\sin\frac A2} gives 2=sin⁡60∘+D2sin⁡30∘\sqrt2=\frac{\sin\frac{60^\circ+D}{2}}{\sin30^\circ}. So sin⁡60∘+D2=22\sin\frac{60^\circ+D}{2}=\frac{\sqrt2}{2}, which means 60∘+D2=45∘\frac{60^\circ+D}{2}=45^\circ and D=30∘D=30^\circ.

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  • 2076 · 1 mark

In young's double slit experiment of interference, the distance between the slits is gradually increased. The width of the fringe system (a) increases (b) decreases (c) remain same (d) first increases and then decreases

Answer

(b) decreases Fringe width β=λDd\beta=\frac{\lambda D}{d}, so it decreases as the slit separation dd increases.

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  • 2076 · 1 mark

2% of light of emitted by a source of luminous intensity 500 cd falls normally on a circular disc of radius 10 cm, calculate the average illumination of surface. (a) 1000 lux (b) 2000 lux (c) 3000 lux (d) 4000 lux

Answer

(d) 4000 lux Total flux =4πI=4π×500=2000π=4\pi I=4\pi\times500=2000\pi lm, and 2% of it is 40π40\pi lm. Area =π(0.1)2=0.01π=\pi(0.1)^2=0.01\pi m2^2, so E=40π0.01π=4000E=\frac{40\pi}{0.01\pi}=4000 lux.

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  • 2076 · 1 mark

A hollow metal sphere of radius 10cm is charged such that the potential at its surface is 80V. The potential at the center of sphere is (a) 8V (b) 80V (c) 800V (d) zero

Answer

(b) 80V Inside a charged hollow conductor the field is zero, so the potential is constant and equals the surface value.

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  • 2076 · 1 mark

Two capacitors of capacitances 2μ\muF and 3μ\muF are connected in parallel. The ratio of charges on them will be (a) 1:1 (b) 2:3 (c) 3:2 (d) 4:9

Answer

(b) 2:3 In parallel the potential difference is the same, so Q=CVQ=CV gives Q1:Q2=C1:C2=2:3Q_1:Q_2=C_1:C_2=2:3.

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  • 2076 · 1 mark

The angle between geographical and magnetic meridian is called (a) angle of dip (b) angle of declination (c) critical angle (d) curie angle

Answer

(b) angle of declination The angle between the geographic meridian and the magnetic meridian is the declination.

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  • 2076 · 1 mark

A charged particle moves along the axis of a current carrying solenoid. It experiences (a) an accelerating force (b) a refracting force (c) a radial force (d) no force

Answer

(d) no force The velocity is parallel to the field along the axis, so F=qvBsin⁡0∘=0F=qvB\sin0^\circ=0.

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  • 2076 · 1 mark

The energy dissipation in a resistor of resistance R in time t is E, then the value of current flowing in the resistance will be (a) ERt\sqrt{\frac{E}{Rt}} (b) (ERt)2\left(\frac{E}{Rt}\right)^2 (c) ERtERt (d) (ERt)1/2(ERt)^{1/2}

Answer

(a) ERt\sqrt{\frac{E}{Rt}} E=I2RtE=I^2Rt, so I=ERtI=\sqrt{\frac{E}{Rt}}.

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  • 2076 · 1 mark

The equation of a wave travelling in a string is given by y=3cos⁡π(100t−x)y=3\cos\pi(100t-x), where x and y are in cm and t is in second. Its wavelength is (a) 3 cm (b) 100 cm (c) 2 cm (d) 5 cm

Answer

(c) 2 cm

Comparing y=3cos⁡(100πt−πx)y=3\cos(100\pi t-\pi x) with y=Acos⁡(ωt−kx)y=A\cos(\omega t-kx) gives k=πk=\pi cm−1^{-1}, so λ=2πk=2\lambda=\dfrac{2\pi}{k}=2 cm.

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  • 2076 · 1 mark

A source of frequency 150 Hz is moving in a direction of a person with a velocity of 110 m/s. The frequency heard by the person will be (Velocity of sound = 330 m/s). (a) 225 Hz (b) 200 Hz (c) 150 Hz (d) 100 Hz

Answer

(a) 225 Hz

For a source approaching a stationary listener:

f′=fvv−vs=150×330330−110=225 Hzf'=f\frac{v}{v-v_s}=150\times\frac{330}{330-110}=225\text{ Hz}

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  • 2076 · 1 mark

In stationary wave (a) strain is max. at nodes (b) strain is max. at antinodes (c) strain is min. at nodes (d) Amplitude is min. at all points

Answer

(a) strain is max. at nodes

In a stationary wave, particles at nodes do not move but the medium on either side moves in opposite directions, so the change in length per unit length (strain) is maximum at nodes. At antinodes the strain is minimum while the displacement is maximum.

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  • 2076 · 1 mark

The de-Broglie wavelength of an electron in nthn^{th} Bohr orbit is related to the radius R of the orbit as, (a) nλ=πRn\lambda=\pi R (b) nλ=32πRn\lambda=\frac32\pi R (c) nλ=2πRn\lambda=2\pi R (d) nλ=4πRn\lambda=4\pi R

Answer

(c) nλ=2πRn\lambda=2\pi R

Bohr's quantisation condition mvR=nh2πmvR=\frac{nh}{2\pi} with λ=hmv\lambda=\frac hmv gives 2πR=nλ2\pi R=n\lambda: the orbit holds a whole number of de-Broglie wavelengths.

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  • 2076 · 1 mark

Which of the following electronic transitions of the hydrogen atom gives an absorption line of highest frequency? (a) n=1n=1 to n=2n=2 (b) n=3n=3 to n=8n=8 (c) n=2n=2 to n=1n=1 (d) n=8n=8 to n=3n=3

Answer

(a) n=1n=1 to n=2n=2

Absorption means the electron moves to a higher level. Of the absorption transitions (a) and (b), the energy change is 13.6(1−14)=10.213.6\left(1-\tfrac14\right)=10.2 eV for (a) and only 13.6(19−164)≈1.313.6\left(\tfrac19-\tfrac1{64}\right)\approx1.3 eV for (b). Options (c) and (d) are emissions.

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  • 2076 · 1 mark

The half-life of a radioactive isotope is 5 yrs. The fraction of atoms decayed in this substance after 15 years will be (a) 1 (b) 34\frac34 (c) 18\frac18 (d) 78\frac78

Answer

(d) 78\frac78

15 years is 3 half-lives, so the fraction remaining is (12)3=18\left(\tfrac12\right)^3=\tfrac18. The fraction decayed is 1−18=781-\tfrac18=\tfrac78.

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  • 2076 · 1 mark

What are β\beta-rays? (a) E.M waves (b) Electron orbiting round the nucleus (c) Charge particle emitted by nucleus (d) Neutral particle

Answer

(c) Charge particle emitted by nucleus

β\beta-rays are fast electrons (or positrons) emitted from the nucleus when a neutron changes into a proton. They are charged particles, not electromagnetic waves.

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  • 2077 · 1 mark

Which of the following pair have same dimension? (a) L/R and CR (b) LR and CR (c) R/L and [LC]1/2[LC]^{1/2} (d) CR and 1/LC

Answer

(a) L/R and CR

Both L/RL/R and CRCR are time constants, so each has the dimension of time [T][T].

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  • 2077 · 1 mark

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm, the further distance travelled before coming to the rest is (a) 4 cm (b) 2 cm (c) 3 cm (d) 1 cm

Answer

(d) 1 cm

For the first 3 cm: (u/2)2=u2−2a(3)(u/2)^2=u^2-2a(3), so a=u2/8a=u^2/8.

Further distance until rest: s=(u/2)22a=u2/4u2/4=1s=\dfrac{(u/2)^2}{2a}=\dfrac{u^2/4}{u^2/4}=1 cm.

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  • 2077 · 1 mark

The 'net force acting on a body is zero' then the wrong statement is (a) body is at rest (b) acceleration is not zero (c) body is in motion (d) acceleration should be zero

Answer

(b) acceleration is not zero

If the net force is zero, then a=0a=0 by Newton's second law. The body is either at rest or moving with constant velocity, so the statement that acceleration is not zero is wrong.

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  • 2077 · 1 mark

A metallic ball is immersed in alcohol. The coefficient of cubical expansion of metal is less than that of alcohol. When the system is heated weight of ball is (a) Increases (b) remains unchanged (c) decreases (d) first increases and then decreases

Answer

(a) Increases

Upthrust =Vρlg=V\rho_l g. On heating, alcohol expands more than the metal ball, so its density falls faster than the ball's volume rises and the upthrust decreases. The apparent weight == true weight −- upthrust therefore increases.

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  • 2077 · 1 mark

The average kinetic energy per degree of freedom per molecule of an ideal gas is (a) KTKT (b) 2KT2KT (c) 12KT\frac12KT (d) 23KT\frac23KT

Answer

(c) 12KT\frac12KT

By the law of equipartition of energy, each degree of freedom carries 12kT\tfrac12 kT of energy per molecule.

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  • 2077 · 1 mark

Two spheres of same material have radii in the ratio 3:2. The heat radiated by them at the same temperature will be (a) 1:1 (b) 4:9 (c) 9:4 (d) 3:2

Answer

(c) 9:4

Heat radiated is proportional to surface area, which is proportional to r2r^2. So the ratio is 32:22=9:43^2:2^2=9:4.

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  • 2077 · 1 mark

A person is in a room whose ceiling and two adjacent walls are mirrors. Number of images formed of an object is (a) 2 (b) 3 (c) 4 (d) 5

Answer

(b) 3

Two adjacent walls that are perpendicular mirrors form 360∘90∘−1=3\frac{360^\circ}{90^\circ}-1=3 images, which is the listed option that matches this setup.

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  • 2077 · 1 mark

The refractive index is 1.414 and refracting angle is 60∘60^\circ, then minimum deviation of light will be (a) 30∘30^\circ (b) 60∘60^\circ (c) 45∘45^\circ (d) 72∘72^\circ

Answer

(a) 30∘30^\circ

n=sin⁡A+D2sin⁡A2⇒sin⁡A+D2=1.414×sin⁡30∘=0.707n=\frac{\sin\frac{A+D}{2}}{\sin\frac A2}\Rightarrow\sin\frac{A+D}{2}=1.414\times\sin30^\circ=0.707

So A+D2=45∘\frac{A+D}{2}=45^\circ, giving D=90∘−60∘=30∘D=90^\circ-60^\circ=30^\circ.

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  • 2077 · 1 mark

Two fixed charge q and 4q are at r distance apart. What will be positive of third charge to be placed so that the system will be in equilibrium? (a) 2r/3 from 4q (b) 2r from q (c) r/2 from q (d) r/2 from 4q

Answer

(a) 2r/3 from 4q

Let the third charge be at distance xx from qq. For equilibrium the forces on it balance:

qx2=4q(r−x)2⇒r−x=2x⇒x=r3\frac{q}{x^2}=\frac{4q}{(r-x)^2}\Rightarrow r-x=2x\Rightarrow x=\frac r3

So it is at r/3r/3 from qq, i.e. 2r/32r/3 from 4q4q.

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  • 2077 · 1 mark

A heater coil is cut into two equal parts and only one part is used in the heater. How will the heat generated vary? (a) one fourth (b) doubled (c) halved (d) four times

Answer

(b) doubled

The heat generated is H=V2RtH=\dfrac{V^2}{R}t. Cutting the coil in half halves the resistance, so the heat generated doubles.

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  • 2077 · 1 mark

A 50 v battery is connected across 10 ohm resistor. The current in the circuit is 4.5 Ampere. The internal resistance of the battery should be (a) zero (b) 5.0 Ohm (c) 0.5 Ohm (d) 1.1 Ohm

Answer

(d) 1.1 Ohm

I=ER+r⇒R+r=504.5=11.1 Ω⇒r=11.1−10≈1.1 ΩI=\frac{E}{R+r}\Rightarrow R+r=\frac{50}{4.5}=11.1\ \Omega\Rightarrow r=11.1-10\approx1.1\ \Omega

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  • 2077 · 1 mark

A circuit contains a capacitor of 420 pf and an inductance L. The value of 'L' to broadcast on Radio at 1020 KHz is (a) 2.8×10−42.8\times10^{-4} H (b) 7.5×10−47.5\times10^{-4} H (c) 5.8×1065.8\times10^{6} H (d) 9.5×1069.5\times10^{6} H

Answer

(c) 5.8×1065.8\times10^{6} H (as listed)

L=14π2f2C=14π2(1.02×106)2(420×10−12)≈5.8×10−5 HL=\frac{1}{4\pi^2f^2C}=\frac{1}{4\pi^2(1.02\times10^6)^2(420\times10^{-12})}\approx5.8\times10^{-5}\text{ H}

The digits 5.8 match option (c), whose exponent appears to be misprinted.

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  • 2077 · 1 mark

Electron accelerated from rest to a potential difference of 100 volt. Its final velocity will be (a) 5×1055\times10^5 m/s (b) 3×1033\times10^3 m/s (c) 4×1044\times10^4 m/s (d) 6×1066\times10^6 m/s

Answer

(d) 6×1066\times10^6 m/s

12mv2=eV⇒v=2eVm=2(1.6×10−19)(100)9.1×10−31≈5.9×106≈6×106 m/s\tfrac12mv^2=eV\Rightarrow v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2(1.6\times10^{-19})(100)}{9.1\times10^{-31}}}\approx5.9\times10^6\approx6\times10^6\text{ m/s}

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  • 2077 · 1 mark

When the proton collides with an electron, which of the following characteristics of proton increases? (a) Energy (b) wavelength (c) Frequency (d) impulse

Answer

(b) wavelength

This is the Compton-type collision, where the incident quantum gives energy and momentum to the electron. The quantum loses energy and frequency, so its wavelength increases.

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  • 2078 · 1 mark

The velocity of a body is given by the equation v=b/t+ct2+dt3v=b/t+ct^2+dt^3. Then dimensional formula of c is (a) [M0LT3][M^0LT^3] (b) [ML0T3][ML^0T^3] (c) [M0L0T][M^0L^0T] (d) [M0LT−3][M^0LT^{-3}]

Answer

(d) [M0LT−3][M^0LT^{-3}]

Each term has the dimension of velocity [LT−1][LT^{-1}], so [c]=[LT−1][T2]=[LT−3][c]=\frac{[LT^{-1}]}{[T^2]}=[LT^{-3}].

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  • 2078 · 1 mark

If magnitudes of two vectors are 3 and 4 and value of their scalar product is 6. The angle between them is (a) 30∘30^\circ (b) 45∘45^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(c) 60∘60^\circ

A⃗⋅B⃗=ABcos⁡θ⇒6=12cos⁡θ⇒cos⁡θ=12\vec A\cdot\vec B=AB\cos\theta\Rightarrow 6=12\cos\theta\Rightarrow\cos\theta=\frac12, so θ=60∘\theta=60^\circ.

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  • 2078 · 1 mark

Stars appears to move from east to west because (a) all stars move from east to west. (b) the earth rotates from west to east. (c) the earth rotates from east to west (d) the back ground of the stars moves from west to east

Answer

(b) the earth rotates from west to east.

The earth spins west to east, so the stars appear to move east to west relative to us.

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  • 2078 · 1 mark

To avoid the slipping while walking on ice one should take smaller steps because of the (a) larger friction (b) smaller friction (c) large normal reaction (d) smaller normal reaction

Answer

(b) smaller friction

Ice offers very little friction, so slipping is likely. Taking smaller steps keeps the body closer to balance and needs less horizontal frictional force at each step.

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  • 2078 · 1 mark

A scooter of mass 120kg is moving with a uniform velocity of 108km/hr. The force required to stop the velocity in 10 sec is (a) 360N (b) 720N (c) 180N (d) 90N

Answer

(a) 360N

v=108 km/h=30 m/sv=108\text{ km/h}=30\text{ m/s}. Deceleration a=3010=3 m/s2a=\frac{30}{10}=3\text{ m/s}^2. F=ma=120×3=360 NF=ma=120\times3=360\text{ N}.

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  • 2078 · 1 mark

In Young's double slit experiment the distance of tenth bright fringe from the center of interference pattern on either side is 3.44cm the slit separation if the screen is at 2m for wave length 589nm is (a) 0.24mm (b) 0.44mm (c) 0.34mm (d) 0.64mm

Answer

(c) 0.34mm

For the nn-th bright fringe y=nλDdy=\frac{n\lambda D}{d}, so d=10×589×10−9×23.44×10−2≈3.4×10−4 m=0.34 mm.d=\frac{10\times589\times10^{-9}\times2}{3.44\times10^{-2}}\approx3.4\times10^{-4}\text{ m}=0.34\text{ mm}.

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  • 2078 · 1 mark

Up thrust depends upon (a) shape of the body (b) depth from the liquid surface (c) volume of liquid displaced (d) mass of the body

Answer

(c) volume of liquid displaced

By Archimedes' principle, upthrust equals the weight of the liquid displaced, VρgV\rho g.

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  • 2078 · 1 mark

Two copper rods of equal length and unequal diameters are heated through the same range of temperature. Its increase in length will be (a) more in thin rod (b) more in thick rod (c) same in both rod (d) uncertain

Answer

(c) same in both rod

ΔL=LαΔT\Delta L=L\alpha\Delta T depends on the length, material and temperature change, not on the diameter.

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  • 2078 · 1 mark

The boiling of water inside the pressure cooker is (a) 0∘C0^\circ C (b) 100∘C100^\circ C (c) 120∘C120^\circ C (d) −120∘C-120^\circ C

Answer

(c) 120∘C120^\circ C

In a pressure cooker the pressure is higher than atmospheric, which raises the boiling point to about 120∘120^\circC.

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  • 2078 · 1 mark

The image of a distant object formed by a far sighted person is (a) at the retina (b) behind the retina (c) in front of retina (d) no image is formed

Answer

(b) behind the retina

A far-sighted (hypermetropic) eye has too weak a converging power or too short an eyeball, so the image falls behind the retina. It is corrected with a convex lens.

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  • 2078 · 1 mark

The charge in uniform motion produces (a) an electric field only (b) a magnetic field only (c) both electric and magnetic field (d) neither electric nor magnetic

Answer

(c) both electric and magnetic field

A charge at rest produces only an electric field. A moving charge also constitutes a current, which produces a magnetic field.

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  • 2078 · 1 mark

In order to increase the capacity of parallel plate capacitor one should introduce between the plates a sheet of (a) tin (b) copper (c) steel (d) mica

Answer

(d) mica

A dielectric between the plates raises the capacitance by the factor KK. Mica is a dielectric (K≈5K\approx5-77), whereas tin, copper and steel are conductors, which would short or reduce the effective plate gap.

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  • 2078 · 1 mark

A 100W/220V bulb is used in a line of 440V. Its power will be (a) 50W (b) 100W (c) 200W (d) 400W

Answer

(d) 400W

R=V2P=2202100=484 ΩR=\frac{V^2}{P}=\frac{220^2}{100}=484\ \Omega. At 440 V, P=4402484=400 WP=\frac{440^2}{484}=400\text{ W} (the bulb would burn out in practice).

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  • 2078 · 1 mark

The electric field intensity at the surface of charged conductor is (a) zero (b) Infinite (c) directed tangentially to the surface (d) directed normally to the surface

Answer

(d) directed normally to the surface

The field just outside a charged conductor is E=σε0E=\frac{\sigma}{\varepsilon_0}, perpendicular to the surface. Any tangential component would move the surface charges.

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  • 2078 · 1 mark

In a non-resonant circuit what will be the nature of the circuit for frequencies higher than the resonant frequency? (a) resistive (b) capacitive (c) inductive (d) passive

Answer

(c) inductive

For f>f0f>f_0, XL=ωL>XC=1ωCX_L=\omega L>X_C=\frac{1}{\omega C}, so the net reactance is inductive.

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  • 2078 · 1 mark

The material used for permanent magnet has (a) high resistivity, high coercivity (b) low retentivity, low coercivity (c) low retentivity, high coercivity (d) high retentivity, high coercivity

Answer

(d) high retentivity, high coercivity

A permanent magnet must retain its magnetism (high retentivity) and resist demagnetisation (high coercivity). Steel is an example.

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  • 2078 · 1 mark

The angle of dip is the angle (a) between the vertical component of earth's magnetic field and magnetic meridian (b) between the vertical component of earth's magnetic field and geographic meridian (c) between Earth's magnetic field's direction and horizontal direction (d) between the earth's magnetic meridian and geographic meridian

Answer

(c) between Earth's magnetic field's direction and horizontal direction

The angle of dip (inclination) is the angle the total magnetic field of the earth makes with the horizontal, measured in the magnetic meridian plane.

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  • 2078 · 1 mark

If an electron has initial velocity in a direction perpendicular to the direction of electric field, the path of an electron is (a) straight line (b) parabola (c) circle (d) ellipse

Answer

(b) parabola The electron feels a constant force along the field, so it moves uniformly in the perpendicular direction and accelerates along the field. This gives y∝x2y \propto x^2, a parabola.

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  • 2078 · 1 mark

The Planck's constant can be experimentally determined by (a) spectroscopic experiments (b) tangent galvanometer experiment (c) compound pendulum experiment (d) photo-electric experiment.

Answer

(d) photo-electric experiment Using eV0=hf−ϕeV_0 = hf - \phi, the slope of the stopping potential versus frequency graph is h/eh/e, which gives Planck's constant.

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  • 2078 · 1 mark

When electrons are accelerated by 50 KV and 0.4% of energy of cathode rays is converted into x-rays and heat is evolved at 600 W. The current is (a) 0.12 A (b) 0.012 A (c) 0.024 A (d) 2.4 A

Answer

(b) 0.012 A 99.6% of the input power appears as heat, so the total power is P=6000.996≈602 WP=\frac{600}{0.996}\approx 602\ \text{W} I=PV=60250×103≈0.012 AI=\frac{P}{V}=\frac{602}{50\times10^{3}}\approx 0.012\ \text{A}

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  • 2078 · 1 mark

The difference between 92U235_{92}U^{235} and 92U238_{92}U^{238} atom is (a) 92U238_{92}U^{238} contain 3 more protons (b) 92U238_{92}U^{238} contain 3 more protons and electrons (c) 92U238_{92}U^{238} contains 3 more neutrons (d) 92U238_{92}U^{238} contain 3 more neutrons and 3 more electrons

Answer

(c) 92U238_{92}U^{238} contains 3 more neutrons Both are isotopes with 92 protons and 92 electrons. Neutrons are 235−92=143235-92=143 and 238−92=146238-92=146, a difference of 3.

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  • 2078 · 1 mark

In the depletion layer of an unbiased p-n junction diode, there are (a) only electrons (b) only holes (c) both electrons and holes (d) only fixed ions

Answer

(d) only fixed ions Free electrons and holes diffuse across the junction and recombine, leaving only immobile ionised donor and acceptor atoms in the depletion layer.

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  • 2079 · 1 mark

A quantity xx is related by x=π3(a2−b2)hx=\frac{\pi}{3}(a^2-b^2)h where a, b and h are lengths, then quantity xx is (a) length (b) velocity (c) area (d) volume

Answer

(d) volume

a2a^2, b2b^2 and hh are lengths, so (a2−b2)h(a^2-b^2)h has dimension L3L^3. The factor π3\frac{\pi}{3} is dimensionless, so xx has dimension L3L^3, which is volume.

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  • 2079 · 1 mark

A stone is dropped from a height of 45 m. The distance travelled by it during its last second is (a) 5 m (b) 10 m (c) 25 m (d) 50 m

Answer

(c) 25 m

Time of fall: 45=12(10)t245=\frac12(10)t^2, so t=3t=3 s. Distance in the first 2 s is 12(10)(2)2=20\frac12(10)(2)^2=20 m.

Distance in the last second is 45−20=2545-20=25 m.

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  • 2079 · 1 mark

A projectile is broken into two pieces at the maximum height. Then its (a) momentum increases (b) momentum decreases (c) kinetic energy increases (d) kinetic energy decreases

Answer

(c) kinetic energy increases

At the highest point the velocity is horizontal. Breaking into two pieces is caused by internal forces, so the total momentum stays the same. The chemical or stored energy released in the break-up is converted into kinetic energy, so the total kinetic energy increases.

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  • 2079 · 1 mark

Jet plane is based on the conservation of (a) mass (b) energy (c) linear momentum (d) angular momentum

Answer

(c) linear momentum

A jet plane expels hot gases backward at high speed. By conservation of linear momentum, the plane gains an equal forward momentum.

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  • 2079 · 1 mark

Which of the following possesses highest viscosity? (a) water (b) blood (c) glycerin (d) honey

Answer

(d) honey

Honey is far more viscous than water or blood, and it is also more viscous than glycerin at room temperature (about 10310^3 to 10410^4 mPa s for honey against about 1.4×1031.4\times10^3 mPa s for glycerin).

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  • 2079 · 1 mark

A metal sheet has a circular hole in it. When it is heated, diameter of the hole (a) increases (b) decreases (c) remains unchanged (d) First increases and then decreases

Answer

(a) increases

On heating, a solid expands uniformly in all directions, so every linear dimension grows in proportion, including the hole. The hole expands as if it were made of the same metal, so its diameter increases.

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  • 2079 · 1 mark

A body initially at 80∘C80^\circ C cools to 64∘C64^\circ C in 5 minutes and to 52∘C52^\circ C in 10 minutes. The temperature of the body after 15 minutes will be (a) 35∘C35^\circ C (b) 30∘C30^\circ C (c) 25∘C25^\circ C (d) 20∘C20^\circ C

Answer

(a) 35∘C35^\circ C (nearest option)

Use Newton's law of cooling with surrounding temperature T0T_0. The excess temperature falls by a constant ratio in equal times:

64−T080−T0=52−T064−T0⇒T0=16∘C\frac{64-T_0}{80-T_0}=\frac{52-T_0}{64-T_0}\Rightarrow T_0=16^\circ C

The excess temperatures are 64,48,3664, 48, 36 and the next is 36×0.75=2736\times0.75=27, so the temperature at 15 min is about 43∘C43^\circ C. This value is not listed, and the nearest option is (a).

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  • 2079 · 1 mark

A completely transparent material will be invisible in vacuum when refractive index μ\mu is (a) unity (b) more than unity (c) less than unity (d) infinity

Answer

(a) unity

A transparent body is visible because of reflection and refraction at its surface, which occur when its refractive index differs from that of the surroundings. In vacuum μ=1\mu=1, so the material becomes invisible when μ=1\mu=1.

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  • 2079 · 1 mark

Sweetness of sound depends upon its (a) wavelength (b) periodicity (c) periodicity and regularity (d) amplitude

Answer

(c) periodicity and regularity

Musical (sweet) sound is produced by regular, periodic vibrations. Noise is irregular and non-periodic, which is why it is unpleasant.

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  • 2079 · 1 mark

A parallel plate capacitor is charged and the battery is then disconnected. If the plates of capacitors are removed further apart by means of insulating handles (a) The charge on capacitors increases (b) The voltage across the plate increases (c) The capacitance increases (d) The energy stored decreases

Answer

(b) The voltage across the plate increases

The battery is disconnected, so the charge QQ stays constant. Moving the plates apart reduces C=ε0AdC=\frac{\varepsilon_0A}{d}. Then V=QCV=\frac{Q}{C} increases, and the stored energy Q22C\frac{Q^2}{2C} also increases.

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  • 2079 · 1 mark

A source of 34 V and 50 Hz is connected in series with coil of 17 mH and resister. If potential difference across resister is 10V. The potential difference across the coil is (a) 12 V (b) 16 V (c) 28 V (d) 32 V

Answer

(d) 32 V

The resistor and coil voltages are 90 degrees out of phase in a series circuit:

V2=VR2+VL2⇒VL=342−102=1056≈32.5 VV^2=V_R^2+V_L^2\Rightarrow V_L=\sqrt{34^2-10^2}=\sqrt{1056}\approx32.5\text{ V}

The nearest option is 32 V.

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  • 2079 · 1 mark

A strong argument for particle nature of cathode rays is that they (a) Produce fluorescence (b) Travel through vacuum (c) Get deflected by electric and magnetic field (d) Cast shadow

Answer

(c) Get deflected by electric and magnetic field

Deflection by electric and magnetic fields shows that cathode rays are streams of negatively charged particles. Fluorescence, travelling through vacuum and casting shadows can also be explained by waves.

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  • 2079 · 1 mark

When objects placed in a room are exposed to X-rays, they appear (a) Invisible (b) Yellow (c) Blue (d) Red

Answer

(a) Invisible

X-rays lie outside the visible range, so objects exposed to them show no colour change and cannot be seen.

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  • 2079 · 1 mark

Which series of Hydrogen atom lies in the infrared region? (a) Lyman (b) Balmer (c) Pfund (d) H-alpha

Answer

(c) Pfund

The Lyman series is in the ultraviolet and the Balmer series is in the visible region. The Paschen, Brackett and Pfund series lie in the infrared. H-alpha is a single line of the Balmer series.

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  • 2079 · 1 mark

When boron 5B10_5B^{10} is bombarded by neutron, alpha particles are emitted. The resulting nucleus has the mass number (a) 11 (b) 7 (c) 6 (d) 15

Answer

(b) 7

510B+01n→37Li+24He{}_5^{10}B+{}_0^1n\rightarrow{}_3^{7}Li+{}_2^{4}He

The mass number is conserved: 10+1=7+410+1=7+4. The resulting nucleus is lithium-7.

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  • 2079 · 1 mark

In P-type semiconductor, the majority and minority charge carries are respectively (a) Protons and electrons (b) Electrons and protons (c) Electrons and holes (d) Holes and electrons

Answer

(d) Holes and electrons

In a P-type semiconductor, trivalent impurity creates holes. Holes are the majority carriers and thermally generated electrons are the minority carriers.

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  • 2080 · 1 mark

Joule x second is the unit of (a) Energy (b) Momentum (c) Angular momentum (d) Power

Answer

(c) Angular momentum

Angular momentum L=IωL=I\omega has unit kg m2^2/s, which equals J s (Planck's constant has the same unit).

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  • 2080 · 1 mark

In equation, F⃗=q(v⃗×B⃗)\vec F=q(\vec v\times\vec B) the quantity F⃗\vec F is (a) Perpendicular to v⃗\vec v only (b) Perpendicular to B⃗\vec B only (c) Perpendicular to both v⃗\vec v and B⃗\vec B (d) Perpendicular to q and B⃗\vec B

Answer

(c) Perpendicular to both v⃗\vec v and B⃗\vec B

A cross product is perpendicular to both vectors that form it.

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  • 2080 · 1 mark

Two balls of the same mass are shot upwards one after another at an interval of 2 seconds along the same vertical line with the same initial velocity of 40 m/s. They were collide at a height of (a) 75 m (b) 80 m (c) 30 m (d) 45 m

Answer

(a) 75 m

Take g=10g=10. Equal heights: 40t−5t2=40(t−2)−5(t−2)240t-5t^2=40(t-2)-5(t-2)^2 gives 40t=60t−10040t=60t-100, so t=5t=5 s. Height =40(5)−5(25)=75=40(5)-5(25)=75 m.

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  • 2080 · 1 mark

Cricketer Sompal Kami hits a cricket ball at 45∘45^\circ to horizontal with a kinetic energy K. The kinetic energy of ball at highest point of path is (a) 0 (b) K/2 (c) K2\frac{K}{\sqrt2} (d) K

Answer

(b) K/2

At the highest point only the horizontal component ucos⁡45∘u\cos45^\circ remains, so KE =12m(ucos⁡45∘)2=Kcos⁡245∘=K2=\frac12m(u\cos45^\circ)^2=K\cos^2 45^\circ=\frac K2.

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  • 2080 · 1 mark

The maximum value of static friction is called (a) Rolling friction (b) Limiting friction (c) Maximum friction (d) Dynamic friction

Answer

(b) Limiting friction

The maximum value of static friction, just before the body starts to slide, is called limiting friction.

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  • 2080 · 1 mark

India has become the first country to successfully land a spacecraft Chandrayaan-3 near the south pole of the Moon. Its rover has confirmed the presence of (a) Oxygen and Sulphur (b) Hydrogen and Oxygen (c) Sulphur and Nitrogen (d) Nitrogen and Water

Answer

(a) Oxygen and Sulphur

The LIBS instrument on the Pragyan rover detected sulphur (S) in the lunar south-pole soil, along with Al, Ca, Fe, Cr, Ti, Mn, Si and oxygen.

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  • 2080 · 1 mark

A girl is sitting in a boat which is floating in Fewa Lake. If she drinks water from the lake, the level of water in the lake will be (a) Increased (b) Decreased (c) Remain unchanged (d) May increase or decrease depending upon the weight of the girl

Answer

(c) Remain unchanged

The boat floats with total weight of boat plus girl balanced by displaced water. Water she drinks is removed from the lake but adds equal weight to the floating system, so displaced volume rises by exactly the volume removed. The lake level stays the same.

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  • 2080 · 1 mark

Fire alarm work on the principle of (a) Volume expansion of solid (b) Volume expansion of water (c) Volume expansion of gas (d) Bimetallic bar which become curve on heating

Answer

(d) Bimetallic bar which become curve on heating

Two metals with different expansion coefficients bend on heating and close the alarm circuit.

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  • 2080 · 1 mark

During 5 - 12 September 2023, temperature at noon is recorded the same at the base camp of Mount Everest. A climber feels hottest when the relative humidity is (a) 99% (b) 50% (c) 30% (d) 10%

Answer

(a) 99%

Sweating cools by evaporation. At high relative humidity evaporation is slowest, so the body feels hottest.

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  • 2080 · 1 mark

A microscope is focused on an 'ink mark'. A glass plate is placed over the spot. In order to focus the spot, microscope is to be raised 1 m. In order to focus scratch on upper surface it must be raised 2 m further. The refractive index of glass is (a) 1.33 (b) 1.50 (c) 1.62 (d) 1.80

Answer

(b) 1.50

Raising the microscope by 1 m for the ink mark gives t−tμ=1t-\frac t\mu=1. The upper surface is a further 2 m, so thickness t=3t=3 m. Then 1=3(1−1μ)1=3\left(1-\frac1\mu\right), giving μ=1.5\mu=1.5.

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  • 2080 · 1 mark

Decrease in aperture of a lens changes (a) Position of the image (b) Size of the image (c) Intensity of the image (d) Both position and size of the image

Answer

(c) Intensity of the image

Reducing the aperture cuts the light passing through, so the image becomes dimmer, while position and size are unchanged.

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  • 2080 · 1 mark

Sound travels in rocks in the form of (a) Longitudinal elastic wave (b) Transverse elastic wave (c) Both longitudinal and transverse elastic waves (d) Stationary wave

Answer

(c) Both longitudinal and transverse elastic waves

Solids have rigidity as well as bulk elasticity, so both wave types propagate.

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  • 2080 · 1 mark

If radii of a solid and hollow sphere are same. Which can hold greater number of charges? (a) Solid sphere (b) Hollow sphere (c) Both will hold equal charges (d) Charge storing capacity depends upon surface temperature

Answer

(c) Both will hold equal charges

Charge resides on the outer surface, and capacity C=4πε0RC=4\pi\varepsilon_0R depends only on the radius.

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  • 2080 · 1 mark

To draw maximum current from a combination of cells, the cells be grouped as (a) series (b) parallel (c) mixed (d) both (a) and (b).

Answer

(c) mixed

Maximum current flows when the external resistance equals the internal resistance of the combination. A mixed grouping can be arranged to achieve this for any load.

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  • 2080 · 1 mark

Two parallel wires carrying currents in the same direction attract each other because of (a) Potential difference between them (b) Mutual inductance between them (c) Electric forces between them (d) Magnetic forces between them

Answer

(d) Magnetic forces between them

Parallel currents in the same direction attract through the magnetic force between them.

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  • 2080 · 1 mark

An emf of 2 volt is produced in a coil when current changes at steady rate from 3 to 2 amperes in 1 milli-second. The value of self-inductance is (a) 0 (b) 2 mH (c) 20 mH (d) 2 H

Answer

(b) 2 mH

∣ε∣=LdIdt|\varepsilon|=L\frac{dI}{dt}, so L=2(1)/(10−3)=2×10−3L=\frac{2}{(1)/(10^{-3})}=2\times10^{-3} H =2=2 mH.

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  • 2080 · 1 mark

In L-C-R circuit, potential difference between terminals of inductance is 60 volts, between terminals of condenser (or capacitor) is 30 volts, and between terminals of resistance is 40 volts. The supply voltage will be equal to (a) 130 volts (b) 70 volts (c) 50 volts (d) 10 volts

Answer

(c) 50 volts

V=VR2+(VL−VC)2=402+302=50V=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{40^2+30^2}=50 V.

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  • 2080 · 1 mark

The magnetic field of Earth is due to (a) Motion and distribution of materials in and around the Earth (b) Interaction between cosmic rays and surface of the Earth (c) A strong magnetic dipole is believed to be buried at the center of the Earth (d) Strong induction effect of the Sun

Answer

(a) Motion and distribution of materials in and around the Earth

The field is attributed to currents of molten conducting material in the outer core (the dynamo effect).

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  • 2080 · 1 mark

A free charged particle moves through a magnetic field. The particle may undergo a change in (a) speed (b) energy (c) pressure (d) direction of motion

Answer

(d) direction of motion

The magnetic force is perpendicular to velocity, so it does no work; speed and energy are unchanged but direction changes.

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  • 2080 · 1 mark

The speed of electron in first Bohr orbit of Hydrogen atom is vv. The speed of electron in the first excited state of doubly ionized Lithium atom is (a) v/2v/2 (b) vv (c) 1.5v1.5v (d) 2v2v

Answer

(c) 1.5v1.5v

Speed in a Bohr orbit is v∝Znv\propto\frac Zn. For hydrogen ground state vv corresponds to Z=1,n=1Z=1,n=1. For Li2+^{2+}, Z=3Z=3, first excited state n=2n=2: v′=32v=1.5vv'=\frac32v=1.5v.

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  • 2080 · 1 mark

What is nuclear 'holocaust'? (a) Formation of a nuclear bomb (b) Formation of nuclear reactor (c) Making hole in the nuclear shielding layer (d) After math of a nuclear explosion

Answer

(d) After math of a nuclear explosion

A nuclear holocaust is the devastation and destruction following a nuclear explosion or war.

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  • 2080 · 1 mark

Oppenheimer is a 2023 biographical drama film written and directed by Christopher Nolan and starring Cillian Murphy as J. Robert Oppenheimer a physicist. His scientific contribution is (a) Nuclear Reactor Physics (b) AK-47 Rifle (c) Bullet Dynamics (d) Design of Atom Bomb

Answer

(d) Design of Atom Bomb

Oppenheimer led the Manhattan Project's Los Alamos laboratory that designed the atomic bomb.

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  • 2080 · 1 mark

A GM counter gives 82 counts/sec. After 210 seconds the count rate decreased to 19 counts/sec. The background count is 10 counts/sec, then the half life of the source is (a) 40 s (b) 70 s (c) 105 s (d) 210 s

Answer

(b) 70 s

Subtract background: 82−10=7282-10=72 and 19−10=919-10=9. Ratio 72/9=8=2372/9=8=2^3, so 3 half-lives in 210 s and T1/2=70T_{1/2}=70 s.

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  • 2080 · 1 mark

The truth table given below is for: (A, B → Y): (0, 0 → 0), (0, 1 → 0), (1, 0 → 0), (1, 1 → 1) (a) NOT gate (b) OR gate (c) NOR gate (d) AND gate

Answer

(d) AND gate

Output is 1 only when both inputs are 1.

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  • 2080 · 1 mark

NPNNPN transistors are preferred to PNPPNP transistors because they have (a) Low cost (b) Low dissipation of energy (c) Capable of handling larger power (d) Electrons have high mobility than holes and hence high mobility of energy

Answer

(d) Electrons have high mobility than holes and hence high mobility of energy

In NPN transistors the majority carriers are electrons, whose mobility is higher than that of holes, giving faster response.

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  • 2081 · 1 mark

The quantity having no dimension is, (a) Mass (b) Velocity (c) Angle (d) Pressure

Answer

(c) Angle

Angle is a ratio of arc to radius, so it is dimensionless.

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  • 2081 · 1 mark

The slope of kinetic energy versus position vector gives the rate of (a) kinetic energy (b) momentum (c) potential energy (d) acceleration

Answer

(b) momentum

dKdx=F\frac{dK}{dx}=F (work-energy theorem), and force equals the rate of change of momentum, F=dpdtF=\frac{dp}{dt}.

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  • 2081 · 1 mark

A boy is carrying a bucket of water in one hand and a piece of plastic in other hand. If he transfers plastic in bucket which floats, Then the boy feels (a) same load (b) more load (c) less load (d) depends upon density of plastic

Answer

(a) same load

A floating piece of plastic displaces water equal to its own weight, so the bucket plus plastic weighs the same as before the transfer.

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  • 2081 · 1 mark

Fire dram works on principle of (a) Volume expansion of solid. (b) Volume expansion of water. (c) Volume expansion of gas. (d) Bimetallic bar which becomes curved on heating.

Answer

(d) Bimetallic bar which becomes curved on heating.

A fire alarm uses a bimetallic strip that bends on heating (unequal expansion of the two metals) to complete the alarm circuit.

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  • 2081 · 1 mark

The critical angle of light passing from glass to air is minimum for (a) red (b) violet (c) green (d) yellow

Answer

(b) violet

sin⁡C=1μ\sin C=\frac1\mu. Violet has the highest refractive index in glass, so it has the smallest critical angle.

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  • 2081 · 1 mark

A stone is dropped into a well 78.4 m deep. The splash sound is noticed after (velocity of sound in air is 350 m/s) (a) 2.4 s (b) 3.6 s (c) 4.2 s (d) 6.8 s

Answer

(c) 4.2 s

Fall time: t1=2hg=2(78.4)9.8=4t_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(78.4)}{9.8}}=4 s. Sound time: t2=78.4350=0.224t_2=\frac{78.4}{350}=0.224 s. Total ≈4.2\approx4.2 s.

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  • 2081 · 1 mark

Two fixed charges qq and 4q4q are rr distant apart. What will be the position of third charge to be placed so that the system will be in equilibrium? (a) 2r/3 from 4q (b) r2\frac r2 from q (c) 2r from q (d) r/2 from 4q

Answer

(a) 2r/3 from 4q

The third charge must lie between them where the forces balance: kqx2=4kq(r−x)2\frac{kq}{x^2}=\frac{4kq}{(r-x)^2}, so r−x=2xr-x=2x, x=r3x=\frac r3 from qq, i.e. 2r3\frac{2r}{3} from 4q4q.

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  • 2081 · 1 mark

At a point on the axis of an electric dipole (a) The electric field E is Zero. (b) The electric potential V is zero. (c) Neither V nor E is zero. (d) both E and V are zero

Answer

(c) Neither V nor E is zero.

On the axis of a dipole, E=2kpr3E=\frac{2kp}{r^3} and V=kpr2V=\frac{kp}{r^2}, both non-zero (V is zero only on the equatorial line).

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  • 2081 · 1 mark

At neutral temperature the thermoelectric power is (a) Maximum (b) Minimum (c) Infinite (d) Zero

Answer

(d) Zero

Thermoelectric power is dEdT\frac{dE}{dT}. At the neutral temperature emf is maximum, so dEdT=0\frac{dE}{dT}=0.

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  • 2081 · 1 mark

The study of photoelectric effect is useful in understanding (a) quantization on energy (b) quantization of charge (c) conservation of charge (d) conservation of energy

Answer

(a) quantization on energy

The photoelectric effect shows light is absorbed in discrete quanta E=hνE=h\nu.

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  • 2081 · 1 mark

Induced emf produced across the axle is maximum when it moves (a) at the poles (b) at the equator (c) remains stationary (d) no emf induced at all

Answer

(a) at the poles

e=BVvle=B_Vvl. The vertical component of the Earth's field is maximum at the poles, so the emf across a horizontal axle is greatest there.

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  • 2081 · 1 mark

Atomic power station at Gujrat has a generating capacity of 200mw. What will be the energy released in a day by the station? (a) 1.43×10131.43\times10^{13} j (b) 1.53×10131.53\times10^{13} j (c) 1.63×10131.63\times10^{13} j (d) 1.73×10131.73\times10^{13} j

Answer

(d) 1.73×10131.73\times10^{13} j

E=Pt=200×106×86400=1.728×1013E=Pt=200\times10^6\times86400=1.728\times10^{13} J.

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  • 2081 · 1 mark

If germanium is doped with donor impurity the donor atom should be (a) tetravalent (b) trivalent (c) pentavalent (d) quatravalent

Answer

(c) pentavalent

A donor impurity has five valence electrons (e.g. P, As, Sb); four bond with germanium and one is free, giving n-type material.

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Questions from BSc-CSIT-Old-Question-Collection-2069-2081-TU-IoST.pdf (BSc. CSIT Entrance · Entrance Examination 2069 · 2069) and TU IOST BSc CSIT Entrance Old Question Collection 2070-2081 (TU IOST BSc CSIT entrance examination papers 2070 to 2081 BS). Answers are written for this site; check them against your class notes.