Measurement, Units, and Dimensional Analysis

(c) $E\cos^2\theta$

The angle with the horizontal is $90^\circ-\theta$, so the horizontal component is $v\sin\theta$ and it is the only velocity left at the top. Kinetic energy at the top is $E\sin^2\theta$. By energy conservation, the potential energy at the top is $E-E\sin^2\theta=E\cos^2\theta$.

(d) 1.8 m³

Work done per second equals power, so $P=\dfrac{mgh}{t}$. Hence $m=\frac{Pt}{gh}=\frac{3000\times 60}{10\times 10}=1800\ \text{kg}$ With $\rho=1000\ \text{kg/m}^3$, $V=1800/1000=1.8\ \text{m}^3$.

(b) (\sqrt{gr})

At the top of the vertical circle, the minimum condition for the string to stay taut is that the tension just becomes zero. Then the weight alone provides the required centripetal force:

$mg = \frac{mv^2}{r} \implies v = \sqrt{gr}$

(d) ω² r

Displacement: (y = r\sin(\omega t))

Velocity: (v = \dfrac{dy}{dt} = r\omega\cos(\omega t))

Acceleration: (a = \dfrac{dv}{dt} = -r\omega^2\sin(\omega t))

Maximum acceleration occurs when (\sin(\omega t) = -1), giving magnitude (a_{\max} = \omega^2 r).

(a) 6 m

The centre of mass of a two-body system lies on the line joining the masses, at a distance from the 1 kg mass given by

$r_1=\frac{m_2}{m_1+m_2}\,d=\frac{2}{1+2}\times 9=6\ \text{m}$

So the centre of mass is 6 m from the 1 kg mass (and 3 m from the 2 kg mass).

(c) ML⁻¹T⁻¹

The coefficient of viscosity η is defined from Newton's law of viscosity, F = ηA(dv/dx), so η = F/(A·dv/dx). Its dimensions are [MLT⁻²] / ([L²]·[LT⁻¹/L]) = ML⁻¹T⁻¹.

(b) 1 m

Both bodies are released from rest and fall freely, so each has the same acceleration $g$. Their relative acceleration is zero, so the initial separation of 1 m stays unchanged at all times — after 3 s the separation is still 1 m.

(a) v α t^1/2

Power P = Fv, and with constant power F = P/v. By Newton's second law:

$m\frac{dv}{dt} = \frac{P}{v} \Rightarrow v\,dv = \frac{P}{m}dt$

Integrating: $\frac{v^2}{2} = \frac{P}{m}t$, so $v^2 \propto t$, giving $v \propto t^{1/2}$.

(a) 1.98 cal

At constant pressure, work done by the gas is

$W = P\Delta V = nR\Delta T$

For 1 mole, $\Delta T = 1\ \text{K}$:

$W = 1 \times 8.31 \times 1 = 8.31\ \text{J}$

Converting to heat units (1 cal = 4.18 J):

$W = \frac{8.31}{4.18} \approx 1.98\ \text{cal}$

(b) $\lambda_m/\lambda'_m$ is inversely proportional to $T/T'$

By Wien's displacement law, $\lambda_m T=b$ (a constant), so $\lambda_m\propto 1/T$. For two bodies, $\frac{\lambda_m}{\lambda'_m}=\frac{T'}{T}=\left(\frac{T}{T'}\right)^{-1}$ so the ratio varies inversely with $T/T'$. The fourth-power relation in (c) belongs to Stefan's law, not Wien's law.

(a) $c_p>c_v$

For any substance, $c_p-c_v=\dfrac{TV\beta^2}{\kappa_T}$, where $\beta$ is the volume expansion coefficient and $\kappa_T$ the isothermal compressibility. Since $\beta^2\ge 0$ and $\kappa_T>0$, $c_p\ge c_v$ always. Between 0 and 4 °C water contracts on heating ($\beta<0$), but $\beta^2>0$, so the difference is still positive and $c_p>c_v$. The relation $c_p-c_v=R$ holds only for an ideal gas.

(c) 3420

The human eye can just resolve an angle of about 1 minute of arc, and $1'=\dfrac{\pi}{10800}\ \text{rad}\approx\dfrac{1}{3420}\ \text{rad}$ (the standard textbook value; exactly it is about 1/3438). Resolving power is the reciprocal of the minimum resolvable angle, so it is nearly $3420$.

(a) Blue

The refractive index of a prism is greatest for the colour of shortest wavelength. Among Green, Blue, Orange and Yellow, blue has the shortest wavelength, so the prism has the maximum refractive index for blue.

(b) 3

The grating element is $d=\frac{1}{5500}\ \text{cm}=\frac{10^{8}}{5500}\ \text{Å}\approx 18182\ \text{Å}$ The highest possible order satisfies $n_{\max}<\dfrac{d}{\lambda}=\dfrac{18182}{6000}\approx 3.03$, so $n_{\max}=3$.

Counting the orders on one side of the central maximum ($n=1,2,3$) gives 3 maxima, which is the option intended here. If both sides and the central maximum were counted, the total would be $2\times3+1=7$, which is not among the options.

(c) 25 unit

Amplitudes are in the ratio of the square roots of intensities, so $A_1:A_2 = \sqrt{4}:\sqrt{9} = 2:3$. For superposition, maximum amplitude $A_{max} = A_1 + A_2 = 2+3 = 5$ units. Since intensity is proportional to the square of amplitude, maximum intensity $= 5^2 = 25$ unit.

(b) 0.7 km

Speed of sound varies as $v\propto\sqrt{T}$, so at 30 °C (303 K), with $v_0=330$ m/s at 273 K: $v=330\sqrt{\frac{303}{273}}\approx 347.7\ \text{m/s}$ Distance in 2 s: $d=347.7\times 2\approx 695\ \text{m}\approx 0.7\ \text{km}$.

(d) 0.6 R

The end correction for a resonance tube (closed pipe) is ( e = 0.6R ), where ( R ) is the radius of the tube. This accounts for the antinode forming slightly outside the open end of the tube.

(c) $1.1\times10^{-10}$ farad

The capacitance of an isolated spherical conductor is $C=4\pi\varepsilon_0 r$. With $\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}$ and $r=1$ m: $C=\frac{1}{9\times10^{9}}\approx 1.1\times10^{-10}\ \text{F}$

(c) 103 V

Dielectric strength is the maximum electric field a dielectric can withstand without breakdown:

$V = E \times d = 10^{6}\ \text{V m}^{-1} \times 1\ \text{mm} = 10^{6} \times 10^{-3}\ \text{m} = 10^{3}\ \text{V}$

So the minimum voltage needed to puncture the 1 mm thick specimen is 10³ V.

(d) Volt/meter

Reason: NC⁻¹ means N/C (newton per coulomb), which is the unit of electric field intensity.
Electric field is also given by (E = V/d), so its unit is volt/meter.
Now,
[ \text{volt/meter} = \frac{\text{joule/coulomb}}{\text{meter}} = \frac{\text{N·m/C}}{\text{m}} = \text{N/C} = \text{NC}^{-1} ] So NC⁻¹ has the same dimension as volt/meter.
Hence, the correct option is (d).

(a) µA

Current $I = \dfrac{n \cdot e}{t} = 6.25 \times 10^{12} \times 1.6 \times 10^{-19} = 1 \times 10^{-6}\ \text{A} = 1\ \mu\text{A}$.

(d) v α E

Reason: Drift velocity is given by ( v = \mu E ), where ( \mu ) is the mobility of the charge carriers. Since mobility is constant for a given conductor at a fixed temperature, drift velocity is directly proportional to the electric field, i.e. ( v \propto E ). This linear relation is exactly what makes Ohm's law hold (( I \propto V )).

(a) A has high coercivity

A permanent magnet must retain its magnetism and resist demagnetisation, so the material needs high coercivity (and high retentivity). High coercivity means a strong reverse field is required to remove its magnetism.

(b) 16:1

Reason: Same mass and material means same volume, so (A_1 L_1 = A_2 L_2), i.e. (L \propto \dfrac{1}{A} \propto \dfrac{1}{d^2}).

Resistance (R = \rho\dfrac{L}{A} \propto \dfrac{L}{A} \propto \dfrac{1}{A^2} \propto \dfrac{1}{d^4}).

In series, current (I) is the same, so heat (H = I^2Rt \propto R).

$\frac{H_1}{H_2} = \frac{R_1}{R_2} = \left(\frac{d_2}{d_1}\right)^4 = \left(\frac{2}{1}\right)^4 = 16:1$

(c) 1 eV

The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is $KE = qV$. For a proton, $q = +e$, so $KE = e \times 1\,\text{V} = 1\,\text{eV}$. The mass of the proton does not affect the energy gained — it only affects the speed.

(b) 5:27

The first Lyman line is $n=2\to1$: $\dfrac{1}{\lambda_L}=R\left(1-\dfrac14\right)=\dfrac{3R}{4}$, so $\lambda_L=\dfrac{4}{3R}$.

The first Balmer line is $n=3\to2$: $\dfrac{1}{\lambda_B}=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}$, so $\lambda_B=\dfrac{36}{5R}$.

$\frac{\lambda_L}{\lambda_B}=\frac{4}{3}\times\frac{5}{36}=\frac{5}{27}$

(c) remains constant

The magnetic force on a moving charge is F = qvB sin θ, which always acts perpendicular to the velocity. A force perpendicular to velocity changes only the direction of motion, not the magnitude of speed. So the electron's speed remains constant (only its direction changes).

(b) 125 gm

After 30 days = 3 half-lives, so mass left = 1 kg × (1/2)³ = 1/8 kg = 125 gm.

(c) 4.5 fm

Nuclear radius is $R=R_0A^{1/3}$ with $R_0\approx1.2$ fm. For $A=27$, $A^{1/3}=3$, so $R=1.2\times3=3.6$ fm. The nearest option is 4.5 fm.

(c) semiconductor

The energy gap of a semiconductor is small — typically between about 0.1 eV and 1.5 eV (silicon ≈ 1.1 eV, germanium ≈ 0.7 eV). Insulators have a much larger gap (above ~3 eV), while conductors have no appreciable gap at all. Since 0.7 eV falls in the semiconductor range, the material is a semiconductor.