MBBS and BDS entrance Physics: NEET-UG questions on Nepal's MBBS syllabus: Current Electricity and Magnetism

2013 · NEET-UG (India) · 4 marks

A wire of resistance $4\ \Omega$ is stretched to twice its original length. The resistance of stretched wire would be:
(a) $2\ \Omega$ (b) $4\ \Omega$ (c) $8\ \Omega$ (d) $16\ \Omega$

Answer

(d) $16\ \Omega$

Stretching keeps volume constant, so doubling the length halves the area. Since $R=\rho\frac{l}{A}$ , resistance becomes $2\times2=4$ times. New resistance $=4\times4=16\ \Omega$ .

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2013 · NEET-UG (India) · 4 marks

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of $10\ \Omega$ is:
(a) $0.2\ \Omega$ (b) $0.5\ \Omega$ (c) $0.8\ \Omega$ (d) $1.0\ \Omega$

Answer

(b) $0.5\ \Omega$

$\varepsilon=I(R+r)$ , so $R+r=\frac{2.1}{0.2}=10.5\ \Omega$ . Therefore $r=10.5-10=0.5\ \Omega$ .

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2013 · NEET-UG (India) · 4 marks

When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are:
(a) $\frac{ma_0}{e}$ west, $\frac{2ma_0}{ev_0}$ up (b) $\frac{ma_0}{e}$ west, $\frac{2ma_0}{ev_0}$ down (c) $\frac{ma_0}{e}$ east, $\frac{3ma_0}{ev_0}$ up (d) $\frac{ma_0}{e}$ east, $\frac{3ma_0}{ev_0}$ down

Answer

(b) $\frac{ma_0}{e}$ west, $\frac{2ma_0}{ev_0}$ down

At rest only the electric force acts: $eE=ma_0$ westward, so $E=\frac{ma_0}{e}$ west (proton is positive). Moving north, the acceleration is $3a_0$ west, so the magnetic force adds $2ma_0$ west: $ev_0B=2ma_0$ . With $\vec F=e\vec v\times\vec B$ , north $\times$ down gives west, so $B$ points down.

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2013 · NEET-UG (India) · 4 marks

A current loop in a magnetic field:
(a) experiences a torque whether the field is uniform or non-uniform in all orientations (b) can be in equilibrium in one orientation (c) can be in equilibrium in two orientations, both the equilibrium states are unstable (d) can be in equilibrium in two orientations, one stable while the other is unstable

Answer

(d) can be in equilibrium in two orientations, one stable while the other is unstable

Torque $\vec\tau=\vec m\times\vec B$ vanishes when $\vec m$ is parallel (stable) or antiparallel (unstable) to $\vec B$ . So there are two equilibrium orientations. A loop does not always experience a torque (it is zero in these two orientations), which rules out (a).

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2013 · NEET-UG (India) · 4 marks

A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of an arc (subtending $60^\circ$ at its centre) as shown in figure. The new magnetic dipole moment will be:
(a) $M$ (b) $\frac{3}{\pi}M$ (c) $\frac{2}{\pi}M$ (d) $\frac{M}{2}$

Answer

(b) $\frac{3}{\pi}M$

Dipole moment is pole strength times distance between poles, $M=ml$ . Bent into an arc of $60^\circ$ , the arc length stays $l$ , so the radius is $r=\frac{l}{\pi/3}=\frac{3l}{\pi}$ . The poles are now separated by the chord $=2r\sin30^\circ=r=\frac{3l}{\pi}$ , so $M'=m\cdot\frac{3l}{\pi}=\frac{3}{\pi}M$ .

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2013 · NEET-UG (India) · 4 marks

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is:
(a) once per revolution (b) twice per revolution (c) four times per revolution (d) six times per revolution

Answer

(b) twice per revolution

The induced emf $e=e_0\sin\omega t$ changes sign each time the loop passes the neutral position (plane perpendicular to the field). That happens twice in each revolution, i.e. once every half cycle.

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2014 · NEET-UG (India) · 4 marks

The resistances in the two arms of the meter bridge are $5\ \Omega$ and $R\ \Omega$ , respectively (the $5\ \Omega$ resistance is in the left gap and $R$ in the right gap, with the balance length $l_1$ measured from end A). When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6\,l_1$ . The resistance $R$ is:
(a) $15\ \Omega$ (b) $20\ \Omega$ (c) $25\ \Omega$ (d) $10\ \Omega$

Answer

(a) $15\ \Omega$

Balance condition: $\frac5R=\frac{l_1}{100-l_1}$ . After shunting, the right arm is $\frac R2$ : $\frac{10}{R}=\frac{1.6l_1}{100-1.6l_1}$ . Dividing: $2=\frac{1.6(100-l_1)}{100-1.6l_1}$ , giving $l_1=25$ cm. Then $\frac5R=\frac{25}{75}$ , so $R=15\ \Omega$ .

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2014 · NEET-UG (India) · 4 marks

A thin semicircular conducting ring (PQR) of radius $r$ is falling with its plane vertical in a horizontal magnetic field $B$ (directed into the plane of the ring), with P and R the ends of the ring and Q the top point. The potential difference developed across the ring when its speed is $v$ , is:
(a) $Bv\pi r^2/2$ and P is at higher potential (b) $\pi r B v$ and R is at higher potential (c) $2rBv$ and R is at higher potential (d) Zero

Answer

(c) $2rBv$ and R is at higher potential

Only the chord joining the ends matters: the motional emf equals that of a straight conductor of length $2r$ moving perpendicular to $B$ , so $\varepsilon = B(2r)v$ . By $\vec v\times\vec B$ , positive charges are pushed towards R, so R is at higher potential. Option (b) wrongly uses the arc length $\pi r$ ; the emf depends on the effective (chord) length.

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2014 · NEET-UG (India) · 4 marks

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that $O$ is their common point for the two. The wires carry $I_1$ and $I_2$ currents, respectively. Point $P$ is lying at distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $P$ will be:
(a) $\frac{\mu_0}{2\pi d}(I_1+I_2)$ (b) $\frac{\mu_0}{2\pi d}(I_1^2-I_2^2)$ (c) $\frac{\mu_0}{2\pi d}(I_1^2+I_2^2)^{1/2}$ (d) $\frac{\mu_0}{2\pi d}\left(\frac{I_1}{I_2}\right)$

Answer

(c) $\frac{\mu_0}{2\pi d}(I_1^2+I_2^2)^{1/2}$

Point P is at distance $d$ from each wire, so each wire gives $B=\frac{\mu_0 I}{2\pi d}$ . The field of each wire at P is perpendicular to the plane containing P and that wire, so the two fields are at right angles to each other. They add as vectors: $B=\sqrt{B_1^2+B_2^2}=\frac{\mu_0}{2\pi d}\sqrt{I_1^2+I_2^2}$ . Simple addition, option (a), ignores the right angle.

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2014 · NEET-UG (India) · 4 marks

In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is $G$ , the resistance of ammeter will be:
(a) $\frac{499}{500}G$ (b) $\frac{1}{500}G$ (c) $\frac{500}{499}G$ (d) $\frac{1}{499}G$

Answer

(b) $\frac{1}{500}G$

Current through the galvanometer is $I_g=\frac{I}{500}$ , so the shunt carries $\frac{499}{500}I$ . Equal potential drops give $I_gG=(I-I_g)S$ , so $S=\frac{G}{499}$ . Ammeter resistance is $G$ and $S$ in parallel: $R_A=\frac{GS}{G+S}=\frac{G\cdot G/499}{G\cdot 500/499}=\frac{G}{500}$ . Option (d) is only the shunt resistance, not the resistance of the whole ammeter.

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2014 · NEET-UG (India) · 4 marks

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is $0.5\ \Omega$ . The power loss in the wires is:
(a) 19.2 kW (b) 19.2 J (c) 12.2 kW (d) 19.2 W

Answer

(a) 19.2 kW

Total potential fall $=8\times150=1200$ V and total resistance $=0.5\times150=75\ \Omega$ . Current $I=\frac{1200}{75}=16$ A, so power loss $=I^2R=16^2\times75=19200$ W $=19.2$ kW. Options (b) and (d) have wrong units: power cannot be in joules, and the value is $19.2$ kW, not $19.2$ W.

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2014 · NEET-UG (India) · 4 marks

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance $R$ , connected across the given cell, has values of (i) infinity, (ii) $9.5\ \Omega$ , the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is:
(a) $0.95\ \Omega$ (b) $0.5\ \Omega$ (c) $0.75\ \Omega$ (d) $0.25\ \Omega$

Answer

(b) $0.5\ \Omega$

The balance length for open circuit gives the emf: potential gradient $=\frac{2.0}{4}=0.5$ V/m, so $E=0.5\times3=1.5$ V; with $R=9.5\ \Omega$ , terminal voltage $V=0.5\times2.85=1.425$ V. $r=R\left(\frac{E}{V}-1\right)=9.5\times\left(\frac{1.5}{1.425}-1\right)=9.5\times0.0526=0.5\ \Omega$ . Option (a) results from wrongly using $E-V=0.075$ directly with $R$ .

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2015 · NEET-UG (India) · 4 marks

A proton and an alpha particle both enter a region of uniform magnetic field $B$ , moving at right angles to the field $B$ . If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:
(a) 1.5 MeV (b) 1 MeV (c) 4 MeV (d) 0.5 MeV

Answer

(b) 1 MeV

Radius $r=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB}$ , so for equal $r$ and $B$ : $K\propto\frac{q^2}{m}$ . For the alpha particle $q=2e$ and $m=4m_p$ , so $\frac{q^2}{m}=\frac{4}{4}=1$ times the proton's value. Hence $K_\alpha=1$ MeV. Option (c) results from forgetting that mass is four times the proton's.

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2015 · NEET-UG (India) · 4 marks

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/ $\mathrm{m^2}$ . The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:
(a) 0.24 N m (b) 0.12 N m (c) 0.15 N m (d) 0.20 N m

Answer

(d) 0.20 N m

The magnitude of the torque on a coil is $\tau = NIAB\sin\theta$ , where $\theta$ is the angle between the coil normal and $B$ . The plane is at $30^\circ$ to the field, so the normal is at $60^\circ$ to it. $NIAB = 50\times2\times(0.12\times0.1)\times0.2 = 0.24$ N m, and $\tau = 0.24\sin60^\circ \approx 0.21 \approx 0.20$ N m. Option (a), 0.24 N m, is only the maximum torque (at $\theta=90^\circ$ ), so it is too large.

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2015 · NEET-UG (India) · 4 marks

An electron moves on a straight line path XY. The abcd is a coil adjacent to the path of electron (the circular coil lies in the plane of the path, with a at the top, b on the left, c at the bottom nearest the line, and d on the right; the electron moves from X to Y along the line below the coil). What will be the direction of current, if any, induced in the coil?
(a) The current will reverse its direction as the electron goes past the coil (b) No current induced (c) abcd (d) adcb

Answer

(a) The current will reverse its direction as the electron goes past the coil

As the electron approaches, the magnetic flux through the coil (due to the moving charge) increases, and as it moves away the flux decreases. By Lenz's law the induced current opposes the change, so its sense is one way during approach and the opposite way after the electron has passed. The current is therefore not zero, and it reverses direction; (b) is wrong because the flux changes with time.

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2015 · NEET-UG (India) · 4 marks

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:
(a) 2 A (b) 1 A (c) 0.5 A (d) 0.25 A

Answer

(c) 0.5 A

The shunt (20 ohm) is in parallel with the coil (480 ohm): $R_{AB}=\dfrac{480\times20}{500}=19.2\ \Omega$ . Total resistance $=40.8+19.2=60\ \Omega$ , so the current in the circuit, which is what the ammeter reads, is $I=30/60=0.5$ A.

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2016 · NEET-UG (India) · 4 marks

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is $4\times10^{-3}$ Wb. The self-inductance of the solenoid is
(a) 1 H (b) 4 H (c) 3 H (d) 2 H

Answer

(a) 1 H

Flux linkage $N\Phi=LI$ , so $L=\dfrac{N\Phi}{I}=\dfrac{1000\times4\times10^{-3}}{4}=1$ H. Remember the flux given is per turn, so multiply by the number of turns.

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2016 · NEET-UG (India) · 4 marks

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is:
(a) $3:2$ (b) $5:1$ (c) $5:4$ (d) $3:4$

Answer

(a) $3:2$

Potential gradient is constant, so balance length is proportional to emf. Cells aiding: $E_1+E_2\propto50$ . Cells opposing: $E_1-E_2\propto10$ . Adding and subtracting: $E_1\propto30$ , $E_2\propto20$ , so $E_1:E_2=3:2$ .

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2016 · NEET-UG (India) · 4 marks

A long straight wire of radius $a$ carries a steady current $I$ . The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$ , at radial distances $\frac a2$ and $2a$ respectively, from the axis of the wire is:
(a) 4 (b) $\frac14$ (c) $\frac12$ (d) 1

Answer

(d) 1

Inside the wire, Ampere's law gives $B=\dfrac{\mu_0Ir}{2\pi a^2}$ , so at $r=\frac a2$ : $B=\dfrac{\mu_0I}{4\pi a}$ . Outside, $B'=\dfrac{\mu_0I}{2\pi r}$ , so at $r=2a$ : $B'=\dfrac{\mu_0I}{4\pi a}$ . The ratio $B/B'=1$ .

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2016 · NEET-UG (India) · 4 marks

The magnetic susceptibility is negative for:
(a) paramagnetic and ferromagnetic materials (b) diamagnetic material only (c) paramagnetic material only (d) ferromagnetic material only

Answer

(b) diamagnetic material only

Diamagnetic materials get a weak magnetisation opposite to the applied field, so $\chi$ is small and negative. Paramagnetic and ferromagnetic materials align with the field, so their $\chi$ is positive (small for para, large for ferro).

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2016 · NEET-UG (India) · 4 marks

A square loop ABCD of side $L$ carrying a current $i$ is placed near and coplanar with a long straight conductor XY carrying a current $I$ . The side AB of the loop is parallel to the conductor and at a distance $L/2$ from it, with the loop lying on the side away from the conductor. The net force on the loop will be:
(a) $\frac{\mu_0 I i L}{2\pi}$ (b) $\frac{2\mu_0 I i}{3\pi}$ (c) $\frac{\mu_0 I i}{2\pi}$ (d) $\frac{2\mu_0 I i L}{3\pi}$

Answer

(b) $\dfrac{2\mu_0Ii}{3\pi}$

Forces on the two sides perpendicular to the wire cancel. Side AB (distance $L/2$ ) is attracted: $F_1=\dfrac{\mu_0Ii\,L}{2\pi(L/2)}=\dfrac{\mu_0Ii}{\pi}$ . The far side (distance $3L/2$ ) is repelled: $F_2=\dfrac{\mu_0Ii\,L}{2\pi(3L/2)}=\dfrac{\mu_0Ii}{3\pi}$ . Net $=F_1-F_2=\dfrac{2\mu_0Ii}{3\pi}$ , towards the wire.

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2017 · NEET-UG (India) · 4 marks

The resistance of a wire is $R$ ohm. If it is melted and stretched to $n$ times its original length, its new resistance will be
(a) $\dfrac{R}{n}$ (b) $n^2R$ (c) $\dfrac{R}{n^2}$ (d) $nR$

Answer

(b) $n^2R$

Melting keeps the volume fixed, $Al=$ constant. If $l\to nl$ then $A\to A/n$ .

$R=\rho l/A$ becomes $\rho(nl)/(A/n)=n^2R$ .

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2017 · NEET-UG (India) · 4 marks

A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of $85\ \mu$ A and is subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by $180^\circ$ against the torque is
(a) 4.55 $\mu$ J (b) 2.3 $\mu$ J (c) 1.15 $\mu$ J (d) 9.1 $\mu$ J

Answer

(d) 9.1 $\mu$ J

Work done in turning the coil from $\theta=0$ to $180^\circ$ is $W=2MB=2NIAB$ .

$A=2.1\times1.25\times10^{-4}=2.625\times10^{-4}\ \mathrm{m^2}$ , so $W=2\times250\times85\times10^{-6}\times2.625\times10^{-4}\times0.85\approx9.5\ \mu$ J, which is nearest to 9.1 $\mu$ J. The 4.55 $\mu$ J option is only $MB$ , the work for $90^\circ$ rotation.

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2017 · NEET-UG (India) · 4 marks

If $\delta_1$ and $\delta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\delta$ is given by
(a) $\tan^2\delta=\tan^2\delta_1+\tan^2\delta_2$ (b) $\cot^2\delta=\cot^2\delta_1-\cot^2\delta_2$ (c) $\tan^2\delta=\tan^2\delta_1-\tan^2\delta_2$ (d) $\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$

Answer

(d) $\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$

In a vertical plane at angle $\alpha$ to the meridian, $\tan\delta_1=\tan\delta/\cos\alpha$ . For the perpendicular plane, $\tan\delta_2=\tan\delta/\sin\alpha$ .

So $\cot^2\delta_1+\cot^2\delta_2=\cot^2\delta(\cos^2\alpha+\sin^2\alpha)=\cot^2\delta$ .

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2017 · NEET-UG (India) · 4 marks

A potentiometer is an accurate and versatile device to make electrical measurements of EMF, because the method involves
(a) potential gradients (b) a condition of no current flow through the galvanometer (c) a combination of cells, galvanometer and resistances (d) cells

Answer

(b) a condition of no current flow through the galvanometer

A potentiometer works on the null (balance) method: at the balance point no current is drawn from the cell being measured, so its EMF is measured without internal-resistance error. Potential gradient is what the wire provides, but the accuracy comes from the zero-current condition.

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2017 · NEET-UG (India) · 4 marks

A long solenoid of diameter 0.1 m has $2\times10^{4}$ turns per metre. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is $10^{2}\ \pi^2\ \Omega$ , the total charge flowing through the coil during this time is
(a) 16 $\mu$ C (b) 32 $\mu$ C (c) 16 $\pi$ $\mu$ C (d) 32 $\pi$ $\mu$ C

Answer

(b) 32 $\mu$ C

Solenoid field $B=\mu_0nI$ , so $\Delta\Phi=N\,\mu_0 n\,\Delta I\,\pi r^2=100\times4\pi\times10^{-7}\times2\times10^4\times4\times\pi\times10^{-4}\approx3.16\times10^{-3}$ Wb.

Charge $q=\dfrac{\Delta\Phi}{R}=\dfrac{3.16\times10^{-3}}{100\pi^2}=32\ \mu$ C. Only the coil area (radius 0.01 m) counts, not the solenoid diameter.

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2018 · NEET-UG (India) · 4 marks

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
(a) 13.89 H (b) 0.138 H (c) 1.389 H (d) 138.88 H

Answer

(a) 13.89 H

Energy stored in an inductor is $U=\tfrac12LI^2$ .

$L=\dfrac{2U}{I^2}=\dfrac{2\times25\times10^{-3}}{(60\times10^{-3})^2}=\dfrac{0.05}{3.6\times10^{-3}}=13.89$ H.

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2018 · NEET-UG (India) · 4 marks

A metallic rod of mass per unit length $0.5\ \mathrm{kg\,m^{-1}}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
(a) 11.32 A (b) 7.14 A (c) 14.76 A (d) 5.98 A

Answer

(a) 11.32 A

The field is vertical, so the force $BIl$ on the rod is horizontal. Balancing along the smooth incline: $BIl\cos30^\circ = mg\sin30^\circ$ . $I = \dfrac{(m/l)\,g\tan30^\circ}{B} = \dfrac{0.5 \times 9.8 \times 0.577}{0.25} \approx 11.32$ A. Using $\sin\theta$ instead of $\tan\theta$ would give a wrong value.

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2018 · NEET-UG (India) · 4 marks

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
(a) the induced electric field due to the changing magnetic field (b) the current source (c) the lattice structure of the material of the rod (d) the magnetic field

Answer

(b) the current source

Pushing the rod out increases gravitational potential energy, and the magnetic field itself does no work. The changing flux induces an emf in the electromagnet circuit (back emf). The source must supply extra energy to maintain the current against this emf, so the work comes from the current source.

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2018 · NEET-UG (India) · 4 marks

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
(a) $500\ \Omega$ (b) $40\ \Omega$ (c) $250\ \Omega$ (d) $25\ \Omega$

Answer

(c) $250\ \Omega$

Current sensitivity $I_s = \theta/I$ and voltage sensitivity $V_s = \theta/V$ , so $V_s = I_s/R_G$ . $R_G = I_s/V_s = \dfrac{5\ \text{div/mA}}{20\ \text{div/V}} = 0.25\ \text{V/mA} = 250\ \Omega$ .

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2018 · NEET-UG (India) · 4 marks

A carbon resistor of $(47\pm4.7)\ \mathrm{k}\Omega$ is to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Green - Orange - Violet - Gold (b) Violet - Yellow - Orange - Silver (c) Yellow - Green - Violet - Gold (d) Yellow - Violet - Orange - Silver

Answer

(d) Yellow - Violet - Orange - Silver

$47\ \text{k}\Omega = 47\times10^{3}$ : first digit 4 = yellow, second digit 7 = violet, multiplier $10^3$ = orange. Tolerance $\pm10\%$ ( $4.7$ k $\Omega$ is 10% of 47 k $\Omega$ ) is silver; gold would be 5%.

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2018 · NEET-UG (India) · 4 marks

A set of $n$ equal resistors, of value $R$ each, are connected in series to a battery of emf $E$ and internal resistance $R$ . The current drawn is $I$ . Now, the $n$ resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes $10I$ . The value of $n$ is
(a) 9 (b) 10 (c) 20 (d) 11

Answer

(b) 10

Series: $I = \dfrac{E}{nR + R} = \dfrac{E}{(n+1)R}$ . Parallel: $I_p = \dfrac{E}{R/n + R} = \dfrac{nE}{(n+1)R}$ . $I_p/I = n$ , and this is given as 10, so $n = 10$ .

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2019 · NEET-UG (India) · 4 marks

Ionized hydrogen atoms and $\alpha$ -particles with the same momenta enter perpendicular to a constant magnetic field, $B$ . The ratio of the radii of their paths $r_H : r_\alpha$ will be
(a) $2:1$ (b) $1:2$ (c) $4:1$ (d) $1:4$

Answer

(a) $2:1$

Radius $r = \dfrac{p}{qB}$ , so for equal momenta $r \propto 1/q$ . $q_H = e$ , $q_\alpha = 2e$ , so $r_H : r_\alpha = 2:1$ . Mass does not matter once momentum is fixed.

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2019 · NEET-UG (India) · 4 marks

In which of the following devices, the eddy current effect is not used?
(a) Induction furnace (b) Magnetic braking in train (c) Electromagnet (d) Electric heater

Answer

(d) Electric heater

Induction furnaces and magnetic braking use eddy currents, and electromagnet cores are laminated precisely to limit them. An electric heater works by Joule heating of a resistive wire carrying a current; it does not use eddy currents.

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2019 · NEET-UG (India) · 4 marks

Which of the following acts as a circuit protection device?
(a) Conductor (b) Inductor (c) Switch (d) Fuse

Answer

(d) Fuse

A fuse has a thin wire that melts when current exceeds a safe limit, breaking the circuit and protecting it. A conductor only carries current, an inductor opposes current changes, and a switch is operated by hand, so none of these protects automatically.

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2019 · NEET-UG (India) · 4 marks

At a point $A$ on the earth's surface the angle of dip, $\delta=+25^\circ$ . At a point $B$ on the earth's surface the angle of dip, $\delta=-25^\circ$ . We can interpret that
(a) $A$ and $B$ are both located in the northern hemisphere. (b) $A$ is located in the southern hemisphere and $B$ is located in the northern hemisphere. (c) $A$ is located in the northern hemisphere and $B$ is located in the southern hemisphere. (d) $A$ and $B$ are both located in the southern hemisphere.

Answer

(c) $A$ is located in the northern hemisphere and $B$ is located in the southern hemisphere.

By convention, the dip angle is positive when the north pole of the needle dips below the horizontal, which happens in the northern hemisphere. It is negative in the southern hemisphere, where the south end dips. So $+25^\circ$ is north and $-25^\circ$ is south.

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2019 · NEET-UG (India) · 4 marks

A 800 turn coil of effective area $0.05\ \mathrm{m^2}$ is kept perpendicular to a magnetic field $5\times10^{-5}$ T. When the plane of the coil is rotated by $90^\circ$ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be
(a) 2 V (b) 0.2 V (c) $2\times10^{-3}$ V (d) 0.02 V

Answer

(d) 0.02 V

Initial flux $=NBA$ (coil perpendicular to field), final flux $=0$ after a $90^\circ$ turn. $\varepsilon=\dfrac{NBA}{t}=\dfrac{800\times5\times10^{-5}\times0.05}{0.1}=0.02$ V.

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2020 · NEET-UG (India) · 4 marks

The solids which have the negative temperature coefficient of resistance are
(a) semiconductors only (b) insulators and semiconductors (c) metals (d) insulators only

Answer

(b) insulators and semiconductors

In insulators and semiconductors, heating frees more charge carriers, so resistance falls as temperature rises (negative $\alpha$ ). In metals, more lattice vibration increases collisions, so resistance rises (positive $\alpha$ ).

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2020 · NEET-UG (India) · 4 marks

A long solenoid of 50 cm length having 100 turns carries a current 2.5 A. The magnetic field at the centre of the solenoid is ( $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ )
(a) $6.28\times10^{-5}$ T (b) $3.14\times10^{-5}$ T (c) $6.28\times10^{-4}$ T (d) $3.14\times10^{-4}$ T

Answer

(c) $6.28\times10^{-4}$ T

$B=\mu_0nI$ with $n=\dfrac{100}{0.5}=200$ turns/m. $B=4\pi\times10^{-7}\times200\times2.5=6.28\times10^{-4}$ T. Using 0.5 m wrongly as the turns-per-length would give the smaller options.

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2020 · NEET-UG (India) · 4 marks

An iron rod of susceptibility 599 is subjected to a magnetising field of $1200\ \mathrm{A\,m^{-1}}$ . The permeability of the material of the rod is: ( $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ )
(a) $2.4\pi\times10^{-5}\ \mathrm{T\,m\,A^{-1}}$ (b) $2.4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ (c) $2.4\pi\times10^{-4}\ \mathrm{T\,m\,A^{-1}}$ (d) $8.0\times10^{-5}\ \mathrm{T\,m\,A^{-1}}$

Answer

(c) $2.4\pi\times10^{-4}\ \mathrm{T\,m\,A^{-1}}$

$\mu=\mu_0(1+\chi)=4\pi\times10^{-7}\times600=2.4\pi\times10^{-4}$ T m A $^{-1}$ . The field $H$ is not needed for $\mu$ ; forgetting the $1$ in $1+\chi$ gives a slightly different, wrong value.

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2020 · NEET-UG (India) · 4 marks

A resistance wire connected in the left gap of a metre bridge balances a $10\ \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3:2$ . If the length of the resistance wire is 1.5 m, then the length of $1\ \Omega$ of the resistance wire is
(a) $1.5\times10^{-1}$ m (b) $1.5\times10^{-2}$ m (c) $1.0\times10^{-2}$ m (d) $1.0\times10^{-1}$ m

Answer

(d) $1.0\times10^{-1}$ m

At balance, $\dfrac{X}{10}=\dfrac{l}{100-l}=\dfrac32$ , so $X=15\ \Omega$ . This wire of 1.5 m has 15 $\Omega$ , so 1 $\Omega$ corresponds to $1.5/15=0.1$ m.

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2020 · NEET-UG (India) · 4 marks

A charged particle having drift velocity of $7.5\times10^{-4}\ \mathrm{m\,s^{-1}}$ in an electric field of $3\times10^{-10}\ \mathrm{V\,m^{-1}}$ has a mobility in $\mathrm{m^2\,V^{-1}\,s^{-1}}$ of
(a) $2.5\times10^{-6}$ (b) $2.25\times10^{-15}$ (c) $2.25\times10^{15}$ (d) $2.5\times10^{6}$

Answer

(d) $2.5\times10^{6}$

Mobility $\mu=\dfrac{v_d}{E}=\dfrac{7.5\times10^{-4}}{3\times10^{-10}}=2.5\times10^{6}\ \mathrm{m^2\,V^{-1}\,s^{-1}}$ . The exponent is $-4-(-10)=+6$ , so the negative-exponent options are wrong.

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2021 · NEET-UG (India) · 4 marks

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is $0.25\ \Omega$ . What will be the effective resistance if they are connected in series?
(a) $1\ \Omega$ (b) $4\ \Omega$ (c) $0.25\ \Omega$ (d) $0.5\ \Omega$

Answer

(b) $4\ \Omega$

Four identical wires of resistance $R$ in parallel give $R/4=0.25\ \Omega$ , so $R=1\ \Omega$ . In series they give $4R=4\ \Omega$ . The tempting $1\ \Omega$ is the resistance of a single wire, not of the series combination.

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2021 · NEET-UG (India) · 4 marks

Column-I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.
Column-I: (A) Drift velocity; (B) Electrical resistivity; (C) Relaxation period; (D) Current density
Column-II: (P) $\frac{m}{ne^2\rho}$ ; (Q) $nev_d$ ; (R) $\frac{eE\tau}{m}$ ; (S) $\frac{E}{J}$
(a) A-R, B-P, C-S, D-Q (b) A-R, B-Q, C-S, D-P (c) A-R, B-S, C-P, D-Q (d) A-R, B-S, C-Q, D-P

Answer

(c) A-R, B-S, C-P, D-Q

Drift velocity $v_d=\frac{eE\tau}{m}$ (R); resistivity $\rho=\frac{E}{J}$ (S); from $\rho=\frac{m}{ne^2\tau}$ the relaxation time is $\tau=\frac{m}{ne^2\rho}$ (P); current density $J=nev_d$ (Q). The other options swap resistivity and relaxation period.

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2021 · NEET-UG (India) · 4 marks

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
(a) 64 cm (b) 62 cm (c) 60 cm (d) 21.6 cm

Answer

(c) 60 cm

In a potentiometer, balance length is proportional to EMF. $\frac{l_2}{l_1}=\frac{E_2}{E_1}$ , so $l_2=36\times\frac{2.5}{1.5}=60$ cm. The value 21.6 cm comes from wrongly multiplying by the inverse ratio.

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2021 · NEET-UG (India) · 4 marks

An infinitely long straight conductor carries a current of 5 A. An electron is moving with a speed of $10^5$ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at the instant.
(a) $4\pi \times 10^{-20}$ N (b) $8 \times 10^{-20}$ N (c) $4 \times 10^{-20}$ N (d) $8\pi \times 10^{-20}$ N

Answer

(b) $8\times10^{-20}$ N

Field of the wire at 20 cm: $B=\frac{\mu_0 I}{2\pi r}=\frac{2\times10^{-7}\times5}{0.2}=5\times10^{-6}$ T. The electron's velocity is parallel to the wire, hence perpendicular to $B$ , so $F=evB=1.6\times10^{-19}\times10^5\times5\times10^{-6}=8\times10^{-20}$ N.

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2021 · NEET-UG (India) · 4 marks

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 \gg R_2$ , the mutual inductance $M$ between them will be directly proportional to:
(a) $\frac{R_1^2}{R_2}$ (b) $\frac{R_2^2}{R_1}$ (c) $\frac{R_1}{R_2}$ (d) $\frac{R_2}{R_1}$

Answer

(b) $\frac{R_2^2}{R_1}$

Take current $I$ in the large loop. The field at the centre is $B=\frac{\mu_0I}{2R_1}$ , nearly uniform over the small loop. Flux $=B\pi R_2^2=\frac{\mu_0\pi IR_2^2}{2R_1}$ , so $M\propto\frac{R_2^2}{R_1}$ .

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2021 · NEET-UG (India) · 4 marks

In the product $\vec{F} = q(\vec{v} \times \vec{B}) = q\vec{v} \times (B\hat{i} + B\hat{j} + B_0\hat{k})$ , for $q = 1$ and $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$ , what will be the complete expression for $\vec{B}$ ?
(a) $8\hat{i} + 8\hat{j} - 6\hat{k}$ (b) $6\hat{i} + 6\hat{j} - 8\hat{k}$ (c) $-8\hat{i} - 8\hat{j} - 6\hat{k}$ (d) $-6\hat{i} - 6\hat{j} - 8\hat{k}$

Answer

(d) $-6\hat i-6\hat j-8\hat k$

With $\vec B=(B,B,B_0)$ and $\vec v=(2,4,6)$ : $\vec v\times\vec B=(4B_0-6B,\ 6B-2B_0,\ 2B-4B)$ . The $\hat k$ term gives $-2B=12$ , so $B=-6$ . The $\hat j$ term gives $-36-2B_0=-20$ , so $B_0=-8$ . Check the $\hat i$ term: $-32+36=4$ . Hence $\vec B=-6\hat i-6\hat j-8\hat k$ .

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2022 · NEET-UG (India) · 4 marks

As the temperature increases, the electrical resistance:
(a) Increases for both conductors and semiconductors (b) Decreases for both conductors and semiconductors (c) Increases for conductors but decreases for semiconductors (d) Decreases for conductors but increases for semiconductors

Answer

(c) Increases for conductors but decreases for semiconductors

In conductors, collisions increase with temperature, so resistance rises (positive temperature coefficient). In semiconductors, heating frees more charge carriers, which outweighs increased collisions, so resistance falls (negative temperature coefficient).

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2022 · NEET-UG (India) · 4 marks

A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
(a) $6.28 \times 10^{-2}$ T (b) $12.56 \times 10^{-2}$ T (c) $12.26 \times 10^{-4}$ T (d) $6.28 \times 10^{-4}$ T

Answer

(b) $12.56 \times 10^{-2}$ T

Inside a long solenoid $B = \mu_0 n I$ . Here $n = 100$ turns/mm $= 10^5$ turns/m, so $B = 4\pi\times10^{-7}\times10^{5}\times1 = 0.1256$ T $= 12.56\times10^{-2}$ T. The radius of 1 mm is not needed.

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2022 · NEET-UG (India) · 4 marks

A copper wire of length 10 m and radius $\left(\frac{10^{-2}}{\sqrt{\pi}}\right)$ m has electrical resistance of 10 $\Omega$ . The current density in the wire for an electric field strength of 10 (V/m) is:
(a) $10^4\ \mathrm{A/m^2}$ (b) $10^6\ \mathrm{A/m^2}$ (c) $10^{-5}\ \mathrm{A/m^2}$ (d) $10^5\ \mathrm{A/m^2}$

Answer

(d) $10^5\ \mathrm{A/m^2}$

Area $A = \pi r^2 = 10^{-4}\ \mathrm{m^2}$ . Resistivity $\rho = RA/L = 10\times10^{-4}/10 = 10^{-4}\ \Omega\,$ m. Current density $J = E/\rho = 10/10^{-4} = 10^5\ \mathrm{A/m^2}$ .

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2022 · NEET-UG (India) · 4 marks

A square loop of side 1 m and resistance 1 $\Omega$ is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is:
(a) 2 weber (b) 0.5 weber (c) 1 weber (d) Zero weber

Answer

(b) 0.5 weber

Flux $\phi = BA\cos\theta$ with the plane perpendicular to the field, so the normal is parallel to $B$ and $\theta=0$ . $\phi = 0.5\times(1\times1) = 0.5$ Wb. The resistance is irrelevant here.

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2022 · NEET-UG (India) · 4 marks

Given below are two statements.
Statement I: Biot-Savart's law gives an expression for the magnetic field strength of an infinitesimal current element ( $Idl$ ) of a current carrying conductor only.
Statement II: Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$ , with the former being related to the field produced by a scalar source, $Idl$ , while the latter being produced by a vector source, $q$ .
In light of above statements choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct (b) Both Statement I and Statement II are incorrect (c) Statement I is correct and Statement II is incorrect (d) Statement I is incorrect and Statement II is correct

Answer

(c) Statement I is correct and Statement II is incorrect

Biot-Savart law gives the field of a current element $Idl$ only, so Statement I is correct. In Statement II the roles are reversed: $Idl$ is a vector source while the charge $q$ is a scalar source, so the statement is wrong.

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2022 · NEET-UG (India) · 4 marks

Two resistors of resistances 100 $\Omega$ and 200 $\Omega$ are connected in parallel in an electrical circuit. The ratio of thermal energy developed in 100 $\Omega$ to that in 200 $\Omega$ in a given time is:
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Answer

(b) 2 : 1

In parallel the voltage is the same, so $H = V^2t/R\propto 1/R$ . Hence $H_{100}:H_{200} = 200:100 = 2:1$ . Using $I^2R$ with the same current is wrong because the currents differ.

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2022 · NEET-UG (India) · 4 marks

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:
(a) Uniform and remains constant for both the regions (b) A linearly increasing function of distance up to the boundary of the wire and then linearly decreasing for the outside region (c) A linearly increasing function of distance $r$ up to the boundary of the wire and then decreasing one with $1/r$ dependence for the outside region (d) A linearly decreasing function of distance up to the boundary of the wire and then a linearly increasing one for the outside region

Answer

(c) A linearly increasing function of distance $r$ up to the boundary of the wire and then decreasing one with $1/r$ dependence for the outside region

Inside, the enclosed current is $I r^2/a^2$ , so $B = \dfrac{\mu_0 I r}{2\pi a^2}\propto r$ . Outside, the full current is enclosed, so $B=\dfrac{\mu_0 I}{2\pi r}\propto 1/r$ .

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2022 · NEET-UG (India) · 4 marks

A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at $2\ \mathrm{rad\,s^{-1}}$ . If the vertical component of earth's magnetic field at that place is $2 \times 10^{-5}$ T and electrical resistance of the coil is $12.56\ \Omega$ , then the maximum induced current in the coil will be:
(a) 0.25 A (b) 1.5 A (c) 1 A (d) 2 A

Answer

(c) 1 A

Maximum emf $= NBA\omega = 1000\times2\times10^{-5}\times\pi(10)^2\times2 = 12.56$ V. Maximum current $= 12.56/12.56 = 1$ A.

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2023 · NEET-UG (India) · 4 marks

The magnetic energy stored in an inductor of inductance 4 H carrying a current of 2 A is:
(a) 4 J (b) 4 mJ (c) 8 mJ (d) 8 J

Answer

(d) 8 J

$U = \tfrac12LI^2 = \tfrac12\times4\times2^2 = 8$ J.

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2023 · NEET-UG (India) · 4 marks

Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$ . The colour of third band must be:
(a) Red (b) Green (c) Orange (d) Yellow

Answer

(c) Orange

$22000\ \Omega = 22\times10^{3}\ \Omega$ . The first two bands give 2 and 2 (red, red), and the third band is the multiplier $10^3$ , which is orange. Yellow would be $10^4$ and green $10^5$ .

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2023 · NEET-UG (India) · 4 marks

The net magnetic flux through any closed surface is:
(a) Zero (b) Positive (c) Infinity (d) Negative

Answer

(a) Zero

Magnetic monopoles do not exist, so field lines form closed loops: as many lines enter a closed surface as leave it. Hence $\oint\vec B\cdot d\vec A=0$ (Gauss's law for magnetism).

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2023 · NEET-UG (India) · 4 marks

10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is:
(a) 10 (b) 100 (c) 1 (d) 1000

Answer

(b) 100

Series: $R_s=10R$ , so $I_s=\frac{E}{10R}$ . Parallel: $R_p=\frac{R}{10}$ , so $I_p=\frac{10E}{R}$ . Ratio $n=\frac{I_p}{I_s}=100$ .

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2023 · NEET-UG (India) · 4 marks

The resistance of platinum wire at $0^\circ$ C is $2\ \Omega$ and $6.8\ \Omega$ at $80^\circ$ C. The temperature coefficient of resistance of the wire is:
(a) $3 \times 10^{-4}\ ^\circ\mathrm{C}^{-1}$ (b) $3 \times 10^{-3}\ ^\circ\mathrm{C}^{-1}$ (c) $3 \times 10^{-2}\ ^\circ\mathrm{C}^{-1}$ (d) $3 \times 10^{-1}\ ^\circ\mathrm{C}^{-1}$

Answer

(c) $3\times10^{-2}\ ^\circ\mathrm{C}^{-1}$

$R_t=R_0(1+\alpha t)$ gives $\alpha=\frac{R_t-R_0}{R_0t}=\frac{6.8-2}{2\times80}=\frac{4.8}{160}=3\times10^{-2}\ ^\circ\mathrm{C}^{-1}$ .

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2023 · NEET-UG (India) · 4 marks

A wire carrying a current $I$ along the positive $x$ -axis has length $L$ . It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})$ T. The magnitude of the magnetic force acting on the wire is:
(a) $3IL$ (b) $\sqrt{5}\,IL$ (c) $5IL$ (d) $\sqrt{3}\,IL$

Answer

(c) $5IL$

$\vec F=I\vec L\times\vec B = IL\,\hat i\times(2\hat i+3\hat j-4\hat k)=IL(3\hat k+4\hat j)$ . Magnitude $=IL\sqrt{9+16}=5IL$ . The $x$ -component of $\vec B$ is parallel to the wire and gives no force.

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2024 · NEET-UG (India) · 4 marks

Match List - I with List - II.
List - I (Material): A. Diamagnetic; B. Ferromagnetic; C. Paramagnetic; D. Non-magnetic
List - II (Susceptibility $\chi$ ): I. $\chi = 0$ ; II. $0 > \chi \ge -1$ ; III. $\chi \gg 1$ ; IV. $0 < \chi < \varepsilon$ ( $\varepsilon$ a small positive number)
Choose the correct answer from the options given below:
(a) A-II, B-I, C-III, D-IV (b) A-III, B-II, C-I, D-IV (c) A-IV, B-III, C-II, D-I (d) A-II, B-III, C-IV, D-I

Answer

(d) A-II, B-III, C-IV, D-I

Diamagnetic: small negative $\chi$ ( $0>\chi\ge-1$ ). Ferromagnetic: $\chi\gg1$ . Paramagnetic: small positive $\chi$ . Non-magnetic: $\chi=0$ .

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2024 · NEET-UG (India) · 4 marks

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi \times 10^{-7}$ SI units):
(a) 4.4 T (b) 4.4 mT (c) 44 T (d) 44 mT

Answer

(b) 4.4 mT

$B=\frac{\mu_0NI}{2R}=\frac{4\pi\times10^{-7}\times100\times7}{2\times0.1}=4.4\times10^{-3}$ T $=4.4$ mT.

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2024 · NEET-UG (India) · 4 marks

The terminal voltage of the battery, whose emf is 10 V and internal resistance $1\ \Omega$ , when connected through an external resistance of $4\ \Omega$ is:
(a) 6 V (b) 8 V (c) 10 V (d) 4 V

Answer

(b) 8 V

Current $I=\frac{E}{R+r}=\frac{10}{4+1}=2$ A. Terminal voltage $V=E-Ir=10-2=8$ V (equivalently $IR=2\times4$ ).

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2024 · NEET-UG (India) · 4 marks

A wire of length $l$ and resistance $100\ \Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
(a) $52\ \Omega$ (b) $55\ \Omega$ (c) $60\ \Omega$ (d) $26\ \Omega$

Answer

(a) $52\ \Omega$

Each part is $10\ \Omega$ . Five in series give $50\ \Omega$ ; five in parallel give $\frac{10}{5}=2\ \Omega$ . Total in series: $50+2=52\ \Omega$ .

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2024 · NEET-UG (India) · 4 marks

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is $9.8 \times 10^{-6}\ \mathrm{kg\,m^2}$ . If the magnitude of the magnetic moment of the needle is $x \times 10^{-5}\ \mathrm{A\,m^2}$ , then the value of $x$ is:
(a) $128\pi^2$ (b) $50\pi^2$ (c) $1280\pi^2$ (d) $5\pi^2$

Answer

(c) $1280\pi^2$

Period $T=\frac{5}{20}=0.25$ s. For a magnetic needle $T=2\pi\sqrt{\frac{I}{MB}}$ , so $M=\frac{4\pi^2 I}{BT^2}$ . $M=\frac{4\pi^2(9.8\times10^{-6})}{0.049\times0.0625}=4\pi^2\times3.2\times10^{-3}=1.28\times10^{-2}\pi^2=1280\pi^2\times10^{-5}\ \mathrm{A\,m^2}$ . So $x=1280\pi^2$ . A common slip is using the frequency in place of the period.

Answer checked by Claude Sonnet · Oct 21, 2026

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2024 · NEET-UG (India) · 4 marks

Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
(a) 2 : 9 (b) 1 : 2 (c) 2 : 3 (d) 1 : 1

Answer

(a) 2 : 9

With supply voltage $V$ , $R_A=\frac{V^2}{1000}$ and $R_B=\frac{V^2}{2000}$ . Series: $P_s=\frac{V^2}{R_A+R_B}=\frac{V^2}{3V^2/2000}=\frac{2000}{3}$ W. Parallel: $P_p=1000+2000=3000$ W. Ratio $P_s:P_p=\frac{2000}{3}:3000=2:9$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2024 · NEET-UG (India) · 4 marks

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic B. hold the sheet there if it is non-magnetic. C. move the sheet away from the pole with uniform velocity if it is conducting. D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
(a) A and C only (b) A, C and D only (c) C only (d) B and D only

Answer

(a) A and C only

A: a magnetic sheet is attracted by the strong pole, so a force is needed to hold it in place. C: moving a conductor away from the pole changes the flux through it, so eddy currents arise and (Lenz's law) oppose the motion, and a force is needed even at uniform velocity. B is wrong because a non-magnetic sheet feels no static force. D is wrong because a non-conducting, non-polar sheet has neither magnetic attraction nor eddy currents.

Answer checked by Claude Sonnet · Oct 21, 2026

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2025 · NEET-UG (India) · 4 marks

A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
(a) $\frac{R}{16}$ (b) $\frac{R}{8}$ (c) $\frac{R}{64}$ (d) $\frac{R}{32}$

Answer

(a) $\frac{R}{16}$

Each of the 8 pieces has resistance $\frac R8$ . Four in parallel: $\frac{R/8}{4}=\frac{R}{32}$ . Two such sets in series: $\frac{R}{32}+\frac{R}{32}=\frac{R}{16}$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2025 · NEET-UG (India) · 4 marks

An electron (mass $9 \times 10^{-31}$ kg and charge $1.6 \times 10^{-19}$ C) moving with speed $c/100$ ( $c$ = speed of light) is injected into a magnetic field $B$ of magnitude $9 \times 10^{-4}$ T perpendicular to its direction of motion. We wish to apply a uniform electric field $E$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8}\ \mathrm{m\,s^{-1}}$ ):
(a) $E$ is parallel to $B$ and its magnitude is $27 \times 10^{2}\ \mathrm{V\,m^{-1}}$ (b) $E$ is parallel to $B$ and its magnitude is $27 \times 10^{4}\ \mathrm{V\,m^{-1}}$ (c) $E$ is perpendicular to $B$ and its magnitude is $27 \times 10^{4}\ \mathrm{V\,m^{-1}}$ (d) $E$ is perpendicular to $B$ and its magnitude is $27 \times 10^{2}\ \mathrm{V\,m^{-1}}$

Answer

(d) $E$ is perpendicular to $B$ and its magnitude is $27\times10^{2}\ \mathrm{V\,m^{-1}}$

For no deflection the electric and magnetic forces balance: $eE=evB$ , so $E=vB$ . The magnetic force is perpendicular to both $v$ and $B$ , so $E$ must also be perpendicular to $B$ (and to $v$ ); a field parallel to $B$ cannot cancel it. $v=\frac{c}{100}=3\times10^6$ m/s, so $E=3\times10^6\times9\times10^{-4}=2.7\times10^3=27\times10^2$ V/m.

Answer checked by Claude Sonnet · Oct 21, 2026

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2025 · NEET-UG (India) · 4 marks

A 2 A current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be:
(a) 2 : 1 (b) 4 : 1 (c) 1 : 4 (d) 1 : 2

Answer

(c) 1 : 4

Magnetic moment of a single-turn coil is $M=IA=I\pi r^2$ . The current is the same (2 A), so $M\propto r^2$ . $\frac{M_1}{M_2}=\left(\frac12\right)^2=\frac14$ , i.e. 1 : 4.

Answer checked by Claude Sonnet · Oct 21, 2026

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Past questions and answers

1.A wire of resistance 4 Ω4\ \Omega4 Ω is stretched to twice its original length. The resistance of stretched wire would be: (a) 2 Ω2\ \Omega2 Ω (b) 4 Ω4\ \Omega4 Ω (c) 8 Ω8\ \Omega8 Ω (d) 16 Ω16\ \Omega16 Ω

2.The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω10\ \Omega10 Ω is: (a) 0.2 Ω0.2\ \Omega0.2 Ω (b) 0.5 Ω0.5\ \Omega0.5 Ω (c) 0.8 Ω0.8\ \Omega0.8 Ω (d) 1.0 Ω1.0\ \Omega1.0 Ω

5.A bar magnet of length lll and magnetic dipole moment MMM is bent in the form of an arc (subtending 60∘60^\circ60∘ at its centre) as shown in figure. The new magnetic dipole moment will be: (a) MMM (b) 3πM\frac{3}{\pi}Mπ3​M (c) 2πM\frac{2}{\pi}Mπ2​M (d) M2\frac{M}{2}2M​

6.A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is: (a) once per revolution (b) twice per revolution (c) four times per revolution (d) six times per revolution

10.In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is GGG, the resistance of ammeter will be: (a) 499500G\frac{499}{500}G500499​G (b) 1500G\frac{1}{500}G5001​G (c) 500499G\frac{500}{499}G499500​G (d) 1499G\frac{1}{499}G4991​G

11.Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω0.5\ \Omega0.5 Ω. The power loss in the wires is: (a) 19.2 kW (b) 19.2 J (c) 12.2 kW (d) 19.2 W

16.A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be: (a) 2 A (b) 1 A (c) 0.5 A (d) 0.25 A

17.A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10−34\times10^{-3}4×10−3 Wb. The self-inductance of the solenoid is (a) 1 H (b) 4 H (c) 3 H (d) 2 H

20.The magnetic susceptibility is negative for: (a) paramagnetic and ferromagnetic materials (b) diamagnetic material only (c) paramagnetic material only (d) ferromagnetic material only

22.The resistance of a wire is RRR ohm. If it is melted and stretched to nnn times its original length, its new resistance will be (a) Rn\dfrac{R}{n}nR​ (b) n2Rn^2Rn2R (c) Rn2\dfrac{R}{n^2}n2R​ (d) nRnRnR

25.A potentiometer is an accurate and versatile device to make electrical measurements of EMF, because the method involves (a) potential gradients (b) a condition of no current flow through the galvanometer (c) a combination of cells, galvanometer and resistances (d) cells

27.The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (a) 13.89 H (b) 0.138 H (c) 1.389 H (d) 138.88 H

30.Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is (a) 500 Ω500\ \Omega500 Ω (b) 40 Ω40\ \Omega40 Ω (c) 250 Ω250\ \Omega250 Ω (d) 25 Ω25\ \Omega25 Ω

33.Ionized hydrogen atoms and α\alphaα-particles with the same momenta enter perpendicular to a constant magnetic field, BBB. The ratio of the radii of their paths rH:rαr_H : r_\alpharH​:rα​ will be (a) 2:12:12:1 (b) 1:21:21:2 (c) 4:14:14:1 (d) 1:41:41:4

34.In which of the following devices, the eddy current effect is not used? (a) Induction furnace (b) Magnetic braking in train (c) Electromagnet (d) Electric heater

35.Which of the following acts as a circuit protection device? (a) Conductor (b) Inductor (c) Switch (d) Fuse

38.The solids which have the negative temperature coefficient of resistance are (a) semiconductors only (b) insulators and semiconductors (c) metals (d) insulators only

45.In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs? (a) 64 cm (b) 62 cm (c) 60 cm (d) 21.6 cm

49.As the temperature increases, the electrical resistance: (a) Increases for both conductors and semiconductors (b) Decreases for both conductors and semiconductors (c) Increases for conductors but decreases for semiconductors (d) Decreases for conductors but increases for semiconductors

52.A square loop of side 1 m and resistance 1 Ω\OmegaΩ is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is: (a) 2 weber (b) 0.5 weber (c) 1 weber (d) Zero weber

54.Two resistors of resistances 100 Ω\OmegaΩ and 200 Ω\OmegaΩ are connected in parallel in an electrical circuit. The ratio of thermal energy developed in 100 Ω\OmegaΩ to that in 200 Ω\OmegaΩ in a given time is: (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

57.The magnetic energy stored in an inductor of inductance 4 H carrying a current of 2 A is: (a) 4 J (b) 4 mJ (c) 8 mJ (d) 8 J

58.Resistance of a carbon resistor determined from colour codes is (22000±5%) Ω(22000 \pm 5\%)\ \Omega(22000±5%) Ω. The colour of third band must be: (a) Red (b) Green (c) Orange (d) Yellow

59.The net magnetic flux through any closed surface is: (a) Zero (b) Positive (c) Infinity (d) Negative

60.10 resistors, each of resistance RRR are connected in series to a battery of emf EEE and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased nnn times. The value of nnn is: (a) 10 (b) 100 (c) 1 (d) 1000

64.A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10−74\pi \times 10^{-7}4π×10−7 SI units): (a) 4.4 T (b) 4.4 mT (c) 44 T (d) 44 mT

65.The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω1\ \Omega1 Ω, when connected through an external resistance of 4 Ω4\ \Omega4 Ω is: (a) 6 V (b) 8 V (c) 10 V (d) 4 V

68.Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is: (a) 2 : 9 (b) 1 : 2 (c) 2 : 3 (d) 1 : 1

72.A 2 A current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be: (a) 2 : 1 (b) 4 : 1 (c) 1 : 4 (d) 1 : 2

A wire of resistance $4\ \Omega$ is stretched to twice its original length. The resistance of stretched wire would be:
(a) $2\ \Omega$ (b) $4\ \Omega$ (c) $8\ \Omega$ (d) $16\ \Omega$

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of $10\ \Omega$ is:
(a) $0.2\ \Omega$ (b) $0.5\ \Omega$ (c) $0.8\ \Omega$ (d) $1.0\ \Omega$

A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of an arc (subtending $60^\circ$ at its centre) as shown in figure. The new magnetic dipole moment will be:
(a) $M$ (b) $\frac{3}{\pi}M$ (c) $\frac{2}{\pi}M$ (d) $\frac{M}{2}$

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is:
(a) once per revolution (b) twice per revolution (c) four times per revolution (d) six times per revolution

In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is $G$ , the resistance of ammeter will be:
(a) $\frac{499}{500}G$ (b) $\frac{1}{500}G$ (c) $\frac{500}{499}G$ (d) $\frac{1}{499}G$

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is $0.5\ \Omega$ . The power loss in the wires is:
(a) 19.2 kW (b) 19.2 J (c) 12.2 kW (d) 19.2 W

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:
(a) 2 A (b) 1 A (c) 0.5 A (d) 0.25 A

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is $4\times10^{-3}$ Wb. The self-inductance of the solenoid is
(a) 1 H (b) 4 H (c) 3 H (d) 2 H

The magnetic susceptibility is negative for:
(a) paramagnetic and ferromagnetic materials (b) diamagnetic material only (c) paramagnetic material only (d) ferromagnetic material only

The resistance of a wire is $R$ ohm. If it is melted and stretched to $n$ times its original length, its new resistance will be
(a) $\dfrac{R}{n}$ (b) $n^2R$ (c) $\dfrac{R}{n^2}$ (d) $nR$

A potentiometer is an accurate and versatile device to make electrical measurements of EMF, because the method involves
(a) potential gradients (b) a condition of no current flow through the galvanometer (c) a combination of cells, galvanometer and resistances (d) cells

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
(a) 13.89 H (b) 0.138 H (c) 1.389 H (d) 138.88 H

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
(a) $500\ \Omega$ (b) $40\ \Omega$ (c) $250\ \Omega$ (d) $25\ \Omega$

Ionized hydrogen atoms and $\alpha$ -particles with the same momenta enter perpendicular to a constant magnetic field, $B$ . The ratio of the radii of their paths $r_H : r_\alpha$ will be
(a) $2:1$ (b) $1:2$ (c) $4:1$ (d) $1:4$

In which of the following devices, the eddy current effect is not used?
(a) Induction furnace (b) Magnetic braking in train (c) Electromagnet (d) Electric heater

Which of the following acts as a circuit protection device?
(a) Conductor (b) Inductor (c) Switch (d) Fuse

The solids which have the negative temperature coefficient of resistance are
(a) semiconductors only (b) insulators and semiconductors (c) metals (d) insulators only

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
(a) 64 cm (b) 62 cm (c) 60 cm (d) 21.6 cm

As the temperature increases, the electrical resistance:
(a) Increases for both conductors and semiconductors (b) Decreases for both conductors and semiconductors (c) Increases for conductors but decreases for semiconductors (d) Decreases for conductors but increases for semiconductors

A square loop of side 1 m and resistance 1 $\Omega$ is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is:
(a) 2 weber (b) 0.5 weber (c) 1 weber (d) Zero weber

Two resistors of resistances 100 $\Omega$ and 200 $\Omega$ are connected in parallel in an electrical circuit. The ratio of thermal energy developed in 100 $\Omega$ to that in 200 $\Omega$ in a given time is:
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

The magnetic energy stored in an inductor of inductance 4 H carrying a current of 2 A is:
(a) 4 J (b) 4 mJ (c) 8 mJ (d) 8 J

Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$ . The colour of third band must be:
(a) Red (b) Green (c) Orange (d) Yellow

The net magnetic flux through any closed surface is:
(a) Zero (b) Positive (c) Infinity (d) Negative

10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is:
(a) 10 (b) 100 (c) 1 (d) 1000

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi \times 10^{-7}$ SI units):
(a) 4.4 T (b) 4.4 mT (c) 44 T (d) 44 mT

The terminal voltage of the battery, whose emf is 10 V and internal resistance $1\ \Omega$ , when connected through an external resistance of $4\ \Omega$ is:
(a) 6 V (b) 8 V (c) 10 V (d) 4 V

Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
(a) 2 : 9 (b) 1 : 2 (c) 2 : 3 (d) 1 : 1

A 2 A current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be:
(a) 2 : 1 (b) 4 : 1 (c) 1 : 4 (d) 1 : 2