JEE Physics Current Electricity MCQs Set 32

Multiple Choice Questions (MCQs) for JEE Physics: Current Electricity

Access targeted multiple-choice questions for Current Electricity designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Current Electricity MCQs for JEE Physics

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Multiple Choice Questions

Question 1: There is a current of 20 A in a copper wire of \( 10^{-6}\ \text{m}^2 \) area of cross-section. If the number of free electrons per cubic metre is \( 10^{29} \), the drift velocity is
(a) \( 125\times10^{-3} \) m/s
(b) \( 12.5\times10^{-3} \) m/s
(c) \( 1.25\times10^{-3} \) m/s
(d) \( 1.25\times10^{-4} \) m/s
Show Answer & Explanation

Answer: (c) \( 1.25\times10^{-3} \) m/s

Explanation:
1. \( v_d=\dfrac{20}{10^{29}\times1.6\times10^{-19}\times10^{-6}} \).
2. \( =1.25\times10^{-3} \) m/s.

Teacher's Note:
1. Drift speeds in metals are of the order of mm/s.

Question 2: The resistance of a wire of uniform diameter \( d \) and length \( L \) is \( R \). The resistance of another wire of the same material but diameter \( 2d \) and length \( 4L \) will be
(a) \( 2R \)
(b) \( R \)
(c) \( R/2 \)
(d) \( R/4 \)
Show Answer & Explanation

Answer: (b) \( R \)

Explanation:
1. \( R=\dfrac{\rho L}{\pi d^2/4} \).
2. Length ×4 and area ×4 cancel.

Teacher's Note:
1. Area depends on the square of the diameter.

Question 3: There is a current of 1.344 A in a copper wire of cross-section \( 1\ \text{mm}^2 \). If the number of free electrons per cm\(^3\) is \( 8.4\times10^{22} \), the drift velocity is
(a) 1.0 mm/s
(b) 1.0 m/s
(c) 0.1 mm/s
(d) 0.01 mm/s
Show Answer & Explanation

Answer: (c) 0.1 mm/s

Explanation:
1. \( n=8.4\times10^{28} \) m\(^{-3}\).
2. \( v_d=\dfrac{1.344}{8.4\times10^{28}\times1.6\times10^{-19}\times10^{-6}}=10^{-4} \) m/s.

Teacher's Note:
1. Convert cm\(^3\) to m\(^3\) carefully.

Question 4: 5 A of current is passed through a metallic conductor. The charge flowing in one minute in coulombs is
(a) 5
(b) 12
(c) 1/12
(d) 300
Show Answer & Explanation

Answer: (d) 300

Explanation:
1. Charge equals current times time in seconds.

Teacher's Note:
1. Charge is current multiplied by time in seconds.
2. 1 minute is 60 s.

Question 5: Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first wire will be
(a) twice that of the second
(b) half that of the second
(c) equal to the second
(d) four times the second
Show Answer & Explanation

Answer: (b) half that of the second

Explanation:
1. \( \dfrac{R_1}{R_2}=\dfrac{L_1}{L_2}\Big(\dfrac{d_2}{d_1}\Big)^2 \).

Teacher's Note:
1. Resistance depends on length directly and on the square of the diameter inversely.
2. Compare ratios instead of computing absolute values.

Question 6: An electric wire of length \( l \) and area of cross-section \( a \) has resistance \( R \) ohm. Another wire of the same material having the same length and area of cross-section \( 4a \) has a resistance of
(a) \( 4R \)
(b) \( R/4 \)
(c) \( R/16 \)
(d) \( 16R \)
Show Answer & Explanation

Answer: (b) \( R/4 \)

Explanation:
1. Area ×4 gives resistance /4.

Teacher's Note:
1. Resistance is inversely proportional to the area of cross-section.
2. Larger area gives more parallel paths for charge.

Question 7: If \( n,e,\tau \) and \( m \) respectively represent the density, charge, relaxation time and mass of the electron, then the resistance of a wire of length \( l \) and area of cross-section \( A \) will be
(a) \( \dfrac{ml}{ne^2\tau A} \)
(b) \( \dfrac{m^2\tau A}{ne^2l} \)
(c) \( \dfrac{ne^2\tau A}{2ml} \)
(d) \( \dfrac{ne^2A}{2m\tau l} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{ml}{ne^2\tau A} \)

Explanation:
1. From Drude's model, \( \sigma=\dfrac{ne^2\tau}{m} \).
2. \( R=\dfrac{l}{\sigma A}=\dfrac{ml}{ne^2\tau A} \).

Teacher's Note:
1. Check dimensions: ohm \( =\) kg m\(^2\) s\(^{-1}\) C\(^{-2}\).

Question 8: A certain piece of silver of given mass is to be made like a wire. Which combination of length \( L \) and area \( A \) will lead to the smallest resistance?
(a) \( L \) and \( A \)
(b) \( 2L \) and \( A/2 \)
(c) \( L/2 \) and \( 2A \)
(d) any of the above, because the volume of silver remains unchanged
Show Answer & Explanation

Answer: (c) \( L/2 \) and \( 2A \)

Explanation:
1. \( R=\rho L^2/V \) for fixed volume.
2. The smallest \( L \) gives the smallest R: \( L/2 \) and \( 2A \).

Teacher's Note:
1. With fixed volume \( R\propto L^2 \).

Question 9: The resistance of a wire is 10 \( \Omega \). Its length is increased by 10% by stretching. The new resistance will now be
(a) 12 \( \Omega \)
(b) 1.2 \( \Omega \)
(c) 13 \( \Omega \)
(d) 11 \( \Omega \)
Show Answer & Explanation

Answer: (a) 12 \( \Omega \)

Explanation:
1. \( R'=R(1.1)^2=12.1\ \Omega \).

Teacher's Note:
1. Use the approximation \( (1+x)^2\approx1+2x \) for small x: 12 \( \Omega \).

Question 10: The resistance of a tungsten wire at \( 150^\circ \)C is 133 \( \Omega \). Its temperature coefficient of resistance is 0.0045 per \( ^\circ \)C. The resistance of this wire at \( 500^\circ \)C will be
(a) 180 \( \Omega \)
(b) 225 \( \Omega \)
(c) 258 \( \Omega \)
(d) 317 \( \Omega \)
Show Answer & Explanation

Answer: (c) 258 \( \Omega \)

Explanation:
1. \( R_t=R_0(1+\alpha t) \) with t in \( ^\circ \)C from \( 0^\circ \)C.
2. \( R_0=\dfrac{133}{1+0.0045\times150}=79.4 \).
3. \( R_{500}=79.4(1+0.0045\times500)=258\ \Omega \).

Teacher's Note:
1. Always reduce to the \( 0^\circ \)C resistance first.

Question 11: A metal wire of specific resistance \( 64\times10^{-6} \) ohm-cm and length 198 cm has a resistance of 7 ohm. The radius of the wire will be
(a) 2.4 cm
(b) 0.24 cm
(c) 0.024 cm
(d) 24 cm
Show Answer & Explanation

Answer: (c) 0.024 cm

Explanation:
1. \( A=\dfrac{64\times10^{-6}\times198}{7}=1.81\times10^{-3}\ \text{cm}^2 \).
2. \( r=\sqrt{A/\pi}=0.024 \) cm.

Teacher's Note:
1. Keep units in cm throughout.

Question 12: A copper wire of length 1 m and radius 1 mm is joined in series with an iron wire of length 2 m and radius 3 mm and a current is passed through the wires. The ratio of the current density in the copper wire to that in the iron wire is
(a) 18 : 1
(b) 9 : 1
(c) 6 : 1
(d) 2 : 3
Show Answer & Explanation

Answer: (b) 9 : 1

Explanation:
1. \( J=I/A \) with the same \( I \).
2. \( J_{Cu}/J_{Fe}=r_{Fe}^2/r_{Cu}^2=9 \).

Teacher's Note:
1. Length and material do not matter for current density in series.

Question 13: If the length of a wire is doubled, then the new resistance will be (volume constant, original resistance R)
(a) \( 2R \)
(b) \( 4R \)
(c) \( R \)
(d) \( R/4 \)
Show Answer & Explanation

Answer: (b) \( 4R \)

Explanation:
1. \( R\propto L^2 \) at constant volume.

Teacher's Note:
1. Stretching keeps the volume constant, so the area also changes.
2. Resistance varies as the square of the length at constant volume.

Question 14: A solenoid is at a potential difference of 60 V and a current of 15 A flows in it. The resistance of the coil is
(a) 4 \( \Omega \)
(b) 8 \( \Omega \)
(c) 0.25 \( \Omega \)
(d) 2 \( \Omega \)
Show Answer & Explanation

Answer: (a) 4 \( \Omega \)

Explanation:
1. Ohm's law.

Teacher's Note:
1. Ohm's law gives resistance as the ratio of potential difference to current.
2. The same relation holds for any ohmic conductor.

Question 15: In a wire of circular cross-section with radius \( r \), free electrons travel with a drift velocity \( v \) when a current \( I \) flows. What is the current in another wire of the same material of half the radius when the drift velocity is \( 2v \)?
(a) \( 2I \)
(b) \( I \)
(c) \( I/2 \)
(d) \( I/4 \)
Show Answer & Explanation

Answer: (c) \( I/2 \)

Explanation:
1. Same material means the same \( n \).
2. \( I'=I\times\dfrac{(r/2)^2}{r^2}\times2=\dfrac I2 \).

Teacher's Note:
1. The area scales with the square of the radius.

Question 16: In a neon discharge tube, \( 2.9\times10^{18} \) Ne\(^+\) ions move to the right each second while \( 1.2\times10^{18} \) electrons move to the left per second. The electron charge is \( 1.6\times10^{-19} \) C. The current in the discharge tube is
(a) 1 A towards the right
(b) 0.66 A towards the right
(c) 0.66 A towards the left
(d) zero
Show Answer & Explanation

Answer: (b) 0.66 A towards the right

Explanation:
1. Ions moving right carry current to the right.
2. Electrons moving left also give current to the right.

Teacher's Note:
1. Add the two contributions.

Question 17: Two wires A and B of the same material and the same mass have radii \( 2r \) and \( r \). If the resistance of wire A is 34 \( \Omega \), then the resistance of B will be
(a) 544 \( \Omega \)
(b) 272 \( \Omega \)
(c) 68 \( \Omega \)
(d) 17 \( \Omega \)
Show Answer & Explanation

Answer: (a) 544 \( \Omega \)

Explanation:
1. Equal volume gives \( L\propto1/r^2 \).
2. \( R\propto L/A\propto1/r^4 \), so \( R_B=34\times2^4=544\ \Omega \).

Teacher's Note:
1. Equal mass of the same material means equal volume.

Question 18: A current of 20 \( \mu \)A flows for 30 seconds in a wire. The charge transferred is
(a) \( 2\times10^{-4} \) C
(b) \( 4\times10^{-4} \) C
(c) \( 6\times10^{-4} \) C
(d) \( 8\times10^{-4} \) C
Show Answer & Explanation

Answer: (c) \( 6\times10^{-4} \) C

Explanation:
1. Convert microampere to ampere.

Teacher's Note:
1. Always convert microampere to ampere before using Q = It.
2. The result is in coulomb.

Question 19: The resistance of a coil is 4.2 \( \Omega \) at \( 100^\circ \)C and the temperature coefficient of resistance of its material is \( 0.004/^\circ \)C. Its resistance at \( 0^\circ \)C is
(a) 6.5 \( \Omega \)
(b) 5 \( \Omega \)
(c) 3 \( \Omega \)
(d) 4 \( \Omega \)
Show Answer & Explanation

Answer: (c) 3 \( \Omega \)

Explanation:
1. \( R_t=R_0(1+\alpha t) \) with \( \alpha t=0.4 \).

Teacher's Note:
1. The linear formula R = R0(1 + alpha t) uses the 0 degree C resistance as the base.
2. Use alpha t = 0.4 to get R0 quickly.

Question 20: Masses of three copper wires are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of their electrical resistances is
(a) 1 : 3 : 5
(b) 5 : 3 : 1
(c) 1 : 15 : 125
(d) 125 : 15 : 1
Show Answer & Explanation

Answer: (d) 125 : 15 : 1

Explanation:
1. Volume \( \propto m \), so \( A\propto m/L \).
2. \( R=\rho L/A\propto L^2/m \).

Teacher's Note:
1. Combine mass and length to get the area.

Question 21: The resistance of a conductor is 5 ohm at \( 50^\circ \)C and 6 ohm at \( 100^\circ \)C. Its resistance at \( 0^\circ \)C is
(a) 1 ohm
(b) 2 ohm
(c) 3 ohm
(d) 4 ohm
Show Answer & Explanation

Answer: (d) 4 ohm

Explanation:
1. \( R_{100}-R_{50}=50R_0\alpha=1 \).
2. \( R_{50}=R_0+50R_0\alpha=R_0+1=5 \), so \( R_0=4\ \Omega \).

Teacher's Note:
1. Resistance is linear in temperature.

Question 22: Equal potentials are applied on an iron and a copper wire of the same length. In order to have the same current in the two wires, the ratio \( r(\text{iron})/r(\text{copper}) \) of their radii must be (specific resistance of iron \( =1.0\times10^{-7} \) ohm-m, of copper \( =1.7\times10^{-8} \) ohm-m)
(a) about 1.2
(b) about 2.4
(c) about 3.6
(d) about 4.8
Show Answer & Explanation

Answer: (b) about 2.4

Explanation:
1. Same V and I means the same R, and the lengths are equal.
2. \( \dfrac{r_{Fe}}{r_{Cu}}=\sqrt{\dfrac{1.0\times10^{-7}}{1.7\times10^{-8}}}\approx2.4 \).

Teacher's Note:
1. Area is proportional to \( r^2 \).

Question 23: A rod of a certain metal is 1.0 m long and 0.6 cm in diameter. Its resistance is \( 3.0\times10^{-3} \) ohm. Another disc made of the same metal is 2.0 cm in diameter and 1.0 mm thick. The resistance between the round faces of the disc is
(a) \( 1.35\times10^{-8} \) ohm
(b) \( 2.70\times10^{-7} \) ohm
(c) \( 4.05\times10^{-6} \) ohm
(d) \( 8.10\times10^{-5} \) ohm
Show Answer & Explanation

Answer: (b) \( 2.70\times10^{-7} \) ohm

Explanation:
1. \( \rho=\dfrac{RA}{L}=\dfrac{3\times10^{-3}\times\pi(0.3\times10^{-2})^2}{1}=8.5\times10^{-8}\ \Omega\,\)m.
2. Disc: \( R=\dfrac{\rho t}{A}=\dfrac{8.5\times10^{-8}\times10^{-3}}{\pi\times10^{-4}}=2.7\times10^{-7}\ \Omega \).

Teacher's Note:
1. Current flows across the thickness between the round faces.

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