NEET Current Electricity MCQs & PYQs

Practice NEET Current Electricity Questions

Question 151:

easy

Two solid conductors are made up of same material have same length and same resistance. One of them has a circular cross section of area $A_{1}$ and the other one has a square cross section of area $A_{2}$. The ratio $A_{1}/A_{2}$ is

(2020-Covid)

Resistance is given by $R = \rho \frac{l}{A}$. Since the material (hence $\rho$), length $l$, and resistance $R$ are the same for both conductors, their cross-sectional areas must be equal. Therefore, $A_{1} = A_{2}$ and the ratio is $1$.

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Question 152:

easy

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:

(2017-Delhi)

The volume of the wire remains constant during stretching. Resistance $R = \rho \frac{l}{A} = \rho \frac{l^{2}}{V}$. If the length becomes $nl$, the new resistance is $R' = \rho \frac{(nl)^{2}}{V} = n^{2}R$.

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Question 153:

easy

A wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire would be:

(2013)

When a wire is stretched, its new resistance becomes $R' = n^{2}R$. Given $n=2$ and $R=4 \Omega$, we have $R' = (2)^{2} \times 4 = 16 \Omega$.

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Question 154:

easy

A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively:

(2008)

Specific resistance is a property of the material and remains the same. When length increases by $10\%$, $l' = 1.1l$. Since $R \propto l^{2}$ (volume constant), new resistance $R' = (1.1)^{2}R = 1.21R$.

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Question 155:

easy

The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then:

(2004)

Specific resistance depends only on the material, so it remains unchanged. Resistance $R = \rho \frac{l}{\pi r^{2}}$. If $l \rightarrow 2l$ and $r \rightarrow 2r$, $R' = \rho \frac{2l}{\pi (2r)^{2}} = \frac{1}{2} \rho \frac{l}{\pi r^{2}} = \frac{R}{2}$.

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Question 156:

easy

A wire $50 \text{ cm}$ long and $1 \text{ mm}^2$ in cross-section carries a current of $4 \text{ A}$ when connected to a $2 \text{ V}$ battery. The resistivity of the wire is:

(1994)

$R = \frac{V}{I} = \frac{2}{4} = 0.5 \Omega$. Using $R = \rho \frac{l}{A}$, we have $\rho = \frac{RA}{l}$. $\rho = \frac{0.5 \times 1 \times 10^{-6}}{0.5} = 1 \times 10^{-6} \Omega \text{ m}$.

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Question 157:

easy

Identify the set in which all the three materials are good conductors of electricity

(1994)

Copper (Cu), Silver (Ag), and Gold (Au) are transition metals. They have an abundance of free electrons available for conduction, making them all excellent conductors of electricity.

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Question 158:

easy

The masses of the wires of copper is in the ratio of $1 : 3 : 5$ and their lengths are in the ratio of $5 : 3 : 1$. The ratio of their electrical resistance is:

(1988)

Resistance $R = \rho \frac{l}{A} = \rho \frac{l^2}{V} = \rho \frac{l^2 d}{m}$. Since the material is the same, $R \propto \frac{l^2}{m}$. The ratio is $R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5} = 25 : 3 : \frac{1}{5} = 125 : 15 : 1$.

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Question 159:

easy

A copper wire of length $10 \text{ m}$ and radius $(10^{-2}/\sqrt{\pi}) \text{ m}$ has electrical resistance of $10 \Omega$. The current density in the wire for an electric field strength of $10 \text{ V/m}$ is:

(2022)

Current density $J = \sigma E = \frac{E}{\rho}$. Since $R = \rho \frac{l}{A}$, $\rho = \frac{RA}{l}$. Thus, $J = \frac{E l}{RA} = \frac{10 \times 10}{10 \times \pi (10^{-2}/\sqrt{\pi})^2} = \frac{100}{10 \times 10^{-4}} = 10^5 \text{ A/m}^2$.

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Question 160:

easy

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.

$$
\begin{array}{ll@{\hspace{2cm}}ll}
\textbf{Column-I} & & \textbf{Column-II} & \\[6pt]
\text{(A)} & \text{Drift Velocity} & \text{(P)} & \dfrac{m}{ne^2\rho} \\[8pt]
\text{(B)} & \text{Electrical Resistivity} & \text{(Q)} & n e v_d \\[8pt]
\text{(C)} & \text{Relaxation Period} & \text{(R)} & \dfrac{eE}{m}\tau \\[8pt]
\text{(D)} & \text{Current Density} & \text{(S)} & \dfrac{E}{J}
\end{array}
$$

2021

Drift velocity $v_d = \frac{eE}{m}\tau$ (A-R). Electrical Resistivity $\rho = \frac{E}{J}$ (B-S). Relaxation Period $\tau = \frac{m}{ne^2\rho}$ (C-P). Current density $J = nev_d$ (D-Q). Hence, the correct matching is (A)-(R), (B)-(S), (C)-(P), (D)-(Q).

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