Resistance & Resistivity — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Resistance & Resistivity MCQs with step-by-step solutions (21 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Resistance & Resistivity · easy · theory
The resistance of a uniform conductor of length $L$ and area of cross-section $A$ is:
A. $\dfrac{\rho A}{L}$
B. $\dfrac{L}{\rho A}$
C. $\dfrac{\rho L}{A}$ ✓ Correct
D. $\rho L A$
Solution: Resistance grows with length and falls with thickness; $\rho$ is the resistivity of the material.
Q2 — Resistance & Resistivity · medium · theory
The resistivity of a material depends on:
A. The area of cross-section
B. The nature of the material and its temperature ✓ Correct
C. The current flowing through it
D. The length of the specimen
Solution: Resistivity is an intrinsic property. A thick short wire and a thin long wire of the same metal have the same $\rho$ but very different $R$.
Q3 — Resistance & Resistivity · easy · theory
The SI unit of resistivity is:
A. $\Omega/\text{m}$
B. $\Omega$
C. $\Omega\cdot\text{m}^2$
D. $\Omega\cdot\text{m}$ ✓ Correct
Solution: From $\rho = \dfrac{RA}{L}$, the units are $\dfrac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega\cdot\text{m}$.
Q4 — Resistance & Resistivity · easy · theory
Electrical conductivity is defined as:
A. The reciprocal of resistivity ✓ Correct
B. The current per unit potential difference
C. The product of resistance and length
D. The reciprocal of resistance
Solution: $\sigma = \dfrac{1}{\rho}$, measured in siemens per metre.
Q5 — Resistance & Resistivity · medium · theory
With rising temperature, the resistance of a metallic conductor:
A. Decreases, because more electrons are freed
B. First decreases then increases
C. Increases, because lattice vibrations scatter electrons more ✓ Correct
D. Remains unchanged
Solution: The number of free electrons in a metal is essentially fixed, so the increased scattering dominates and the resistance rises.
Q6 — Resistance & Resistivity · medium · theory
With rising temperature, the resistance of a semiconductor:
A. Becomes infinite
B. Decreases, because many more charge carriers are generated ✓ Correct
C. Increases, because of lattice vibrations
D. Remains unchanged
Solution: In a semiconductor the sharp rise in carrier concentration outweighs the increased scattering, giving a negative temperature coefficient.
Q7 — Resistance & Resistivity · medium · theory
The temperature coefficient of resistance $\alpha$ is defined by the relation:
A. $R_t = \dfrac{R_0}{1 + \alpha\Delta T}$
B. $R_t = R_0 + \Delta T$
C. $R_t = R_0(1 + \alpha\Delta T)$ ✓ Correct
D. $R_t = R_0\alpha\Delta T$
Solution: It measures the fractional change in resistance per degree rise of temperature, and is positive for metals.
Q8 — Resistance & Resistivity · medium · theory
A superconductor is a material whose resistance:
A. Increases without limit at low temperature
B. Is very large at all temperatures
C. Falls abruptly to zero below a critical temperature ✓ Correct
D. Is independent of temperature
Solution: Below the critical temperature a current once established persists indefinitely without any applied voltage.
Q9 — Resistance & Resistivity · easy · theory
Of resistance and resistivity, the quantity that depends on the dimensions of the specimen is:
A. Resistivity only
B. Resistance only ✓ Correct
C. Both of them
D. Neither of them
Solution: Cutting a wire changes its resistance but leaves the resistivity of the metal untouched.
Q10 — Resistance & Resistivity · medium · numerical
A wire is stretched so that its length is doubled, its volume remaining constant. Its resistance becomes:
A. Twice
B. Half
C. One-fourth
D. Four times ✓ Correct
Solution: At constant volume, doubling $L$ halves $A$, so $R = \dfrac{\rho L}{A}$ increases by a factor of $2 \times 2 = 4$.
Q11 — Resistance & Resistivity · hard · numerical
A wire of resistance $16\,\Omega$ is stretched uniformly to double its length at constant volume. Its new resistance is:
A. $32\,\Omega$
B. $16\,\Omega$
C. $8\,\Omega$
D. $64\,\Omega$ ✓ Correct
Solution: $R \propto L^2$ when the volume is fixed, so $R' = 4 \times 16 = 64\,\Omega$.
Q12 — Resistance & Resistivity · medium · numerical
A wire of resistivity $1.7 \times 10^{-8}\,\Omega\cdot\text{m}$, length $10\text{ m}$ and area $10^{-6}\text{ m}^2$ has resistance:
A. $1.7\,\Omega$
B. $0.017\,\Omega$
C. $0.17\,\Omega$ ✓ Correct
D. $17\,\Omega$
Solution: $R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 10}{10^{-6}} = 0.17\,\Omega$.
Q13 — Resistance & Resistivity · easy · numerical
A wire of resistance $R$ is cut into two equal halves. The resistance of each half is:
A. $\dfrac{R}{4}$
B. $2R$
C. $R$
D. $\dfrac{R}{2}$ ✓ Correct
Solution: Resistance is proportional to length at fixed cross-section, so halving the length halves the resistance.
Q14 — Resistance & Resistivity · medium · numerical
If the radius of a wire is doubled while its length is unchanged, its resistance becomes:
A. One-half
B. One-fourth ✓ Correct
C. Twice
D. Four times
Solution: Area $\propto r^2$, so doubling the radius quadruples $A$ and divides the resistance by $4$.
Q15 — Resistance & Resistivity · medium · numerical
A wire has both its length and its area of cross-section doubled. Its resistance:
A. Becomes four times
B. Doubles
C. Remains unchanged ✓ Correct
D. Halves
Solution: $R = \dfrac{\rho L}{A}$, and doubling both $L$ and $A$ leaves the ratio, and hence $R$, the same.
Q16 — Resistance & Resistivity · hard · numerical
A conductor of resistance $100\,\Omega$ at $0^\circ\text{C}$ has $\alpha = 0.004\,^\circ\text{C}^{-1}$. Its resistance at $50^\circ\text{C}$ is:
A. $110\,\Omega$
B. $140\,\Omega$
C. $120\,\Omega$ ✓ Correct
D. $102\,\Omega$
Solution: $R_t = R_0(1 + \alpha\Delta T) = 100(1 + 0.004 \times 50) = 100 \times 1.2 = 120\,\Omega$.
Q17 — Resistance & Resistivity · medium · numerical
A wire is stretched to three times its original length at constant volume. Its resistance becomes:
A. Nine times ✓ Correct
B. One-third
C. Six times
D. Three times
Solution: $R \propto L^2$ at constant volume, so $R' = 3^2R = 9R$.
Q18 — Resistance & Resistivity · hard · numerical
Two wires of the same material have lengths in the ratio $1 : 2$ and areas of cross-section in the ratio $2 : 1$. The ratio of their resistances is:
A. $1 : 2$
B. $1 : 4$ ✓ Correct
C. $1 : 1$
D. $4 : 1$
Solution: $R \propto \dfrac{L}{A}$, so the ratio is $\dfrac{1/2}{2/1} = \dfrac{0.5}{2} = \dfrac{1}{4}$.
Q19 — Resistance & Resistivity · hard · numerical
A conductor of length $4\text{ m}$ and area $10^{-6}\text{ m}^2$ has resistance $2\,\Omega$. Its resistivity is:
A. $5 \times 10^{-6}\,\Omega\cdot\text{m}$
B. $8 \times 10^{-6}\,\Omega\cdot\text{m}$
C. $2 \times 10^{-7}\,\Omega\cdot\text{m}$
D. $5 \times 10^{-7}\,\Omega\cdot\text{m}$ ✓ Correct
Solution: $\rho = \dfrac{RA}{L} = \dfrac{2 \times 10^{-6}}{4} = 5 \times 10^{-7}\,\Omega\cdot\text{m}$.
Q20 — Resistance & Resistivity · medium · numerical
A material has resistivity $2 \times 10^{-8}\,\Omega\cdot\text{m}$. Its conductivity is:
A. $2 \times 10^8\text{ S/m}$
B. $5 \times 10^{-7}\text{ S/m}$
C. $5 \times 10^7\text{ S/m}$ ✓ Correct
D. $2 \times 10^{-8}\text{ S/m}$
Solution: $\sigma = \dfrac{1}{\rho} = \dfrac{1}{2 \times 10^{-8}} = 5 \times 10^7\text{ S/m}$.
Q21 — Resistance & Resistivity · hard · numerical
A wire has resistance $50\,\Omega$ at $20^\circ\text{C}$ with $\alpha = 0.005\,^\circ\text{C}^{-1}$. Its resistance at $100^\circ\text{C}$ is:
A. $70\,\Omega$ ✓ Correct
B. $60\,\Omega$
C. $90\,\Omega$
D. $75\,\Omega$
Solution: $R = 50[1 + 0.005(100 - 20)] = 50(1 + 0.4) = 70\,\Omega$.