A current of 0.01 A passes through the potentiometer wire of resistivity 10⁶ Ω cm and area of cross- section 10⁻² cm. The potential gradient is (in V/m) (a) 10⁹ (b) 10¹¹ (c) 10¹⁰ (d) 10⁸

Answer: (d). Potential gradient k = IR/L = (Iρ)/(A) (area taken as 10⁻² cm²). ρ/A = 10⁶ Ω cm10⁻² cm² = 10⁸ Ω/cm = 10¹⁰ Ω/m. k = 0.01 × 10¹⁰ = 10⁸ V/m. So the answer is (d) 10⁸.

A current of 0.01 A passes through the potentiometer wire of… | MHT CET 2026 (13 May, Shift 2)

Q2MHT CET 2026 · 13 MAY · SHIFT 2Easy

Potentiometer

A current of $0.01$ A passes through the potentiometer wire of resistivity $10^{6}\ \Omega$ cm and area of cross- section $10^{-2}$ cm. The potential gradient is (in V/m)

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Answer: (d)

Potential gradient $k = \dfrac{IR}{L} = \dfrac{I\rho}{A}$ (area taken as $10^{-2}\text{ cm}^2$ ).
$\dfrac{\rho}{A} = \dfrac{10^{6}\ \Omega\text{ cm}}{10^{-2}\text{ cm}^2} = 10^{8}\ \Omega/\text{cm} = 10^{10}\ \Omega/\text{m}$ .
$k = 0.01 \times 10^{10} = 10^{8}\text{ V/m}$ .
So the answer is (d) $10^{8}$ .

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Potentiometer question from MHT CET 2026 (13 May, Shift 2)

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