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The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 \Omega is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. The internal resistance of the cell is:

A

1.68 \Omega

B

0.13 \Omega

C

0.31 \Omega

D

1.12 \Omega

Step-by-Step Solution

The internal resistance rr of a cell determined using a potentiometer is given by the formula r=R(l1l21)r = R \left( \frac{l_1}{l_2} - 1 \right), where l1l_1 is the balancing length in open circuit, l2l_2 is the balancing length when external resistance RR is connected, and RR is the external resistance.

Given: l1=76.3 cml_1 = 76.3 \text{ cm} l2=64.8 cml_2 = 64.8 \text{ cm} R=9.5 ΩR = 9.5 \ \Omega

Substituting these values: r=9.5×(76.364.81)r = 9.5 \times \left( \frac{76.3}{64.8} - 1 \right) r=9.5×(1.17751)r = 9.5 \times (1.1775 - 1) r=9.5×0.17751.686 Ωr = 9.5 \times 0.1775 \approx 1.686 \ \Omega.

Rounding to two decimal places gives 1.69 \Omega , but 1.68 \Omega is the closest valid option provided.

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