A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 10 Ω resistor: The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of X is

Question

A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 10 Ω resistor: The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of X is (A) 10 Ω (B) 10.6 Ω (C) 10.8 Ω (D) 11.1 Ω

Tardigrade Admin · Accepted Answer

Using the concept of balanced Wheat stone bridge, we have, (P/Q) = (R/S) ∴ (X/(52 + 1)) = (10/(48 + 2)) ∴ X = (10 × 53/50) = 10.6 Ω

$10Ω$

$10.6Ω$

$10.8Ω$

$11.1Ω$

Using the concept of balanced Wheat stone bridge, we have,
$\frac{P}{Q} = \frac{R}{S}$
$∴ \frac{X}{( 52 + 1 )} = \frac{10}{( 48 + 2 )}$
$∴ X = \frac{10 \times 53}{50} = 10.6Ω$

Q. A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 10Ω resistor: The galvanometer shows null point when tapping-key is at 52cm mark. The end-corrections are 1cm and 2cm respectively for the ends A and B. The determined value of X is

10Ω

10.6Ω

10.8Ω

11.1Ω

Using the concept of balanced Wheat stone bridge, we have, QP​=SR​ ∴(52+1)X​=(48+2)10​ ∴X=5010×53​=10.6Ω

$10Ω$

$10.6Ω$

$10.8Ω$

$11.1Ω$

Using the concept of balanced Wheat stone bridge, we have,
$\frac{P}{Q} = \frac{R}{S}$
$∴ \frac{X}{( 52 + 1 )} = \frac{10}{( 48 + 2 )}$
$∴ X = \frac{10 \times 53}{50} = 10.6Ω$