Question

A resistance of 2 W is connected across one gap of a meter–bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 W, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20cm. Neglecting any correction, the unknown resistance is –

• A. 3 W
• B. 4 W
• C. 5 W
• D. 6 W

Answer 1

Answer: (A)

3 W

Answer 2

$\frac { \ell } { ( 100 - \ell ) }$= ….(1)

$\frac { \ell + 20 } { 100 - ( \ell + 20 ) }$=

$\frac { \ell + 20 } { 80 - \ell }$= …..(2)

by (1) × (2)

$\left( \frac { \ell } { 100 - \ell } \right)$ × $\left( \frac { \ell + 20 } { 80 - \ell } \right)$ = 1

̃ l² + 20 l = 8000 – 180 l + l²

̃ 200 l = 8000

l = 40

So S = $\frac { 2 \times 60 } { 40 }$ = 3 W