In a meter bridge experiment, the balance point is obtained at length ' l 1 ' cm from left hand when resistances in the left gap and right gap are 15 Ω and R Ω respectively. When the resis

Question

In a meter bridge experiment, the balance point is obtained at length ' l 1 ​ ' cm from left hand when resistances in the left gap and right gap are 15 Ω and R Ω respectively. When the resistance R is shunted with equal resistance the new balance point is at ( 1.6 l 1 ​ ) . The resistance R in ohm is

TestHub · Accepted Answer

For a meter bridge, the balance condition is given by $\frac{R _{1}}{R _{2}} = \frac{l}{100 - l}$ .

Initially, the resistances are $15 Ω$ and $R Ω$ , and the balance point is at $l_{1}$ .

$\frac{15}{R} = \frac{l _{1}}{100 - l _{1}}$

When $R$ is shunted with an equal resistance $R$ , the equivalent resistance in the right gap becomes $\frac{R \times R}{R + R} = \frac{R}{2}$ . The new balance point is at $1.6 l_{1}$ .

$\frac{15}{\frac{R}{2}} = \frac{1.6 l _{1}}{100 - 1.6 l _{1}}$

$\frac{30}{R} = \frac{1.6 l _{1}}{100 - 1.6 l _{1}}$

Dividing the second equation by the first equation:

$\frac{\frac{30}{R}}{\frac{15}{R}} = \frac{\frac{1.6 l _{1}}{100 - 1.6 l _{1}}}{\frac{l _{1}}{100 - l _{1}}}$

$2 = \frac{1.6 ( 100 - l _{1} )}{100 - 1.6 l _{1}}$

$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$

$1.6 l_{1} = 40 \Rightarrow l_{1} = 25$

Substituting $l_{1} = 25$ into the first equation:

$\frac{15}{R} = \frac{25}{100 - 25}$

$\frac{15}{R} = \frac{25}{75} = \frac{1}{3}$

$R = 15 \times 3 = 45 Ω$

Answer: $45$

Physics - Current Electricity Question with Solution | TestHub

Options:

Doubts & Discussion