Question

The resistance in the two arms of the meter bridge are $5\Omega$ and $R\Omega$, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at $1.6l_1$ . The resistance 'R' is

• A. $10\Omega$
• B. $15\Omega$
• C. $20\Omega$
• D. $25\Omega$

Answer 1

The correct option is B $15\Omega$

$\text{Step 1: Initial Balance condition for meter bridge[Refer Fig. in Question]}$

As shown in the figure in Question, initial balancing length is $l_1$

So Applying balancing condition of meter bridge:

$\frac{P}{S}=\frac{{\ell }_1}{100-{\ell }_1}$

$\Rightarrow \frac{5}{R}=\frac{{\ell }_1}{100-{\ell }_1}$ $....\left(1\right)$

$\text{Step 2: New balance condition after shunting R with equal resistance R}$ $\text{[Ref. Fig.]}$

$S^{\prime }=\frac{R\times R}{R+R}=\frac{R}{2}$

Let the new balancing length be $L=1.6{\ell }_1$

So we have:

$\frac{5}{S^{\prime }}=\frac{L}{100-L}$

$\Rightarrow \frac{5}{R/2}=\frac{10}{R}=\frac{1.6{\ell }_1}{100-1.6{\ell }_1}$ $....\left(2\right)$

$\text{Step 3: Solving equations}$

Divide equation $\left(1\right)$ by $\left(2\right)$

$\frac{1}{2}=\frac{{\ell }_1(100-1.6{\ell }_1)}{(100-{\ell }_1)\times 1.6{\ell }_1}$

$\Rightarrow 0.8=\frac{100-1.6{\ell }_1}{100-{\ell }_1}$

$\Rightarrow 80-0.8{\ell }_1=100-1.6{\ell }_1$

$\Rightarrow {\ell }_1=25cm$

Put the value of ${\ell }_1$ in eq $\left(1\right)$

$\frac{5}{R}=\frac{25}{75}$ $\Rightarrow R=15\Omega$

Hence Option $B$ is correct.