Question

$n$ identical resistance are taken in which $\frac{n}{2}$ resistors are joined in series in the left gap and the remaining $\frac{n}{2}$ resistances are joined in parallel in the right gap of a metre bridge. Balancing length in $cm$ is

• A. $100\cdot \frac{n^2}{n^2+4}$
• B. $100\cdot \frac{n^2}{n^2+1}$
• C. $400\cdot \frac{1}{n^2+4}$
• D. $400\cdot \frac{1}{n^2+1}$

Answer 1

Answer: (A)

Meter bridge is shown in the figure below,

When $\frac{n}{2}$ resistances are joined in series in left gap
each of resistance $R_1$, then equivalent resistance in left gap.
$R=\frac{R_1}{2}+\frac{R_1}{2}+\frac{R_1}{2}+\dots \frac{n}{2}.\text{ times }=\frac{R_{1}n}{2}$
When $\frac{n}{2}$ resistors are joined in parallel in right
gap, then the equivalent resistance in right gap.
$\frac{1}{S}=\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\dots \cdot \frac{n}{2}\text{ times }=\frac{n}{2R_1}$
$\Rightarrow S=\frac{2R_1}{n}$
If $l$ be the balancing length in the meter bridge wire, then
$\frac{R}{S}=\frac{l}{100-l}\Rightarrow \frac{\frac{R_{1}n}{2}}{\frac{2R_1}{n}}=\frac{l}{100-l}$
$\frac{n^2}{4}=\frac{l}{100-l}\Rightarrow l=\frac{100n^2}{n^2+4}$