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.
$R=\frac{R_{1}}{2}+\frac{R_{1}}{2}+\frac{R_{1}}{2}+\ldots \frac{n}{2} . \text { times }=\frac{R_{1} n}{2}$
When $\frac{n}{2}$ resistors are joined in parallel in right
gap, then the equivalent resistance in right gap.
$\frac{1}{S}=\frac{1}{R_{1}}+\frac{1}{R_{1}}+\frac{1}{R_{1}}+\ldots \cdot \frac{n}{2} \text { times }=\frac{n}{2 R_{1}}$
$\Rightarrow S=\frac{2 R_{1}}{n}$
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{2 R_{1}}{n}}=\frac{l}{100-l} $
$\frac{n^{2}}{4}=\frac{l}{100-l} \Rightarrow l=\frac{100 \,n^{2}}{n^{2}+4}$