Question

A copper wire of resistance $R$ is cut in to ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance

• A. $R$
• B. $\frac{R}{4}$
• C. $\frac{R}{5}$
• D. $\frac{R}{25}$

Answer 1

The correct option is D $\frac{R}{25}$

Resistance of metallic wire$R$

When cut into $10$ equal parts, each part will have resistance $=\frac{R}{10}$

When two pieces are joined in series, then $R^{\prime }=\frac{R}{10}+\frac{R}{10}=\frac{2R}{10}=\frac{R}{5}$

Now each such combinations are joined in parallel then

$\frac{1}{R^{\prime \prime }}=\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}=\frac{5}{\frac{R}{5}}=\frac{25}{R}$

$\Rightarrow R^{\prime \prime }=\frac{R}{25}\Omega$