Question

A resistance wire connected in the left gap of a meter bridge balances a $10\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3:2$. If the length of the resistance wire is $1.5\text{m}$, then the length of $1\Omega$ of the resistance wire is:

• A. $1.0\times {10}^{-2}\text{m}$
• B. $1.5\times {10}^{-2}\text{m}$
• C. $1.5\times {10}^{-1}\text{m}$
• D. $1.0\times {10}^{-1}\text{m}$

Answer 1

The correct option is D $1.0\times {10}^{-1}\text{m}$

Line diagram of a meter bridge:

Referring the figure,

$R:S=3:2$

$S=10\Omega$

Now, the ratio of resistances can be written as,

$\frac{R}{S}=\frac{3}{2}$

​$R=\frac{3}{2}\times 10$

$\Rightarrow R=15\Omega$

If the length of the resistance wire having resistance $15\Omega$ is $1.5\text{m}$, so the length of the wire having resistance $1\Omega$ will be,

$L=\frac{1.5}{15}\times 1$

$\therefore L=1\times {10}^{-1}\text{m}$

Hence, option (d) is correct answer.