Question

A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length $l$. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be

• A. $\frac{1}{8\ell }$
• B. $\frac{1}{4\ell }$
• C. $8\ell$
• D. $16\ell$

Answer 1

Answer: (C)

In a meter bridge the ratio of two resistances is $\frac{R}{R^{{}^{\prime }}}=\frac{l}{l^{{}^{\prime }}}$
where $l$ and $l^{{}^{\prime }}$ are balancing lengths Resistance $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$
If material remains same $\rho ={\rho }^{{}^{\prime }}$
Given $l^{{}^{\prime }}=2l$
$r^{{}^{\prime }}=\frac{r}{2}$
$\therefore R^{{}^{\prime }}=\frac{\rho l^{{}^{\prime }}}{A^{{}^{\prime }}}$
$=\frac{\rho 2l}{\pi {\left(\frac{r}{2}\right)}^2}$
$=\frac{8\rho l}{\pi r^2}$
$R^{{}^{\prime }}=8R$
Therefore, the new balancing point is expected to be $8l$.