Question

In a metre bridge set up with null deflection in galvanometer, find $\alpha$ of material of wire.

• A. $4.2\times {10}^{-5}\text{per}{}^{\circ }\text{C}$
• B. $3.4\times {10}^{-5}\text{per}{}^{\circ }\text{C}$
• C. $4.6\times {10}^{-3}\text{per}{}^{\circ }\text{C}$
• D. $3.2\times {10}^{-3}\text{per}{}^{\circ }\text{C}$

Answer 1

The correct option is C $4.6\times {10}^{-3}\text{per}{}^{\circ }\text{C}$

From meter bridge set up with null deflection we can say at $10{}^{o}C$,
$\frac{40}{R}=\frac{20}{80}$
$R=\frac{\text{/}40\times 80}{\text{/}20}=160\Omega$
And also at $40{}^{o}C$,
$\frac{40}{R^{{}^{\prime }}}=\frac{18}{82}$
$R^{{}^{\prime }}=\frac{82\times 40}{18}\approx 182\Omega$
Now we know that
$R^{\prime }=R(1+\alpha \Delta T)$
where $\Delta T$ is the temperature difference and is given by
$\Delta T=40-10=30{}^{o}C$
$182=160\{1+30\alpha \}$
$1.14=1+30\alpha$
$30\alpha =0.14$
$\alpha =\frac{0.14}{30}=4.6\times {10}^{-3}\text{per}{}^{o}C$