Question

A conductor at ${10}^{\circ }$C connected in the left gap of metre bridge gives balancing length 40 cm. When the conductor is at ${60}^{\circ }$C, the balance point shifts by 5cm. The temperature coefficient of resistance of the material of the conductor is

• A. $\frac{1}{{210}^{\circ }C}$
• B. $\frac{1}{{220}^{\circ }C}$
• C. $\frac{1}{{200}^{\circ }C}$
• D. $\frac{1}{{201}^{\circ }C}$

Answer 1

The correct option is A $\frac{1}{{210}^{\circ }C}$

$\frac{R_1}{R}=\frac{l}{100-l}=\frac{40}{60}=\frac{2}{3}{R}_1=\frac{2}{3}R\frac{R_2}{R}=\frac{45}{100-45}=\frac{45}{55}=\frac{9}{11}{R}_2=\frac{9}{11}R{R}_1=R_0[1+\alpha t_1];R_2=R_0[1+\alpha t_2]\frac{2}{3}R=R_0[1+10\alpha ]\to \left(1\right)\frac{9}{11}R=R_0[1+60\alpha ]\to \left(2\right)\frac{\left(1\right)}{\left(2\right)}\frac{2}{3}\times \frac{11}{9}=\frac{1+10\alpha }{1+60\alpha }$
By showing
$\alpha =\frac{1}{210}{C}^{-1}$