Physics - Rotation Question with Solution | TestHub

If the wire is replaced by another wire of same material but with double the length and half the thickness, the resistance of the wire as a whole will change. Let us calculate this change. Initial resistance, $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}}{\pi {\mathrm{r}}^2}$ Fubak resistance, ${\mathrm{R}}^{\prime }=\rho \frac{2\mathrm{L}}{{\mathrm{A}}^{\prime }}=\rho \frac{2\mathrm{L}}{\pi {\left(\frac{\mathrm{r}}{2}\right)}^2}$ $=\rho \frac{8\mathrm{L}}{\pi {\mathrm{r}}^2}=8\left(\rho \frac{\mathrm{L}}{\pi {\mathrm{r}}^2}\right)=8\mathrm{R}$ Therefore, the resistance will increase by eight times. In a meter-bridge, we have $\begin{array}{l}\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}\\ \text{ where }\mathrm{R}=\text{ unknown resistor }\\ \mathrm{S}=\text{ known resistor }\\ l=\text{ balancing length }\\ \therefore l=\frac{100\mathrm{R}}{\mathrm{S}+\mathrm{R}}\end{array}$ When the resistance of the new wire is $8\mathrm{R}$ then the new balancing length will be $\begin{array}{rl}l& =\frac{100\times 8\mathrm{R}}{\mathrm{S}+8\mathrm{R}}=\frac{100\times 8\mathrm{R}}{(\mathrm{S}+\mathrm{R})+7\mathrm{R}}\\ & =\frac{100\times 8\mathrm{R}}{\left(\frac{100\mathrm{R}}{l}\right)+7\mathrm{R}}\therefore l^{\prime }=\frac{800l}{100+7l}\end{array}$ No option is correct.