Q188 / 200 · Manipal (MET) 2014 · Physics

Consider the cylindrical element as shown in the figure. Current flowing the through element is $I$ and resistivity of material of the cylinder is $\rho$. Choose the correct option out the following.

A
Power loss in first half is four times the power loss in second half
Correct answer
B
Voltage drop in first half is twice of voltage drop in second half
C
Current density in both halves are equal
D
Electric field in both halves is equal

Solution

$\begin{aligned} \frac{R_1}{R_2}=\frac{A_1}{A_2}= & \frac{4}{1} \\ & \frac{P_1}{P_2}=\frac{I^2 R_1}{I^2 R_2}=\frac{4}{1} \Rightarrow \frac{V_1}{V_2}=\frac{I R_1}{I R_2}=\frac{4}{1} \\ & \frac{J_1}{J_2}=\frac{1}{4}\end{aligned}$

Hence, the correct answer is option: 'B'.

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