Q149 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Physics
A fuse wire of radius $0.2 \mathrm{~mm}$ blows off with a current of $5 \mathrm{~A}$. The fuse wire of same material but of radius $0.3 \mathrm{~mm}$ will blow off with a current of
- A$\frac{15}{2} \mathrm{~A}$
- B$\frac{5 \sqrt{3}}{2} \mathrm{~A}$
- Correct answerC$5 \sqrt{\frac{27}{8}} \mathrm{~A}$
- D5 A
Solution
If $I=$ current rating of fuse wire and $r=$ radius of fuse wire, then relation, $I^2 \propto r^3$
$$
\begin{aligned}
& \frac{I_2}{I_1}=\left(\frac{r_2}{r_1}\right)^{3 / 2}, \text { given } r_1=0.2 \mathrm{~mm}, I_1=5 \mathrm{~A} \\
& r_2=0.3 \mathrm{~mm}, I_2=? \\
& I_2=5 \times\left(\frac{0.3}{0.2}\right)^{3 / 2}=5 \sqrt{\frac{27}{8}} \mathrm{~A}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.