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Q.
When a current is passed in a conductor, 5 $^{\circ}$C rise in temperature is observed. If the strength of current is made thrice, rise in temperature will be
Joule heating in a wire,
H = I2 R t
For given R and t, $H \infty I^{2}$
$\therefore \frac{H_{1}}{H_{2}}\frac{I^{2}_{1}}{I^{2}_{2}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \ldots\left(i\right)$
Also, $H = m s \Delta T$
For given m and s, $H \infty \Delta T$
$\therefore \frac{H_{1}}{H_{2}}=\frac{\Delta T_{1}}{\Delta T_{2}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \ldots\left(ii\right)$
For eqn. $\left(i\right)$ and $\left(ii\right)$
$\frac{I_{1}^{2}}{I_{2}^{2}}=\frac{\Delta T_{1}}{\Delta T_{2}}$
Here,$ I_{1}=I, \Delta T_{1}=5 ^{\circ}C, I_{2}=3I$
$\Delta T_{2}=?$
$\therefore \left(\frac{I}{3I}\right)^{2}=\frac{5}{\Delta T_{2}} $Or $\Delta T_{2}=45^{\circ}C$