The temperature gradient in a rod of length $75 \mathrm{~cm}$ is $40^{\circ} \mathrm{C} / \mathrm{m}$. If…
The temperature gradient in a rod of length $75 \mathrm{~cm}$ is $40^{\circ} \mathrm{C} / \mathrm{m}$. If the temperature of cooler end of the rod is $10^{\circ} \mathrm{C}$, then the temperature of hotter end is
$50^{\circ} \mathrm{C}$
$40^{\circ} \mathrm{C}$
$35^{\circ} \mathrm{C}$
$25^{\circ} \mathrm{C}$
Solution
We know
$\begin{aligned}
& \mathrm{T}_{\mathrm{g}}=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{x}} \\
& \Rightarrow \frac{\mathrm{T}_1-10}{0 \cdot 75}=40
\end{aligned}$
$\therefore \quad$ Temperature of the hotter end is:
$\begin{aligned}
& \mathrm{T}_1=(40 \times 0.75)+10 \\
& \mathrm{~T}_1=40^{\circ} \mathrm{C}
\end{aligned}$