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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
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Verified Answer
The correct answer is:
$40^{\circ} \mathrm{C}$
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:
$$
\begin{aligned}
& \mathrm{T}_1=(40 \times 0.75)+10 \\
& \mathrm{~T}_1=40^{\circ} \mathrm{C}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{T}_{\mathrm{g}}=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{x}} \\
& \Rightarrow \frac{\mathrm{T}_1-10}{0 \cdot 75}=40
\end{aligned}
$$
$\therefore \quad$ Temperature of the hotter end is:
$$
\begin{aligned}
& \mathrm{T}_1=(40 \times 0.75)+10 \\
& \mathrm{~T}_1=40^{\circ} \mathrm{C}
\end{aligned}
$$
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