MHT CET · Physics · Thermal Properties of Matter
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
- A \(50^{\circ} \mathrm{C}\)
- B \(40^{\circ} \mathrm{C}\)
- C \(35^{\circ} \mathrm{C}\)
- D \(25^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(40^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
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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