MHT CET · Physics · Thermal Properties of Matter
A metal rod cools at the rate of \(4{ }^{\circ} \mathrm{C} / \mathrm{min}\) when its temperature is \(90^{\circ} \mathrm{C}\) and the rate of \(1{ }^{\circ} \mathrm{C} / \mathrm{min}\) when its temperature is \(30^{\circ} \mathrm{C}\). The temperature of the surrounding is
- A \(20^{\circ} \mathrm{C}\)
- B \(15^{\circ} \mathrm{C}\)
- C \(10^{\circ} \mathrm{C}\)
- D \(5^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(10^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
According to Newton's law of cooling,
\(\begin{array}{ll}
& \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\theta_1-\theta_0}{\theta_2-\theta_0} \\
\therefore \quad & \frac{4}{1}=\frac{90-\theta_0}{30-\theta_0} \\
\therefore \quad & 120-90=4 \theta_0-\theta_0 \\
\therefore \quad & 3 \theta_0=30
\end{array}\)
\(\therefore \quad\) Temperature of the surroundings is,
\(\theta_0=10^{\circ} \mathrm{C}\)
\(\begin{array}{ll}
& \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\theta_1-\theta_0}{\theta_2-\theta_0} \\
\therefore \quad & \frac{4}{1}=\frac{90-\theta_0}{30-\theta_0} \\
\therefore \quad & 120-90=4 \theta_0-\theta_0 \\
\therefore \quad & 3 \theta_0=30
\end{array}\)
\(\therefore \quad\) Temperature of the surroundings is,
\(\theta_0=10^{\circ} \mathrm{C}\)
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