MHT CET · Physics · Thermal Properties of Matter
A conducting rod of length \(1 \mathrm{~m}\) has area of cross-section \(10^{-3} \mathrm{~m}^2\). One end is immersed in boiling water \(\left(100^{\circ} \mathrm{C}\right)\) and the other end in Ice \(\left(0^{\circ} \mathrm{C}\right)\). If coefficient of thermal conductivity of rod is \(96 \mathrm{cal} / \mathrm{s} . \mathrm{m}^{\circ} \mathrm{C}\) and latent heat for ice is \(8 \times 10^4 \mathrm{cal} / \mathrm{kg}\) then the amount of ice which will melt in one minute is
- A \(5.4 \times 10^{-3} \mathrm{~kg}\)
- B \(7.2 \times 10^{-3} \mathrm{~kg}\)
- C \(1.8 \times 10^{-3} \mathrm{~kg}\)
- D \(3.6 \times 10^{-3} \mathrm{~kg}\)
Answer & Solution
Correct Answer
(B) \(7.2 \times 10^{-3} \mathrm{~kg}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{Q}=\frac{\mathrm{KA}(\Delta \theta) \mathrm{t}}{\ell}=\mathrm{mL} \\ & \mathrm{m}=\frac{\mathrm{KA}(\Delta \theta) \mathrm{t}}{\ell \mathrm{L}}=\frac{96 \times 10^{-3} \times 100 \times 60}{1 \times 8 \times 10^4} \\ & =7.2 \times 10^{-3} \mathrm{~kg}\end{aligned}\)
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