AP EAMCET · PHYSICS · Thermal Properties of Matter
A block of steel of mass \(2 \mathrm{~kg}\) slides down a rough inclined plane of inclination of \(\sin ^{-1}\left(\frac{3}{5}\right)\) at a constant speed. The temperature of the block as it slides through \(80 \mathrm{~cm}\), assuming that the mechanical energy lost is used to increase the temperature of the block is nearly
(Specific heat capacity of steel \(=420 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\) and Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(0.0190^{\circ} \mathrm{C}\)
- B \(0.0114^{\circ} \mathrm{C}\)
- C \(0.0152^{\circ} \mathrm{C}\)
- D \(0.0952^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(0.0152^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Work done due to sliding down the block = heat gain by block of steel…
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