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A block of steel of mass 2 kg slides down a rough inclined plane of inclination of $\sin^{- 1} (\frac{3}{5})$ at a constant speed. The increase in temperature of the block as it slides through 80 cm, assuming that the mechanical energy lost is used to increase the temperature of the block is nearly (Specific heat capacity of steel $= 420 {Jkg}^{- 1} K^{- 1}$ and Acceleration due to gravity $= 10 {ms}^{- 2}$ )

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${0.0190}^{\circ} C$

${0.0152}^{\circ} C$

${0.0952}^{\circ} C$

${0.0114}^{\circ} C$

answer is B.

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Detailed Solution

Acceleration of block is zero.

$m g \sin θ = f_{k}$

Energy loss due to friction = Gain in heat energy of block

$f_{k} d = m s △ θ$

$\Rightarrow (2) (10) (\frac{3}{5}) (0.8) = (2) (420) △ θ \Rightarrow △ θ = 0.0114 {}^{o}C$

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