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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 $=10 \mathrm{~ms}^{-2}$ )
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(Specific heat capacity of steel $=420 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ and Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$0.0152^{\circ} \mathrm{C}$
Work done due to sliding down the block = heat gain by block of steel
$\begin{aligned} & \therefore \Delta \mathrm{T}=\frac{\mathrm{g} \cos \theta(\mathrm{s})}{\mathrm{C}}=\frac{10 \times \frac{4}{5} \times \frac{80}{100}}{420} \\ & =\frac{10 \times 4 \times 80}{5 \times 100 \times 420}=\frac{16}{5 \times 210}=0.0152^{\circ} \mathrm{C}\end{aligned}$
$\begin{aligned} & \therefore \Delta \mathrm{T}=\frac{\mathrm{g} \cos \theta(\mathrm{s})}{\mathrm{C}}=\frac{10 \times \frac{4}{5} \times \frac{80}{100}}{420} \\ & =\frac{10 \times 4 \times 80}{5 \times 100 \times 420}=\frac{16}{5 \times 210}=0.0152^{\circ} \mathrm{C}\end{aligned}$
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