A copper block of mass 1 kg slides down on a rough inclined plane of inclination 37° at a constant speed. Find the increase in the temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (Specific heat of copper= 420 J kg^-1K^-1, g = 10ms^-2)

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7.6 × 10^-3°C

8.6 × 10^-3°C

6.6 × 10^-3°C

9.6 × 10^-3°C

answer is C.

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Detailed Solution

$\begin{array}{l} m = 1 kg, θ = 37^{\circ}; s = 60 \times 10^{- 2} m = 0 .6 \\ l. (mgsin θ) = Jm s Δθ \\ \begin{aligned} Δθ = \frac{(gsin 37^{0}) 0 .6}{Js} \\ Δθ & = \frac{10 \times 3 \times 0 .6}{5 \times 420} = \frac{0 .3}{3 .5} f = mgsin θ \\ 0 {.008571}^{\circ} C = 8 .6 \times 10^{- 3} C \end{aligned} \end{array}$ Question Image

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