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A metal box is dragged on the ground through a distance of 42m. Coefficient of friction between the box and the ground is 0.2. Then the rise in temperature of the box if specific heat of material of the box is 0.98 C.G.S units and J = 4.2 J/cal

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0.04°C

0.05°C

0.01°C

0.02°C

answer is B.

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Detailed Solution

Work done = Heat developed
$42 \times 0 .2 \times 9 .8 \times m = m \times 0 .98 \times 4200 ΔT ΔT = \frac{42 \times 0 .2 \times 9 .8}{4200 \times 0 .98} = 0 {.02}^{\circ} C$

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