A 0.1kg steel ball falls from a height of 10m and bounces to a height of 7m. The rise in temperature of the ball is:
(Given, $C = 0.1 k c a l / k g / c^{0}$ )

{0.05}^{0} C

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{0.064}^{0} C

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{0.06}^{0} C

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{0.07}^{0} C

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Solution

The correct option is A ${0.07}^{0} C$
Change in potential energy of the ball $=$ Energy absorbed by the ball
$m g (h_{1} - h_{2}) = m c Δ T$
$Δ T = \frac{(10) (10 - 7)}{0.1 \times 4.2 \times 1000} = {0.071}^{o} C$

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A 0.1kg steel ball falls from a height of 10m and bounces to a height of 7m. The rise in temperature of the ball is:(Given, C=0.1kcal/kg/c0)

The correct option is A 0.070CChange in potential energy of the ball=Energy absorbed by the ballmg(h1−h2)=mcΔTΔT=(10)(10−7)0.1×4.2×1000=0.071oC

A 0.1kg steel ball falls from a height of 10m and bounces to a height of 7m. The rise in temperature of the ball is:
(Given, $C = 0.1 k c a l / k g / c^{0}$ )

The correct option is A ${0.07}^{0} C$
Change in potential energy of the ball $=$ Energy absorbed by the ball
$m g (h_{1} - h_{2}) = m c Δ T$
$Δ T = \frac{(10) (10 - 7)}{0.1 \times 4.2 \times 1000} = {0.071}^{o} C$