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A small ball of mass m is dropped from an aeroplane moving at 50 m height above the ground with a speed of $25 \sqrt{2}$ meter/sec. If half of the mechanical energy of ball with respect to ground is lost as a thermal energy due to air friction. The change in the temperature of the ball as it lands on the ground is. Specific heat capacity of ball is 56.25 J/kg. (g = 10 $m / \sec^{2}$ )

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$5^{0} C$

$10^{0} C$

$20^{0} C$

$500 / 56 . 25^{0} C$

answer is B.

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Detailed Solution

$\frac{1}{2} (M \times g \times h + \frac{1}{2} {MV}^{2}) = Ms ∆ T$

$\frac{1}{2} (10 \times 50 + \frac{1}{2} \times {(25 \sqrt{2})}^{2}) = 56.25 \times ∆ T$

$∆ T = \frac{500 + 625}{56.25 \times 2} = 10^{0} C$

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