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Thermodynamics
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A ball is dropped from a height of $10\; m$ & it rises to a height of $5\; m$ . Assuming $\;50 \%\;$ of the energy goes into increasing temperature of ball , the temperature rise is (specific heat of ball's material = $\;1000 JK^{-1}$)

$(a)\;10^{-3} \times 100^{\circ}C\qquad(b)\;10^{-3} \times 50^{\circ}C\qquad(c)\;{10^{-3}}^{\circ}C\qquad(d)\;10^{-2} \times 50^{\circ}C$

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Answer : $\;10^{-3} \times 50^{0}C$
Explanation :
Energy lost =mg (10-5)
Since $\;50\%\;$ of the energy lost = $\;mC \bigtriangleup T$
$mg\times5=m \times 10^{3} \times \bigtriangleup T$
$\bigtriangleup T=50 \times {10^{-3}}^{0}C$
answered Mar 12, 2014 by
edited Mar 25, 2014 by yamini.v

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