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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A ball is dropped from a height of 1000 Km .Due to heat from air friction & melts just before reaching the ground . The latent heat of fusion of ball's material is - (specific heat of material =100 $\;{J kg^{-1}}^{0}C$ Air tempeature=$\;25^{0}C\;$ , Melting point =$\;{J kg^{-1}}^{0}C\;$)

$(a)\;9.1\;MJkg^{-1}\qquad(b)\;9.3\;MJkg\qquad(c)\;9.7\;MJkg^{-1}\qquad(d)\;None\;of\;these$

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Answer : (c) $\;9.7\;MJkg^{-1}$
Explanation :
Initial energy of ball = $\;mgh=m \times 9.8 \times 1000 \times 10^{3}$
$=9.8 m \times 10^{6} J$
This energy heats up the ball to $\;1125^{0}C\;$ & meets it
$9.8 \times 10^{6}=m \times 100 \times (1125-25)+m \times L$
$9.8 \times 10^{6}=100 \times 1100 +mL$
$9.8 \times 10^{6}=0.11\times10^{6}+L$
$L=9.69 \times 10^{6} J kg^{-1}\;.$
answered Mar 12, 2014 by yamini.v
 

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