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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A bullet is fired into a box & penetrates it . The penetration process melts it . Assuming that $\;25 \%\;$ of its K.E . goes in heating & the initial and final temperature are $\;27^{0}C\;$ & $\;327^{0}C\;$, the initial velocity is - (specific heat capacity = $\;100 Jkg^{-1}K^{-1}\;$ Latent heat of fusion = $\;2 \times 10^{4} Jkg^{-1}$)

$(a)\;6.3 \times 10^{2}\;m/s\qquad(b)\;6.1 \times 10^{2}\;m/s\qquad(c)\;6.5\times 10^{2}\;m/s\qquad(d)\;6.6 \times 10^{2}\;m/s$

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Answer : (a) $\;6.3 \times 10^{2}\;m/s$
Explanation :
Let the mass of bullet be 'm' kg
Heat used in increasing temp = $\;m \times 100 \times (327-27)J$
Heat used in melting = $m \times 2 \times 10^{4}J$
Since $\;25 \%\;$ of KE goes into heat -
$\large\frac{1}{4}(\large\frac{1}{2}mv^2)=m\times 100 \times 300 + m \times 2 \times 10^{4}$
$\large\frac{v^2}{8}=3 \times 10^{4} + 2 \times 10^{4}$
$v=6.3 \times 10^{2} \;m/s\;.$
answered Mar 12, 2014 by yamini.v
 

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