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

Thermodynamics

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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 -

$\large\frac{1}{4}(\large\frac{1}{2}mv^2)=m\times 100 \times 300 + m \times 2 \times 10^{4}$

$\large\frac{v^2}{8}=3 \times 10^{4} + 2 \times 10^{4}$

$v=6.3 \times 10^{2} \;m/s\;.$

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