$(a)\;2 \times 10^{3} \;cals^{-1}\qquad(b)\;3 \times 10^{3} \;cals^{-1}\qquad(c)\;4 \times 10^{3} \;cals^{-1}\qquad(d)\;5 \times 10^{3} \;cals^{-1}$

Thermodynamics

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Answer : (b) $\;3 \times 10^{3} \;cals^{-1}$

Explanation :

K.E lost = $\;\large\frac{1}{2} \times 1000 \times (36 \times \large\frac{5}{18})^2 J$

$=5 \times 10^{4}J$

$50 \%\;$ of KE lost in 2 secs

rate of production in $\;cal\;s^{-1}$

$=\large\frac{50}{100}\times \large\frac{5 \times 10^{4}}{2} \times \large\frac{1}{4.18}$

$=2.99 \times 10^{3}\;cal s^{-1}\;.$

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