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5.Work, Energy, Power and Collision
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If momentum of a body is increased by $20 \%$, then its kinetic energy increases by$....\%$
A
$36$
B
$40$
C
$44$
D
$48$
(JEE MAIN-2022)
Solution
$P ^{\prime}= P +\frac{20}{100} P =1.2 P$
$\%$ change in $KE =\frac{ K ^{\prime}- K }{ K } \times 100$
$=\left(\frac{\frac{ P ^{2}}{2 m }-\frac{ P ^{2}}{2 m }}{\frac{ P ^{2}}{2 m }}\right) \times 100$
$= {\left[(1.2)^{2}-1\right] \times 100 }$
$= 44\%$
Standard 11
Physics
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