Work Power Energy Question 6
Question 6 - 2024 (29 Jan Shift 2)
A bob of mass ’ $m$ ’ is suspended by a light string of length ’ $L$ ‘. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position $\mathrm{B}$. The ratio of kinetic energies $\frac{(\text { K.E. }){A}}{(\text { K.E. }){B}}$ is :
(1) $3: 2$
(2) $5: 1$
(3) $2: 5$
(4) $1: 5$
Show Answer
Answer: (2)
Solution:
Apply energy conservation between A $\backslash$ & B
$\frac{1}{2} \mathrm{mV}{\mathrm{L}}^{2}=\frac{1}{2} \mathrm{mV}{\mathrm{H}}^{2}+\mathrm{mg}(2 \mathrm{~L})$
$\because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}$
So, $\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$
$\frac{(\mathrm{K} . \mathrm{E}){\mathrm{A}}}{(\mathrm{K} . \mathrm{E}){\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^{2}}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^{2}}=\frac{5}{1}$
