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6.System of Particles and Rotational Motion
hard
A rolling wheel of $12 \,kg$ is on an inclined plane at position $P$ and connected to a mass of $3 \,kg$ through a string of fixed length and pulley as shown in figure. Consider $PR$ as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom $Q$ of the inclined plane $P Q$ will be $\frac{1}{2} \sqrt{ xgh } \,m / s$. The value of $x$ is.............
A
$5$
B
$6$
C
$1$
D
$3$
(JEE MAIN-2022)
Solution
Net loss in $PE =$ Gain in $KE$
$12 gh -3 gh =\frac{1}{2} 3 v ^{2}+\frac{1}{2} 12 v ^{2}+\frac{1}{2}\left[12 r ^{2}\right]\left(\frac{ v }{ r }\right)^{2}$
$9 gh =\frac{1}{2}[3+12+12] v ^{2}$
$v ^{2}=\frac{2 gh }{3} \Rightarrow v =\frac{1}{2} \sqrt{\frac{8}{3} gh }$
$x =\frac{8}{3} \simeq 3$
Standard 11
Physics
