JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
Net loss in \(PE =\) Gain in \(KE\) \(12 gh -3 gh =\frac{1}{2} 3 v ^{2}+\frac{1}{2} 12 v ^{2}+\frac{1}{2}\left[12 r ^{2}\right]\left(\frac{ v }{ r }\right)^{2}\) \(9 gh =\frac{1}{2}[3+12+12] v ^{2}\) \(v ^{2}=\frac{2 gh }{3} \Rightarrow v =\frac{1}{2} \sqrt{\frac{8}{3} gh }\)…
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