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Question: A smooth chain PQ of mass M rest against a \(\frac { 1 } { 4 }\) th circular and smooth surface of ...

A smooth chain PQ of mass M rest against a 14\frac { 1 } { 4 } th circular and smooth surface of radius r. If released, its velocity to come over the horizontal part of the surface is –

A

2gr×14\sqrt { 2 \mathrm { gr } } \times \frac { 1 } { 4 }

B

2gr⁡(1−1π)\sqrt { 2 \operatorname { gr } \left( 1 - \frac { 1 } { \pi } \right) }

C

2gr⁡(1−2π)\sqrt { 2 \operatorname { gr } \left( 1 - \frac { 2 } { \pi } \right) }

D

gr⁡(1−2π)\sqrt { \operatorname { gr } \left( 1 - \frac { 2 } { \pi } \right) }

Answer

2gr⁡(1−2π)\sqrt { 2 \operatorname { gr } \left( 1 - \frac { 2 } { \pi } \right) }

Explanation

Solution

Q mg hcm = 12\frac { 1 } { 2 } mv2

Ž v = 2ghCM\sqrt { 2 \mathrm { gh } _ { \mathrm { CM } } } = 2g(r−2rπ)\sqrt { 2 g \left( r - \frac { 2 r } { \pi } \right) } = 2gr⁡(1−2π)\sqrt { 2 \operatorname { gr } \left( 1 - \frac { 2 } { \pi } \right) }