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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2013 · Paper 1 · Physics · Laws of Motion

A bob of mass m , suspended by a string of length l _ 1 , is given...

A bob of mass mm, suspended by a string of length l1{l}_{1}, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass mm, suspended by a string of length l2{l}_{2}, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l1l2\dfrac{{l}_{1}}{{l}_{2}} isA bob of mass m , suspended by a string of length l _ 1 , is given a minimum velocity required to complete a fullA bob of mass m , suspended by a string of length l _ 1 , is given a minimum velocity required to complete a full

Correct answer

5

Step-by-step solution

A bob of mass m , suspended by a string of length l _ 1 , is given a minimum velocity required to complete a full
As shown in the figure shown above, the minimum speed required by the bob of mass mm attached to the string of length l1{l}_{1} in order to complete the full circle is 5gl1\sqrt{5g{l}_{1}}.
A bob of mass m , suspended by a string of length l _ 1 , is given a minimum velocity required to complete a full
When the bob goes to the highest point, its speed becomes gl1\sqrt{g{l}_{1}}. Here, it strikes with the identical mass, which is tied to the string of length l2{l}_{2}. The speed delivered to the object is gl1\sqrt{g{l}_{1}}. This should be equal to 5gl2\sqrt{5g{l}_{2}} so that this bob could complete a full circle.
gl1=5gl2\sqrt{g{l}_{1}} = \sqrt{5g{l}_{2}}
⇒gl1=5gl2⇒ g{l}_{1} = 5g{l}_{2}
⇒l1l2=5⇒ \dfrac{{l}_{1}}{{l}_{2}} = 5

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