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Question

A bob of mass m is suspended by a light string to length L. It is imparted a horizontal velocity v0 at the lowest point A such that is completes a circle in the vertical plane.
Match the Column I with Column II.

Column IColumn II
(A)Velocity v0 is(p)3
(B)Velocity at point B is(q)qL
(C)Velocity of point C is(r)5gL
(D)Ratio of kinetic energy at B and C is(s)3gL

941071_58aa1e3346c94b4fbe0bb87e573ff0f1.png

A
Ap,Bq,Cs,Dr
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B
Aq,Br,Cp,Ds
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C
Ar,Bs,Cq,Dp
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D
As,Bp,Cr,Dq
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Solution

The correct option is A Ar,Bs,Cq,Dp
There are two external forces on the bob: gravity (mg( and tension (T) in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy (E) of the system is conserved.
If we take potential energy of the system to be zero at the lowest point A, then
At A,E=12mv20....(i)
From Newton's second law
TAmg=mv20L....(ii)
where TA is the tension in the string at A.
At the highest point C, to complete the circle tension in string will be minimum (zero).
At C,E=12mv2C+mg(2L).....(iii)
From Newton's second law
mg=mv2CL....(iv)
From (iv), vC=gL;Cq
From (iii), E=12m(gL)+2mgL=52mgL.....(v)
Using (i), 12mv50=52mgL,
v0=5gL;Ar....(vi)
At B,E=12mv2B+mg(L)
or 12mv2B=Emg(L)=52mgmgL (Using (v))
12mvB2=32mgL
vB=3gL;Bs
The ratio of kinetic energies at B and C is
KBKC=12mv2B12mv2C=3gLgL=31;Dp.

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