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Question

A ball is held at rest in position A by two light cords. The horizontal cord is now cut and the ball swings to position B. What is the ratio of the tension in the cord at position B to that at position A originally?


A
3
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B
3/4
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C
1/2
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D
1
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Solution

The correct option is D 3/4
FBD of ball at position A:


For equilibrium of ball:
Tcos30=mg
T=23mg ... (i)
and T1=Tsin30 ... (ii)

FBD of ball at position B:


Applying equilibrium condition for ball along the length of the string
T1=mgcos30=32mg ... (iii)

From eq (i) and (iii)
T1T=(32mg)(23mg)=34

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