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Question

A simple pendulum with bob of mass m and length x is held in position at an angle θ1, and then angle θ2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are v1 and v2 respectively. Then v1v2 is

A
1cosθ11cosθ2
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B
1cosθ11cosθ2
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C
2gx(1cosθ1)1cosθ2
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D
1cosθ12gx(1cosθ2)
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Solution

The correct option is B 1cosθ11cosθ2
By energy conservation,
v=2g(xxcosθ)
Similarly, v1=2gx(1cosθ1)
v2=2gx(1cosθ2)
v1v2=1cosθ11cosθ2

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