From a fixed support, two small identical spheres are suspended by means of strings of length $1$ m each. They are pulled aside as shown and then released. B is the mean position. Then the two spheres collide.

At B after

0.25

second

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At B after

0.5

second

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On the right side of B after some time

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On the right side of B when the strings are inclined at

15^{o}

with B

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Solution

The correct option is B At B after $0.5$ second
Regardless of what their amplitudes are they would take same time to reach B i.e quarter of time period (imagine if each of them are released from the fixed support, Point B is located at one quarter of the total circle).
Length of string, $l = 1 m$
Time period, $T = 2 π \sqrt{\frac{l}{g}} = 2 s$
Therefore they will meet at point B after $\frac{T}{4} s = 0.5 s$

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From a fixed support, two small identical spheres are suspended by means of strings of length 1m each. They are pulled aside as shown and then released. B is the mean position. Then the two spheres collide.

From a fixed support, two small identical spheres are suspended by means of strings of length $1$ m each. They are pulled aside as shown and then released. B is the mean position. Then the two spheres collide.