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(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Given $l = 1 \,m$ and taking $g = 10 \,m/s^2$ (or $\pi^2 \approx 1

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at $B$ after $0.5 \,s$

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(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Given $l = 1 \,m$ and taking $g = 10 \,m/s^2$ (or $\pi^2 \approx 10$),we have $T = 2 \pi \sqrt{\frac{1}{10}} \approx 2 \,s$.For a simple pendulum,the time taken to travel from the extreme position to the mean position $(B)$ is $\frac{T}{4}$.Since both spheres are released from their respective extreme positions,they will both reach the mean position $B$ at time $t = \frac{T}{4} = \frac{2}{4} = 0.5 \,s$.Therefore,the two spheres will collide at the mean position $B$ after $0.5 \,s$.

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

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