Q167 / 180 · KCET 2013 · Physics

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

  • A
    at $B$ after $0.25 \mathrm{~s}$
  • Correct answer
    B
    at $B$ after $0.5 \mathrm{~s}$
  • C
    on the right side of $B$ after some time
  • D
    on the right side of $B$ when the strings are inclined at $15^{\circ}$ with $B$

Solution

Irrespective of amplitude, both spheres take same time to reach the mean position, i.e., $\frac{T}{4} \mathrm{~S}$ $\therefore \quad T=2 \pi \sqrt{\frac{l}{g}}=2 \times 3.14 \times \sqrt{\frac{1}{10}}=2 \mathrm{~s}$ So, they will collide at $\frac{T}{4}=\frac{2}{4}=0.5 \mathrm{~s}$.

Hence, the correct answer is option: 'B'.

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