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.

Question

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. (A) at B after 0.25 second (B) at B after 0.5 second (C) on the right side of B after some time (D) on the right side of B when the strings are inclined at 15° with B

Tardigrade Admin · Accepted Answer

Irrespective of amplitude, both spheres take same time to reach the mean position, i.e., (T/4) s ∴ T=2 π √(l/g)=2 × 3.14 × √(1/10)=2 s So, they will collide at (T/4)=(2/4)=0.5 s

Q. 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

at $B$ after $0.5$ second

on the right side of $B$ after some time

on the right side of $B$ when the strings are inclined at $1 5^{\circ}$ with $B$

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

Q. 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.

at B after 0.25 second

at B after 0.5 second

on the right side of B after some time

on the right side of B when the strings are inclined at 15∘ with B