Home
Class 12
PHYSICS
In the system shown in figure m(1) gt m(...

In the system shown in figure `m_(1) gt m_(2)`. System is held at rest by thread BC. Just after the thread BC is burnt.

A

acceleration of `m_(1)` will be equal to zero.

B

acceleration of `m_2` will be downwards

C

magnitude of acceleration of two blocks will be non -zero and unequal

D

maginitude of acceleration of both the blocks will be `((m_(1) + m_(2))/(m_(1) + m_(2)))g`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

The system shown in the figure is in equilibrium at rest. The spring and string are massless Now the string is cut. Find the acceleration of the blocks just after the string is cut.

In the pulley system shown in the figure, the mass of A is half of that of rod B. The rod length is 500 cm. The mass of pulleys and the threads may be neglected The mass A is set at the same level as the lower end of the rod and then released After releasing the mass A, it would reach the top end of the rod B in time (Assume, g = 10 m//s^2 )

Two blocks asre connected by a string as shown in figure. They are released from rest. Show that after they moved a distance L, their common speed is given by sqrt((2(m_(2)-mu m_(1))gL)/((m_(1)+m_(2)))) where mu is the coefficient of friction.