A block ‘A’ of mass ‘m’ is attached at one end of a light spring and the other end of the spring is connected to another block ‘B’ of mass 2 m through a light string as shown in the figure. ‘A’ is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is ‘a’ In the next case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is ‘b’. The value of a/b is: (Pulley, string and the spring are massless)

Question

A block ‘A’ of mass ‘m’ is attached at one end of a light spring and the other end of the spring is connected to another block ‘B’ of mass 2 m through a light string as shown in the figure. ‘A’ is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is ‘a’ In the next case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is ‘b’. The value of a/b is: (Pulley, string and the spring are massless) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/13cf29f18282969210c37743b102faa3-.png" /> (A) 0 (B) undefined (C) 2 (D) (1/2)

Tardigrade Admin · Accepted Answer

For first case tension in spring will be Ts=2mg just after ‘A’ is released 2mg-mg=ma ⇒ a=g In second case Ts=mg

2mg-mg=2mb b=g/2 a/b=2

$0$

undefined

$2$

$\frac{1}{2}$

For first case tension in spring will be

$T_{s} = 2 m g$ just after ‘A’ is released

$2 m g - m g = ma$

$\Rightarrow a = g$

In second case $T_{s} = m g$

$2 m g - m g = 2 mb$

$b = g /2$

$a / b = 2$

0

undefined

2

21​

For first case tension in spring will be Ts​=2mg just after ‘A’ is released 2mg−mg=ma ⇒a=g In second case Ts​=mg 2mg−mg=2mb b=g/2 a/b=2

$0$

$2$

$\frac{1}{2}$

For first case tension in spring will be

$T_{s} = 2 m g$ just after ‘A’ is released

$2 m g - m g = ma$

$\Rightarrow a = g$

In second case $T_{s} = m g$

$2 m g - m g = 2 mb$

$b = g /2$

$a / b = 2$