Two masses, (M+m) and (M-m) are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

gm/M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2gM/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

gm/2M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

g (M^{2} - m^{2}) / 2 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A gm/M
Net force, $F = (M + m) g - (M - m) g = 2 m g$
If a is the acceleration of the system, then:
$F = (M + m + M - m) a = 2 M a$
$\Rightarrow 2 M a = 2 m g$
$\Rightarrow a = \frac{m g}{M}$

Suggest Corrections

Similar questions

Q. Two masses

m_{1}

and

m_{2}

(m_{1} > m_{2})

are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the center of mass of the system will be (

g =

acceleration due to gravity)

Two blocks of masses $m_{1} = 3 m$ and $m_{2} = m$ are attached to the ends of a string which passes over a frictionless fixed pulley (which is a uniform disc of mass $M = 2 m$ and radius R) as shown in Fig. The masses are then released. The acceleration of the system is