Two masses M and M/2 are joined together by means of light inextensible string passed over a ffictionless pulley as shown in figure. When the bigger mass is released, the smaller one will ascend with an acceleration of

Question

Two masses M and M/2 are joined together by means of light inextensible string passed over a ffictionless pulley as shown in figure. When the bigger mass is released, the smaller one will ascend with an acceleration of <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e32d6f596d6f54a9db678b3ff64e8821-.png" /> (A) (g/3) (B) (3 g/2) (C) g (D) (g/2)

Tardigrade Admin · Accepted Answer

Here, m1=M, m2=(M/2) Acceleration of the system. a=((m1-m2) g/m1+m2)=((M-(M/2)) g/(M+(M)2))=(g/3)

Q. Two masses $M$ and $M /2$ are joined together by means of light inextensible string passed over a ffictionless pulley as shown in figure. When the bigger mass is released, the smaller one will ascend with an acceleration of

$\frac{g}{3}$

$\frac{3 g}{2}$

$g$

$\frac{g}{2}$

Here, $m_{1} = M, m_{2} = \frac{M}{2}$
Acceleration of the system.
$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( M - \frac{M}{2} ) g}{( M + \frac{M}{2} )} = \frac{g}{3}$

Q. Two masses M and M/2 are joined together by means of light inextensible string passed over a ffictionless pulley as shown in figure. When the bigger mass is released, the smaller one will ascend with an acceleration of

3g​

23g​

g

2g​

Here, m1​=M,m2​=2M​ Acceleration of the system. a=m1​+m2​(m1​−m2​)g​=(M+2M​)(M−2M​)g​=3g​

Q. Two masses $M$ and $M /2$ are joined together by means of light inextensible string passed over a ffictionless pulley as shown in figure. When the bigger mass is released, the smaller one will ascend with an acceleration of

$\frac{g}{3}$

$\frac{3 g}{2}$

$g$

$\frac{g}{2}$

Here, $m_{1} = M, m_{2} = \frac{M}{2}$
Acceleration of the system.
$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( M - \frac{M}{2} ) g}{( M + \frac{M}{2} )} = \frac{g}{3}$