Question

As shown in figure, all the surfaces are frictionless and the pulleys and the string are light. The magnitude of acceleration of block $2M$ is

• A. $\frac{g}{3}$
• B. $\frac{g}{6}$
• C. $\frac{2g}{3}$
• D. $\frac{g}{4}$

Answer 1

The correct option is B $\frac{g}{6}$


From string constraint(string should remain taut), or length of string $\left(l\right)$ will be constant.
$l=x_A+2x_B+C$
$\Rightarrow 0=\frac{d^2{x}_A}{d{t}^2}+2\frac{d^2{x}_B}{d{t}^2}$
$\because a_A=-\frac{d^2{x}_A}{d{t}^2}{a}_B=\frac{d^2{x}_B}{d{t}^2}$
$\Rightarrow 0=-a_A+2a_B$

$\therefore a_A=2a_B....\left(i\right)$
F.B.D. of $\left(A\right)$, forces shown along the inclined plane as forces $\perp$ to inclined plane forces will be balanced.


Applying newton's $2^{nd}$ law along direction of motion:
$T-Mg\sin {30}^{\circ }=M{a}_A$
$\Rightarrow T-\frac{Mg}{2}=M{a}_A$
$a_A=\frac{T}{M}-\frac{g}{2}....\left(ii\right)$

From F.B.D. of $\left(B\right)$,


$2Mg-2T=2M{a}_B$
$\Rightarrow Mg-T=M{a}_B$
$a_B=g-\frac{T}{M}....\left(iii\right)$
Adding Eq $\left(ii\right)\&\left(iii\right)$,
$a_A+a_B=\frac{g}{2}$
$2a_B+a_B=\frac{g}{2}$
$\therefore a_B=\frac{g}{6}$