Question

Two pulley arrangements of figure given are identical. The mass of the rope in negligible. In fig(a), the mass $m$ is lifted by attaching a mass $2m$ to the other end of the rope. In fig(b), $m$ is lifted up by pulling the other end of the rope with a constant downward force $F=2mg$. The acceleration of $m$ in the two cases are respectively

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

Answer 1

The correct option is B $\frac{g}{3},g$

Let a and a' be the acceleration in both cases respectivlely. Then for fig (a),

$T-mg=ma....\left(i\right)$
and $2mg-T=2ma...\left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$mg=3ma$
$\therefore a=\frac{g}{3}$
For fig (b),

$T^{\prime }-mg=m{a}^{\prime }......\left(iii\right)$
and $2mg-T^{\prime }=0...\left(iv\right)$
Solving $\left(iii\right)$ and $\left(iv\right)$
$a^{\prime }=g$
$\therefore a=\frac{g}{3}$ and $a^{\prime }=g$