Laws Of Motion Ques 10
- The pulley arrangements of figures (a) and (b) are identical. The mass of the rope is negligible. In Fig. (a), the mass $m$ is
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Answer:
Correct Answer: 10.$F$
Solution:
- In these cases, $a=\frac{\text { Net pulling force }}{\text { Total mass }}$
Net pulling force in both the cases is
$$ 2 m g-m g=m g $$
But mass (to be pulled) in case (a) is $3 m$ and in case (b) is $m$.
$$ \begin{array}{lll} \text { Therefore, } & a _1=\frac{m g}{3 m}=\frac{g}{3} & \text { [in case (a)] } \\ \text { and } & a _2=\frac{m g}{m}=g & \text { [in case } \\ \text { or } & a _1<a _2 & \end{array} $$
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