Question

Two bodies of equal masses are connected by a light inextensible string passing over a smooth frictionless pulley. The amount of mass that should be transferred from one to another so that both the masses mole 50 m in 5 s is

• A. 30%
• B. 40%
• C. 70%
• D. 50%

Answer 1

Answer: (B)

Since each mass is moving 50 m in 5 s, therefore using the relation $ s=ut+\frac{1}{2}a{{t}^{2}}, $ we have $ 50=0\times 5+\frac{1}{2}\times a\times {{5}^{2}} $ or $ a=\frac{100}{25}=4\,m{{s}^{2}} $ Let mass of one become $ {{m}_{1}} $ and that of other $ {{m}_{2}}, $ where $ {{m}_{1}}>{{m}_{2}} $ . As $ {{m}_{1}} $ moves downwards with acceleration $ a=4\,m{{s}^{-2}} $ $ \therefore $ $ a=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g $ So, $ 4=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)(10) $ $ \left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)=\frac{4}{10}=\frac{2}{5} $ $ \therefore $ % of mass transferred $ =\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\times 100 $ $ =\frac{2}{5}\times 100 $ $ =40% $