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Question

Physics Question on laws of motion

A light string passing over a smooth light pulley connects two blocks of masses m1m_{1} and m2m_{2} (vertically). If the acceleration of the system is g/8g / 8, then the ratio of the masses is :

A

8:01

B

9:07

C

4:03

D

5:03

Answer

9:07

Explanation

Solution

As the string is inextensible, both masses have the same acceleration aa. Also, the pulley is massless and frictionless, hence, the tension at both ends of the string is the same. Suppose the mass m2m_{2} is greater than mass m1,m_{1}, so, the heavier mass m2m_{2} is accelerated downward and the lighter mass m1m_{1} is accelerated upwards. Therefore, by Newton's 2 nd law Tm1g=m1aT-m_{1} g=m_{1} a...(1) m2gT=m2am_{2} g-T=m_{2} a...(2) After solving Eqs. (1) and (2) a=(m2m1)(m1+m2)g=g8a=\frac{\left(m_{2}-m_{1}\right)}{\left(m_{1}+m_{2}\right)} \cdot g=\frac{g}{8}(given) so,g8=m2(1m1/m2)m2(1+m1/m2)g\frac{g}{8}=\frac{m_{2}\left(1-m_{1} / m_{2}\right)}{m_{2}\left(1+m_{1} / m_{2}\right)} \cdot g Let m1m2=x\frac{m_{1}}{m_{2}}=x Thus E (3) becomes 1x1+x=18\frac{1-x}{1+x}=\frac{1}{8} or x=79x=\frac{7}{9} or m2m1=97\frac{m_{2}}{m_{1}}=\frac{9}{7} So, the ratio of the masses is 9: 7 .