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Consider the case of two bodies of masses `m_(1)` and `m_(2)` which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations :
(`i`) when `m_(1) gt m_(2)`. In this case
`a=((m_(1)-m_(2))/(m_(1)+m_(2)))g` and `T=((2m_(1)m_(2))/(m_(1)+m_(2)))g`
(`ii`) when `m_(2) gt m_(1)`. In this case
`a=((m_(2)-m_(1))/(m_(1)+m_(2)))g` and `T=((2m_(1)m_(2))/(m_(1)+m_(2)))g`
(`iii`) When `m_(1)=m_(2)=m`. In this case `a=0`, `T=mg`
If `m_(1)=10kg`, `m_(2)=6kg` and `g=10m//s^(2)` then what is the acceleration of masses ?

A

`2.5 m//s^(2)`

B

`5m//s^(2)`

C

`20m//s^(2)`

D

`40m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=((m_(1)-m_(2))/(m_(1)+m_(2)))g" " T-((2m_(1)m_(2))/(m_(1)+m_(2)))g`
`a=((10-6)/(16))g=2.5m//s^(2)`
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