Question

Two particles of mass `m_(1)` and `m_(2)(m_(1) gt m_(2))` are attached to the ends of a light inextensible string passing over a smooth fixed pulley. When the level of `m_(1)` is higher than that of `m_(2)` by 49 cm and string is just taut `m_(2)` is released. 0.5 s after motion started the masses are at the same level. the ratio of `m_(1)` to `m_(2)` is (Take `g=9.8ms^(-2)`)
A. `7:3`
B. `4:3`
C. `3:2`
D. `2:1`

Answer 1

Correct Answer - 3
`((m_(1)-m_(2))/(m_(1)+m_(2)))g =a`, where `1/2(2a)(1/2)^(2)`
`=49/100=g/20`
Therefore `m_(1):m_(2)=3:2`