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Two masses M(1) and M(2) connected by me...

Two masses `M_(1)` and `M_(2)` connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If `theta = 60^(@) and alpha = 30^(@)`, then the ratio of `M_(1)` to `M_(2)` is

A

`1:2`

B

`2:sqrt3`

C

`1:sqrt3`

D

`sqrt3:1`

Text Solution

Verified by Experts

The correct Answer is:
C
`m_(1)g sin theta=m_(2)g sin alpha`
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Knowledge Check

  • Two masses m_(1) and m_(2) are attached to a string which passes over a frictionless smooth pulley. When m_(1)=10kg,m_(2)=6kg , the acceleration of masses is

    A
    `20m//s^(2)`
    B
    `5m//s^(2)`
    C
    `2.5m//s^(2)`
    D
    `10m//s^(2)`

  • Two masses m_(1) and m_(2) are connected by a light string passing over a smooth pulley. When set free m_(1) moves downward by 2 m in 2 s. The ratio m_(1)//m_(2) is

    A
    `9//7`
    B
    `11//9`
    C
    13/11
    D
    15/13

  • Two masses m_(1) and m_(2) are connected by light inextensible string passing over a smooth pulley P . If the pulley moves vertically upwards with an acceleration equal to g then .

    A
    Tension on the string is `(4m_(1)m_(2)g)/(m_(1)+m_(2))`
    B
    Tension on the string is `(2m_(1)m_(2)g)/(m_(1)+m_(2))`
    C
    The acceleration of mass `m_(1)` with respect to ground is `(3m_(2)-m_(1))/(m_(1)+m_(2))g` .
    D
    The acceleration of mass `m_(1)` with respect to ground is `(2(m_(2)-m))/(m_(1)+m_(2))g`