A block of 50 kg on a smooth plane inclined at `60^(@)` and another block of 30 kg on a smooth plane inclined at `30^(@)` with horizontal are connected by a light string passing over a frictionless pully as shown in the figure. Take g = `10 ms^(-2) , sqrt(3) =1.7`. The acceleration of the blocks and tension in the string are.

Options

A. `3.437 ms^(-2)` and `25.3 11N`
B. `34.37 ms^(-2)` and `253.11N`
C. `3.437 cms^(-2)` and `25.3 11 dyne `
D. `3.437 ms^(-2)` and `253. 11N`

Answer

D

Explanation

Here magnitude of acceleration and tension on both block is equal.
As per equation of motion we get
`T- m_(2) g sin 30^(@) = m_(2) a `
`m_(1) g sin60^(@) - m_(2) g sin 30^(@) = m_(1) a + m_(2) a`
`50xx 10 xx (sqrt(3))/( 2) 30 xx 1 xx (1)/(2) = (50 + 30) a `
`80 a =- 275`
Now putting value of a in eq (i)
`50 xx 10 xx (sqrt(3))/( 2) -t =50 (3.437)`
`425 -171 .85 =T`
`T = 253 .15 N = 253 .11 N`