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In the arrangement shown in the all cont...

In the arrangement shown in the all contact surfaces are smooth strings and pulleys are massless
Given `M_(1) =1kg, M_(2) =2kg, M_(3) =4kg` and `g =10ms^(-2)`
.

A

The acceleration of block of mass `M_(3)` is `4ms^(-2)`

B

The acceleration of block of mass `M_(1)` is `4ms^(-2)`

C

The tension `(T)` in the string connecting blocks of masses `M_(3)` and `M_(2)` is `24N` .

D

The tension `(T)` in the string connecting blocks of masses `M_(1)` and `M_(2)` is `24N` .

Text Solution

Verified by Experts

The correct Answer is:
A, C
`x_(2) + (x_(2) -x) =1, 2x_(2) -x_(1) =1, 2v_(2) =v_(1)`
`2a_(2)=a_(1)`
But acceleration of `M_(3)` =acceleration of `M_(2) =a`
So acceleration of `M_(1) =2xa =2a`
`M_(3) g -T = M_(3)a…(1) T -2T' =M_(2) a…..(2)`
`T' =M_(1) (2a)…(3) M_(3) g = (M_(3) + 4M_(1) +M_(2))a`
acceleration of `M_(3)`
`a=(M_(3)g)/(M_(3)+M_(2)+4M_(1))=(4x10)/(4+2+4)=4ms^(-2)`
Acceleration of `M_(1) =2a =2x4 =8 ms^(-2)`
From (1) `T' =M_(3) g - M_(3) a =4xx6 =24N`
From (3)`T =M_(1) (2a) =1 xx 8 = 8N`
.
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