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Block A of weight 100N rests on a fricti...

Block `A` of weight `100N` rests on a frictionless inclined plane of slope angle `30^(@)`(Fig. 5.7). `A` flexible cord attached to `A` passes over a frictonless pulled and is connected to block `B` of weight `W`. Find the weight `W` for which the system in equilibrium.

A

25N

B

50N

C

75N

D

100N

Text Solution

Verified by Experts

The correct Answer is:
B
As the system is in equilibrium
`therefore T=W….(i)`
and `T=mg sin 30^(@) …(ii)`
From (i) and (ii), we get
`W=mg sin 30^(@)=(100N)xx(1)/(2)=50N`
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NCERT FINGERTIPS ENGLISH-LAWS OF MOTION-Assertion And Reason
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