Question

If pulley and all strings as shown are massless,
then tension T, = 100 N. Calculate weight of the
block (g = 10 m/s2)
(1) 50 N
(3) 50√3 N
(2) 100 N
(4) 100√3N​

Answer 1

such an easy question of board level if u really want to clear jee u must try to answer higher level problems

Answer 2

The weight of the block is (2)100N.

Given,

Tension,T2 in the string=100N

Pulley and string are massless.

To find,

the weight of the block.

Solution:

• This question is from the chapter "Laws of motion".
• The concept that will be used here is that of equilibrium position.
• It suggests that when a body is in an equilibrium state, the vector sum of all the forces on it is equal to 0.

The weight of the block, W=mg.

This weight of W N will be downwards and will create a tension, T in the string.

T=W.

The tension in the string that is connected to the fixed support will also be T.

The total tension in the pulley will be 2T.

Now, taking the components of T1 and T2.

Vertical component of T1, Tv=T1cos30°

Horizontal component of T1, Th=T1sin30°

Vertical component of T2, Tv=T2cos60°

Horizontal component of T2, Th=T2sin60°.

As the system is in equilibrium, the horizontal components and the vertical components will be equal.

$T1sin30=T2sin60\\T1 X\frac{1}{2} =100 X\frac{\sqrt{3} }{2} \\T1=100\sqrt{3} N.$

As there are two vertical components they will be added and will be made equal to 2T.

$T1cos30+T2cos60=2T\\100\sqrt{3} X \frac{\sqrt{3} }{2} +100X\frac{1}{2} =2T\\\frac{100X3}{2} +\frac{100}{2}=2T\\\frac{300+100}{2} =2T\\2T=\frac{400}{2} \\2T=200\\T=100 N.$

As T=W,

W=100N.

Hence, the weight of the block will be 100N.

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