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Question

Three blocks of masses 2 kg, 4 kg and 6 kg arranged as shown in figure connected by string on a frictionless incline of 37. A force of 120 N is applied upward along the incline to the uppermost block. The cords are light. The tensions T1 and T2 in the strings are (g=10 m/s2, sin37=35)


A
T2=60 N
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B
T2=36 N
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C
T1=8 N
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D
T1=20 N
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Solution

The correct option is D T1=20 N
Given,
Force applied along the inclined plane, F=120 N.

Considering the blocks 2 kg, 4 kg, 6 kg and string as a part of single system.


Net upward force acting along the inclined plane is,

Fnet=F12gsin37

Fnet=120120×35=48 N

So, the net upward acceleration,

a=FnetMsystem=4812

a=4 m/s2

All the blocks will have some acceleration a such that length of every string remains constant.

F.B.D of 4 kg:


Balancing force along the inclined plane:

T2T14gsin37=4a ...(1)

Similarly for 6 kg block we can write,

FT26gsin37=6a

120T26×10×35=6×4

120T236=24

T2=60 N

Substituting the value of T2 in equation (1), we get

60T14×10×(35)=4×4

60T124=16

T1=20 N

Hence, options (c) and (d) are the correct alternatives.

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