Question

Three equal weights of mass 2 kg each are
hanging on a string passing over a fixed
frictionless pulley. The tension in the string AB
is nearly

(1) Zero
(?) 15 N
(2) 13 N
​

Answer 1

Answer

Correct option is

B

13N

The Free body diagram of the system is shown.

For block A: T−mg=ma

T−2g=2a

We get T=2a+2g

For block B + C system : mg+mg−T=(m+m)a

2g+2g−T=(2+2)a

⟹4g−T=4a

Or 4g−2a−2g=4a

Or 2g=6a

This gives a=g/3

For block C: mg−T

′

=ma

2g−T

′

=2a

Or 2g−T

′

=

3

2g

Thus we get T

′

=4g/3=

3

4×10

=13N

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