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Question

# Three equal weights A, B and C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is

A
Zero
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B
13 N
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C
3.3 N
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D
19.6 N
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Solution

## The correct option is B 13 NFBD of each mass Let ′a′ be acceleration of each block then T−mg=ma T1+mg−T=ma mg−T1=ma Adding ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯mg=3ma a=g3 Now mg−T1=ma=mg3 ⇒T1=2mg3=2×2×9.83 ≈13 N Alternative Method; a=Net unbalanced forcetotal mass As on right side total mass is 4 kg and on left side mass is 2 kg a=(4−2)g6=g3m/s2 Taking FBD of C block T1=2(g−a)=2(g−g3)=4g3=13 N

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