Three equal weights of 3 kg each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses II and III will be (Take g=10 m / sec 2 )

Question

Three equal weights of 3 kg each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses II and III will be (Take g=10 m / sec 2 ) (A) 5 N (B) 6 N (C) 10 N (D) 20 N

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/14f443012010f9ce549355340b83a800-.png" /> Three equal weights of 3 kg each are hanging on a string passing over a pulley as shown in figure. If a is the common acceleration. T-3 g=3 a (For mass I) 3 g+T1- T=3 a (For mass II) 3 g-T1=3 a (For mass III) Adding, above three equations 3 g=9 a, a=g / 3 T1=3 g-3 a=3 g-g=2 g=20 N

Q. Three equal weights of 3kg each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses II and III will be (Take g=10m/sec2 )

5 N

6 N

10 N

20 N

Three equal weights of 3kg each are hanging on a string passing over a pulley as shown in figure. If a is the common acceleration. T−3g=3a (For mass I) 3g+T1−​T=3a (For mass II) 3g−T1​=3a (For mass III) Adding, above three equations 3g=9a,a=g/3 T1​=3g−3a=3g−g=2g=20N

Q. Three equal weights of $3 k g$ each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses II and III will be (Take $g = 10 m / se c^{2}$ )