Questions

The ratio of tensions in the string connected to the block of mass m₂ in Figure (a) and Figure (b) respectively, is (friction is absent everywhere): $[m_{1} = 50 kg, m_{2} = 80 kg$ $and F = 1000 N]$

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detailed solution

Correct option is C

In Figure (a): Let tension in the string is T1F−T1=m1a⇒1000−T1=50a ......(i)T1−m2g=m2a⇒T1−80g=80a .......(ii) From (i) and (ii) T1=1200013NIn Figure (b): Let tension in the string is T2. F+m1g−T2=m1a⇒ 1000+50g−T2=50a .......(iii)T2−80g=80aT2−80g=80a .......(iv)From (iii) and (iv),T2=1600013N⇒T1T2=1200016000=34

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