Dear Student, Please find below the solution to the asked query: The tension in the string between A & B is T1 and between B & C is T2 Then the strings are inextensible So, three blocks should have same acceleration Let the acceleration of the blocks is a For A,ma=T1-mg⇒T1=ma+mgFor B,ma=mg+T2-T1For C,ma=mg-T2⇒T2=mg-maSo,ma=mg+mg-ma-ma+mg⇒ma=mg+mg-ma-ma-mg⇒ma=mg-2ma⇒3ma=mg⇒a=g3So,T1=mg+a=2g+g3=8g3 and T2=mg-a=2g-g3=4g3 Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

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Dear Student,

Please find below the solution to the asked query:

The tension in the string between A & B is T₁ and between B & C is T₂. Then the strings are inextensible. So, three blocks should have same acceleration. Let the acceleration of the blocks is a.

$F o r A, m a = T_{1} - m g \Rightarrow T_{1} = m a + m g F o r B, m a = m g + T_{2} - T_{1} F o r C, m a = m g - T_{2} \Rightarrow T_{2} = m g - m a S o, m a = m g + (m g - m a) - (m a + m g) \Rightarrow m a = m g + m g - m a - m a - m g \Rightarrow m a = m g - 2 m a \Rightarrow 3 m a = m g \Rightarrow a = \frac{g}{3} S o, T_{1} = m (g + a) = 2 (g + \frac{g}{3}) = \frac{8 g}{3} a n d T_{2} = m (g - a) = 2 (g - \frac{g}{3}) = \frac{4 g}{3}$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

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