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Four blocks of same mass connected by strings are pulled by a force F on a smooth horizontal surface as
shown in figure.
The tension T₁, T₂ and T₃ will be [AIMS 2017]

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$T_{1} = \frac{1}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F$

$T_{1} = \frac{1}{4} F, T_{2} = \frac{3}{2} F, T_{3} = \frac{1}{4} F$

$T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F$

$T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{4} F$

answer is C.

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Detailed Solution

Given situation can be represented as

$F \leftarrow m_{1} \to T_{1} \leftarrow m_{2} \to T_{2} \leftarrow m_{3} \to T_{3} \leftarrow m_{4}$

$\Rightarrow T_{1} = \frac{(m_{2} + m_{3} + m_{4})}{m_{1} + m_{2} + m_{3} + m_{4}} F$

Given, $m_{1} = m_{2} = m_{3} = m_{4} = m$

$∴ T_{1} = \frac{3}{4} F$

Similarly, $T_{2} = \frac{(m_{3} + m_{4}) F}{m_{1} + m_{2} + m_{3} + m_{4}}$

$∴ T_{2} = \frac{1}{2} F$

Also, $T_{3} = \frac{m_{4} F}{m_{1} + m_{2} + m_{3} + m_{4}} \Rightarrow T_{3} = \frac{1}{4} F$

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