Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F \leftarrow M T_{1} \leftarrow - M T_{2} \leftarrow - M T_{3} \leftarrow - M$

T_{1} = \frac{1}{4} F; T_{2} = \frac{3}{2} F; T_{3} = \frac{1}{4} F

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T_{1} = \frac{1}{4} F; T_{2} = \frac{1}{2} F; T_{3} = \frac{1}{2} F

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T_{1} = \frac{3}{4} F; T_{2} = \frac{1}{2} F; T_{3} = \frac{1}{4} F

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T_{1} = \frac{3}{4} F; T_{2} = \frac{1}{2} F; T_{3} = \frac{1}{2} F

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Solution

The correct option is C $T_{1} = \frac{3}{4} F; T_{2} = \frac{1}{2} F; T_{3} = \frac{1}{4} F$
Force balance on first block
$F = 4 M a$ where is the acceleration of system.
Force balance on second block
$F - T_{1} = M a$
$T_{1} = \frac{3}{4}$ F
Force balance on third block
$T_{1} - T_{2}$ =Ma
$T_{2}$ = $\frac{3}{4}$ F- $\frac{3}{4}$ F= $\frac{1}{2}$ F
Force balance on third block
$T_{2} - T_{3}$ =Ma
$T_{3}$ =Ma= $\frac{1}{4}$ F

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Four blocks of same mass connected by cords are pulled by a force F on a smooth horizontal surface, as shown in figure. The tensions T1,T2 and T3 will beF←MT1←−MT2←−MT3←−M

The correct option is C T1=34F;T2=12F;T3=14FForce balance on first blockF=4Ma where is the acceleration of system. Force balance on second blockF−T1=MaT1=34FForce balance on third block T1−T2=MaT2=34F-34F=12FForce balance on third block T2−T3=MaT3=Ma=14F

Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F \leftarrow M T_{1} \leftarrow - M T_{2} \leftarrow - M T_{3} \leftarrow - M$