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The force F in figure is just sufficient to hold the 100 N block and weightless pulley in equilibrium. There is no appreciable friction. Then:

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$T_{1} = \frac{100}{7} N$

$T_{1} = \frac{100}{3} N$

$T_{2} = 50 N$

$T_{3} = \frac{400}{7} N$

answer is A, D.

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Detailed Solution

$T_{2} = 2 T_{1}$ and $T_{3} = 2 T_{2}$

$∴ T_{3} = 4 T_{1}$ and $T_{1} = F$

Now $T_{1} + T_{2} + T_{3} = 100$

or $T_{1} + 2 T_{1} + 4 T_{1} = 100$

$∴ T_{1} = \frac{100}{7} N$

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