A lift of mass $100 k g$ starts moving from rest in the upward direction. Fig. shows the variation of speed of the lift $T_{1} T_{2}$ and $T_{3}$ stand for tension in the rope from zero to two seconds, two to six seconds, six to seven seconds respectively, then:

T_{1} : T_{2} : T_{3} :: 1 : 1 : 1

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T_{1} : T_{2} : T_{3} :: 6 : 5 : 3

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T_{1} : T_{2} : T_{3} :: 3 : 5 : 6

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T_{1} : T_{2} : T_{3} :: 6 : 5 : 6

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Solution

The correct option is B $T_{1} : T_{2} : T_{3} :: 6 : 5 : 3$
force is change in momentum $F = m a = m \times \frac{v}{t}$
therefore $T_{1} - 100 g = 100 \times \frac{4 - 0}{2 - 0} => 1200 N$
$T_{2} - 100 g = 0 =>= 1000 N$
$T_{3} = 100 g + 100 \times \frac{0 - 4}{7 - 6} = 600$

$T 1 : T 2 : T 3 :: 1200 : 1000 : 600 = 6 : 5 : 3$

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A lift of mass 100kg starts moving from rest in the upward direction. Fig. shows the variation of speed of the lift T1T2 and T3 stand for tension in the rope from zero to two seconds, two to six seconds, six to seven seconds respectively, then:

The correct option is B T1:T2:T3::6:5:3force is change in momentum F=ma=m×vttherefore T1−100g=100×4−02−0=>1200NT2−100g=0=>=1000NT3=100g+100×0−47−6=600T1:T2:T3::1200:1000:600=6:5:3

A lift of mass $100 k g$ starts moving from rest in the upward direction. Fig. shows the variation of speed of the lift $T_{1} T_{2}$ and $T_{3}$ stand for tension in the rope from zero to two seconds, two to six seconds, six to seven seconds respectively, then: