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Standard XII

Physics

Acceleration

A lift of mas...

Question

A lift of mass $^{'} m^{'}$ is connected to a rope which is moving upward with the maximum acceleration $^{'} a^{'}$ . For maximum safe stress, the elastic limit of the rope is $^{'} T^{'}$ . The minimum diameter of the rope is $(g =$ gravitational acceleration):

{[\frac{2 m (g + a)}{π T}]}^{\frac{1}{2}}

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2 {[\frac{m (g + a)}{π T}]}^{\frac{1}{2}}

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{[\frac{m (g + a)}{π T}]}^{\frac{1}{2}}

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{[\frac{m (g + a)}{2 π T}]}^{\frac{1}{2}}

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Solution

The correct option is B $2 {[\frac{m (g + a)}{π T}]}^{\frac{1}{2}}$
Let the tension be F So $F - m g = m a \to F = m (g + a)$
Now $T = \frac{F}{π r^{2}}$
$r = (\frac{m (g + a)}{π T})^{\frac{1}{2}}$
$D = 2 r$

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A lift of mass ′m′ is connected to a rope which is moving upward with the maximum acceleration ′a′. For maximum safe stress, the elastic limit of the rope is ′T′. The minimum diameter of the rope is (g= gravitational acceleration):

The correct option is B 2[m(g+a)πT]12Let the tension be F So F−mg=ma→F=m(g+a)Now T=Fπr2r=(m(g+a)πT)12D=2r

A lift of mass $^{'} m^{'}$ is connected to a rope which is moving upward with the maximum acceleration $^{'} a^{'}$ . For maximum safe stress, the elastic limit of the rope is $^{'} T^{'}$ . The minimum diameter of the rope is $(g =$ gravitational acceleration):

The correct option is B $2 {[\frac{m (g + a)}{π T}]}^{\frac{1}{2}}$
Let the tension be F So $F - m g = m a \to F = m (g + a)$
Now $T = \frac{F}{π r^{2}}$
$r = (\frac{m (g + a)}{π T})^{\frac{1}{2}}$
$D = 2 r$