A rod of length 10m is hinged at the centre.A force of 20 N is applied at one end. At what distance should another force of 25 N be applied from the centre so that the rod is in equilibrium?

70 cm from point

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30 cm from

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40 cm from

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60 cm from

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Solution

The correct option is B
30 cm from

For equilibrium, net torque on the rod should be zero. Let us apply 30 N at a distance R from the centre.

So, 20 X 5 = 25X R
R = 4 m

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A rod of length 10m is hinged at the centre.A force of 20 N is applied at one end. At what distance should another force of 25 N be applied from the centre so that the rod is in equilibrium?

The correct option is B 30 cm from For equilibrium, net torque on the rod should be zero. Let us apply 30 N at a distance R from the centre. So, 20 X 5 = 25X R R = 4 m

The correct option is B
30 cm from

For equilibrium, net torque on the rod should be zero. Let us apply 30 N at a distance R from the centre.

So, 20 X 5 = 25X R
R = 4 m