The roller of diameter 0.6 m has to be raised on the pavement XY by forces $F_{1}$ and $F_{2}$ each of magnitude 15 N. The ratio of torques produced by the two forces is __________.

1:3

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1:2

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3:1

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2:1

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Solution

The correct option is D 2:1
We know that:
Torque = Force $\times$ Perpendicular distance
Now,
Torque due to $F_{1}$ = 15 N $\times$ 0.6 m = 9 N m
Torque due to $F_{2}$ = 15 N $\times$ (0.6/2) m = 4.5 N m
Thus,
Ratio of both the torques = 9/4.5 = 2:1

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The roller of diameter 0.6 m has to be raised on the pavement XY by forces F1 and F2 each of magnitude 15 N. The ratio of torques produced by the two forces is __________​.

The correct option is D 2:1We know that: Torque = Force × Perpendicular distance Now, Torque due to F1 = 15 N × 0.6 m = 9 N m Torque due to F2 = 15 N × (0.6/2) m = 4.5 N m Thus, Ratio of both the torques = 9/4.5 = 2:1

The roller of diameter 0.6 m has to be raised on the pavement XY by forces $F_{1}$ and $F_{2}$ each of magnitude 15 N. The ratio of torques produced by the two forces is __________.

The correct option is D 2:1
We know that:
Torque = Force $\times$ Perpendicular distance
Now,
Torque due to $F_{1}$ = 15 N $\times$ 0.6 m = 9 N m
Torque due to $F_{2}$ = 15 N $\times$ (0.6/2) m = 4.5 N m
Thus,
Ratio of both the torques = 9/4.5 = 2:1