A $1 m$ long uniform beam of $2 k g$ mass is being lifted vertically up by a force F at the $100 c m$ mark. What is the minimum force required to do so?
[ $g = 10 m / s^{2}$ ]

1 N

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2 N

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10 N

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20 N

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Solution

The correct option is C $10 N$
The force F is minimum when moment of force F is equal to moment of force of centre of gravity.
So, $F \times 100 c m = 2 g \times 50 c m$ or $F = 10 N$ where $(g = 10 m / s^{2})$

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A 1 m long uniform beam of 2 kg mass is being lifted vertically up by a force F at the 100 cm mark. What is the minimum force required to do so?[ g=10 m/s2]

The correct option is C 10 NThe force F is minimum when moment of force F is equal to moment of force of centre of gravity.So, F×100cm=2g×50cm or F=10N where (g=10m/s2)

A $1 m$ long uniform beam of $2 k g$ mass is being lifted vertically up by a force F at the $100 c m$ mark. What is the minimum force required to do so?
[ $g = 10 m / s^{2}$ ]

The correct option is C $10 N$
The force F is minimum when moment of force F is equal to moment of force of centre of gravity.
So, $F \times 100 c m = 2 g \times 50 c m$ or $F = 10 N$ where $(g = 10 m / s^{2})$