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Standard XII

Physics

Newton's Second Law: Rotatory Version

A horizontal ...

Question

A horizontal board of negligible mass supports a mass weighing $20 N$ at a distance of $2 m$ from the point $A$ . Calculate the weight which must be placed at point $B$ , which is $5 m$ from point $A$ , in order to keep the board in rotational equilibrium?

4 N

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8 N

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10 N

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40 N

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60 N

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Solution

The correct option is C $8 N$
Given : $W_{1} = 20$ N $r_{1} = 2$ m $r_{2} = 5$ m
As the system is in rotational equilibrium, thus torque about point A is zero i.e $τ_{A} = 0$
$∴$ $W_{1} \times r_{1} - W_{2} \times r_{2} = 0$
OR $20 \times 2 = W_{2} \times 5$ $⟹ W_{2} = 8$ N

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A horizontal board of negligible mass supports a mass weighing 20N at a distance of 2m from the point A. Calculate the weight which must be placed at point B, which is 5m from point A, in order to keep the board in rotational equilibrium?

The correct option is C 8NGiven : W1=20 N r1=2 m r2=5 mAs the system is in rotational equilibrium, thus torque about point A is zero i.e τA=0∴ W1×r1−W2×r2=0OR 20×2=W2×5 ⟹W2=8 N

A horizontal board of negligible mass supports a mass weighing $20 N$ at a distance of $2 m$ from the point $A$ . Calculate the weight which must be placed at point $B$ , which is $5 m$ from point $A$ , in order to keep the board in rotational equilibrium?

The correct option is C $8 N$
Given : $W_{1} = 20$ N $r_{1} = 2$ m $r_{2} = 5$ m
As the system is in rotational equilibrium, thus torque about point A is zero i.e $τ_{A} = 0$
$∴$ $W_{1} \times r_{1} - W_{2} \times r_{2} = 0$
OR $20 \times 2 = W_{2} \times 5$ $⟹ W_{2} = 8$ N