Two stationary objects are in an equilibrium position on a seesaw. The first object has a mass of $3 k g$ and rests $10 m$ from the pivot. The other rests $1 m$ from the pivot. Find out the mass of the second object .

0.3 k g

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3 k g

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10 k g

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30 k g

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50 k g

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Solution

The correct option is B $30 k g$
Given : $m_{1} = 3 k g$ $r_{1} = 10$ m $r_{2} = 1$ m
Let the mass of the 2nd object be $m$ .
As the system is in equilibrium, thus net torque of the system about the pivot is zero i.e $τ = 0$
$∴$ $m_{1} r_{1} - m_{2} r_{2} = 0$
OR $3 \times 10 - m \times 1 = 0$ $⟹ m = 30 k g$

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Two stationary objects are in an equilibrium position on a seesaw. The first object has a mass of 3kg and rests 10m from the pivot. The other rests 1m from the pivot. Find out the mass of the second object .

The correct option is B 30kgGiven : m1=3kg r1=10 m r2=1 mLet the mass of the 2nd object be m.As the system is in equilibrium, thus net torque of the system about the pivot is zero i.e τ=0∴ m1r1−m2r2=0OR 3×10−m×1=0 ⟹m=30kg

Two stationary objects are in an equilibrium position on a seesaw. The first object has a mass of $3 k g$ and rests $10 m$ from the pivot. The other rests $1 m$ from the pivot. Find out the mass of the second object .