A table has a heavy circular top of radius 1 | vedclass

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(C) To prevent toppling,the table will rotate about the line joining two adjacent legs. Let the center of the table be $C$ and the point where the mas

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$47$

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(C) To prevent toppling,the table will rotate about the line joining two adjacent legs. Let the center of the table be $C$ and the point where the mass $m$ is placed at the edge be $A$. The line joining two adjacent legs acts as the pivot axis,passing through point $B$.In the critical condition,the torque due to the mass $m$ about the pivot line must be balanced by the torque due to the weight of the table $(M)$ about the same line.Let $R$ be the radius of the table. The distance from the center $C$ to the pivot line $B$ is $BC = R \cos 45^{\circ} = \frac{R}{\sqrt{2}}$.The distance from the edge $A$ to the pivot line $B$ is $AB = R - BC = R - \frac{R}{\sqrt{2}} = R(1 - \frac{1}{\sqrt{2}})$.Equating the torques about the pivot line:$m g (AB) = M g (BC)$$m (R(1 - \frac{1}{\sqrt{2}})) = M (\frac{R}{\sqrt{2}})$$m = M \frac{1/\sqrt{2}}{1 - 1/\sqrt{2}} = M \frac{1}{\sqrt{2} - 1}$Given $M = 20 \,kg$ and $\sqrt{2} \approx 1.414$:$m = \frac{20}{1.414 - 1} = \frac{20}{0.414} \approx 48.3 \,kg$.The closest value among the options is $47 \,kg$.

$A$ table has a heavy circular top of radius $1 \,m$ and mass $20 \,kg$ placed on four light (considered massless) legs placed symmetrically on its circumference. The maximum mass that can be kept anywhere on the table without toppling it is close to ............. $kg$.

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