In an experiment with a beam balance,an unkno | vedclass

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(C) For the first case,the torque balance equation about the pivot is: $16 \ell_{1} = m \ell_{2}$  From this,we get: $m = \frac{16 \ell_{1}}{\ell_{2}}

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$8$

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(C) For the first case,the torque balance equation about the pivot is: $16 \ell_{1} = m \ell_{2}$  From this,we get: $m = \frac{16 \ell_{1}}{\ell_{2}}$  For the second case,the torque balance equation about the pivot is: $m \ell_{1} = 4 \ell_{2}$  From this,we get: $m = \frac{4 \ell_{2}}{\ell_{1}}$  Equating the two expressions for $m$:  $\frac{16 \ell_{1}}{\ell_{2}} = \frac{4 \ell_{2}}{\ell_{1}}$  $\ell_{1}^2 = \frac{4}{16} \ell_{2}^2 = \frac{1}{4} \ell_{2}^2$  $\ell_{1} = \frac{\ell_{2}}{2}$  Substituting $\ell_{1} = \frac{\ell_{2}}{2}$ into the second equation:  $m = \frac{4 \ell_{2}}{\ell_{2}/2} = 4 	imes 2 = 8 \, kg$  Therefore,the value of the unknown mass $m$ is $8 \, kg$.

In an experiment with a beam balance,an unknown mass $m$ is balanced by two known masses of $16 \, kg$ and $4 \, kg$ as shown in the figure. The value of the unknown mass $m$ is ....... $kg$.

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