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A uniform meter scale is supported from its $20\ cm$ mark. A body suspended from $10\ cm$ mark keeps the scale horizontal. However, the scale gets unbalanced if the body is completely immersed in water. To regain the balance the body is shifted to the $8\ cm$ mark. Therefore, the specific gravity of the material of the body is
$5$
$6$
$7$
$4$
Solution
$\tau_{p}=0$
$\mathrm{Mg} \times 30=\mathrm{mg} \times 10$
$\Rightarrow \mathrm{m}=3 \mathrm{M} \ldots(1)$
$\mathrm{F}_{\mathrm{b}}=(\mathrm{m} / \rho) \cdot \rho_{\mathrm{ \omega}} \cdot \mathrm{g}$
$\tau_{\text {pnet }}=0$
$\Rightarrow \mathrm{Mg} \times 30+\frac{\mathrm{m}}{\rho} \cdot \rho_{\mathrm{ \omega}} \cdot \mathrm{g} \times 12=\mathrm{mg} \times 12$
$\mathrm{M} \times 30+\frac{3 \mathrm{M}}{\rho_{\mathrm{r}}} \times 12=(3 \mathrm{M}) \times 12$
$\rho_{r}=6$
