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An object weighs \(\mathrm{m}_1\) in a liquid of density \(\mathrm{d}_1\) and that in liquid of density \(\mathrm{d}_2\) is \(\mathrm{m}_2\). The density \(d\) of the object is
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The correct answer is:
\(\mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)
Hints: \(V\left(d-d_1\right) g=m_1 g\)
\(\mathrm{V}\left(\mathrm{d}-\mathrm{d}_2\right) \mathrm{g}=\mathrm{m}_2 \mathrm{~g}\)
\(\frac{\mathrm{d}-\mathrm{d}_1}{\mathrm{~d}-\mathrm{d}_2}=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \quad \therefore \mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)
\(\mathrm{V}\left(\mathrm{d}-\mathrm{d}_2\right) \mathrm{g}=\mathrm{m}_2 \mathrm{~g}\)
\(\frac{\mathrm{d}-\mathrm{d}_1}{\mathrm{~d}-\mathrm{d}_2}=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \quad \therefore \mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)
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