NEET Physics Mechanical Properties of Fluids Questions Solved

In making an alloy, a substance of specific gravity ${\mathrm{s}}_{1}$ and mass ${\mathrm{m}}_{1}$ is mixed with another substance of specific gravity ${\mathrm{s}}_{2}$ and mass ${\mathrm{m}}_{2}$ ; then the specific gravity of the alloy is

(a) $\left(\frac{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{{\mathrm{s}}_{1}+{\mathrm{s}}_{2}}\right)$                               (b) $\left(\frac{{\mathrm{s}}_{1}{\mathrm{s}}_{2}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}\right)$

(c) $\frac{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\left(\frac{{\mathrm{m}}_{1}}{{s}_{1}}+\frac{{m}_{2}}{{\mathrm{s}}_{2}}\right)}$                            (d) $\frac{\left(\frac{{\mathrm{m}}_{1}}{{\mathrm{s}}_{1}}+\frac{{\mathrm{m}}_{2}}{{\mathrm{s}}_{2}}\right)}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}$

Specific gravity of alloy =

$=\frac{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\left(\frac{{\mathrm{m}}_{1}}{{\mathrm{\rho }}_{1}}+\frac{{\mathrm{m}}_{2}}{{\mathrm{\rho }}_{2}}\right)×{\mathrm{\rho }}_{\mathrm{w}}}=\frac{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\frac{{\mathrm{m}}_{1}}{{\mathrm{\rho }}_{1}}{{\mathrm{\rho }}_{\mathrm{w}}}}+\frac{{\mathrm{m}}_{2}}{{\mathrm{\rho }}_{2}}{{\mathrm{\rho }}_{\mathrm{w}}}}}=\frac{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\frac{{\mathrm{m}}_{1}}{{\mathrm{s}}_{1}}+\frac{{\mathrm{m}}_{2}}{{\mathrm{s}}_{2}}}$

[ As specific gravity of substance = ]

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