Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$, respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2} .$ The ratio of their terminal velocities would be, (1) $\frac{79}{36}$ (2) $\frac{79}{72}$ (3) $\frac{19}{36}$ (4) $\frac{39}{72}$

Home » Two small spherical metal balls, having equal masses, are made from materials of densities and and have radii of and , respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals and whose density is The ratio of their terminal velocities would be, (1) (2) (3) (4)

Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$ , respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2} .$ The ratio of their terminal velocities would be, (1) $\frac{79}{36}$ (2) $\frac{79}{72}$ (3) $\frac{19}{36}$ (4) $\frac{39}{72}$

Physics | Physics

Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$ , respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2} .$ The ratio of their terminal velocities would be, (1) $\frac{79}{36}$ (2) $\frac{79}{72}$ (3) $\frac{19}{36}$ (4) $\frac{39}{72}$

Answer (1)
Solution:

As $V_{T}=\frac{2 a^{2}}{9 \eta}(\rho-\sigma) g$
$\begin{array}{l} V_{T_{1}}=\frac{2 \times(1)^{2}}{9 \eta}\left(\rho_{1}-0.1 \rho_{2}\right) g \\ V_{T_{1}}=\frac{2 \times 1}{9 \eta}\left(8 \rho_{2}-0.1 \rho_{2}\right) g \\ \cdot \\ V_{T_{2}}=\frac{2 \times(2)^{2}}{9 \eta}\left(\rho_{2}-0.1 \rho_{2}\right) g \\ \therefore \frac{V_{T_{1}}}{V_{T_{2}}}=\frac{7.9}{4(0.9)}=\frac{79}{36} \end{array}$