Mercury is filled in a tube of radius $2 \mathrm{~cm}$ up to a height of $30 \mathrm{~cm}$. The force exerted by mercury on the bottom of the tube is. . . . . . $\mathrm{N}$.

(Given, atmospheric pressure $=10^5 \mathrm{Nm}^{-2}$, density of mercury $=1.36 \times 10^4 \mathrm{~kg} \mathrm{~m}^3, \mathrm{~g}=10 \mathrm{~ms}^2$, $\left.\pi=\frac{22}{7}\right)$

A small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta $ , then the sphere encounters an opposing force of

A table tennis ball has radius $(3 / 2) \times 10^{-2} m$ and mass $(22 / 7) \times 10^{-3} kg$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 m$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is (are) correct?

[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]

$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.

$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.

$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.

$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.

As the temperature of water increases, its viscosity

A spherical solid ball of volume $V$ is made of a material of density $\rho_1$ . It is falling through a liquid of density $\rho_2 (\rho_2 < \rho_1 )$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{viscous}= -kv^2 (k >0 )$,The terminal speed of the ball is