Why is it easier to swim in sea water than in river water ?
It is easier to swim in sea water than in a river water because, the sea water contains salt which increases the density of water and also increases its upthrust $\left[\mathrm{F}=m_{0} g=\right.$ (Volume $\times$ density)g], so the body sinks less in it and we swim easily.
A hemispherical bowl just floats without sinking in a liquid of density $1.2 × 10^3kg/m^3$. If outer diameter and the density of the bowl are $1 m$ and $2 × 10^4 kg/m^3$ respectively, then the inner diameter of the bowl will be........ $m$
A concrete sphere of radius $R$ has a cavity of radius $ r$ which is packed with sawdust. The specific gravities of concrete and sawdust are respectively $2.4$ and $0.3$ for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be
A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material is $\frac{27}{8}$ $w.r.t.$ water, the value of $r$ is$......R$
A fire hydrant delivers water of density $\rho $ at a volume rate $L$. The water travels vertically upward through the hydrant and then does $90^o$ turn to emerge horizontally at speed $V$. The pipe and nozzle have uniform cross-section throughout. The force exerted by the water on the corner of the hydrant is
A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_{\ell}$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is $\mathrm{T}_0$, the height of the liquid is $\mathrm{H}$ and the atmospheric pressure is $\mathrm{P}_0$ (Neglect surface tension).
Figure: $Image$
$1.$ As the bubble moves upwards, besides the buoyancy force the following forces are acting on it
$(A)$ Only the force of gravity
$(B)$ The force due to gravity and the force due to the pressure of the liquid
$(C)$ The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid
$(D)$ The force due to gravity and the force due to viscosity of the liquid
$2.$ When the gas bubble is at a height $\mathrm{y}$ from the bottom, its temperature is
$(A)$ $\mathrm{T}_0\left(\frac{\mathrm{P}_0+\rho_0 \mathrm{gH}}{\mathrm{P}_0+\rho_t \mathrm{gy}}\right)^{2 / 5}$
$(B)$ $T_0\left(\frac{P_0+\rho_t g(H-y)}{P_0+\rho_t g H}\right)^{2 / 5}$
$(C)$ $\mathrm{T}_0\left(\frac{\mathrm{P}_0+\rho_t \mathrm{gH}}{\mathrm{P}_0+\rho_t \mathrm{gy}}\right)^{3 / 5}$
$(D)$ $T_0\left(\frac{P_0+\rho_t g(H-y)}{P_0+\rho_t g H}\right)^{3 / 5}$
$3.$ The buoyancy force acting on the gas bubble is (Assume $R$ is the universal gas constant)
$(A)$ $\rho_t \mathrm{nRgT}_0 \frac{\left(\mathrm{P}_0+\rho_t \mathrm{gH}\right)^{2 / 5}}{\left(\mathrm{P}_0+\rho_t \mathrm{gy}\right)^{7 / 5}}$
$(B)$ $\frac{\rho_{\ell} \mathrm{nRgT}_0}{\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gH}\right)^{2 / 5}\left[\mathrm{P}_0+\rho_{\ell} \mathrm{g}(\mathrm{H}-\mathrm{y})\right]^{3 / 5}}$
$(C)$ $\rho_t \mathrm{nRgT} \frac{\left(\mathrm{P}_0+\rho_t g \mathrm{H}\right)^{3 / 5}}{\left(\mathrm{P}_0+\rho_t g \mathrm{~g}\right)^{8 / 5}}$
$(D)$ $\frac{\rho_{\ell} \mathrm{nRgT}_0}{\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gH}\right)^{3 / 5}\left[\mathrm{P}_0+\rho_t \mathrm{~g}(\mathrm{H}-\mathrm{y})\right]^{2 / 5}}$
Give the answer question $1,2,$ and $3.$