Two identical cylindrical vessels with their bases at the same level, each contains a liquid of density $d$. The height of the liquid in one vessel is $h_1$ and that in the other vessel is $h_2$. The area of either base is $A$. The work done by gravity in equalizing the levels when the two vessels are connected is

$A$ cylindrical tank of height $1 \ m$ and cross-sectional area $A = 4000 \ cm^2$ is initially empty. It is kept under a tap of cross-sectional area $a_1 = 1 \ cm^2$. Water starts flowing from the tap at $t = 0$ with a speed $v_1 = 2 \ m/s$. There is a small hole in the base of the tank of cross-sectional area $a_2 = 0.5 \ cm^2$. The variation of the height of water in the tank (in meters) with time $t$ is best depicted by:

$A$ liquid kept in a cylindrical vessel is rotated about a vertical axis passing through the center of the circular base. The difference in the heights of the liquid at the center of the vessel and its edge is ($R=$ radius of vessel,$\omega=$ angular velocity of rotation,$g=$ acceleration due to gravity).

The Karman line is a theoretical construct that separates the Earth's atmosphere from outer space. It is defined as the height at which the lift on an aircraft flying at the speed of a polar satellite $(8 \, km/s)$ is equal to its weight. Taking a fighter aircraft of wing area $30 \, m^2$ and mass $7500 \, kg$,the height of the Karman line above the ground will be in the range of .............. $km$. (Assume the density of air at height $h$ above the ground to be $\rho(h) = 1.2 e^{-h/10} \, kg/m^3$,where $h$ is in $km$,and the lift force to be $\frac{1}{2} \rho v^2 A$,where $v$ is the speed of the aircraft and $A$ is its wing area.)

$A$ bucket contains water filled up to a height $h = 15 \ cm$. The bucket is tied to a rope which is passed over a frictionless light pulley,and the other end of the rope is tied to a weight of mass which is half of that of the (bucket $+$ water). The water pressure above atmospheric pressure at the bottom is ....... $kPa$.

Two solid spheres $A$ and $B$ of equal volumes but of different densities $d_A$ and $d_B$ are connected by a string. They are fully immersed in a fluid of density $d_F$. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if:
$(A)$ $d_A < d_F$
$(B)$ $d_B > d_F$
$(C)$ $d_A + d_B = 2d_F$
$(D)$ $d_A > d_F$