There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is
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A log of wood of mass $120 Kg$ floats in water. The weight that can be put on the raft to make it just sink, should be ....... $Kg$ (density of wood = $600 Kg/m^3$)
A fluid container is containing a liquid of density $\rho $ is accelerating upward with acceleration a along the inclined place of inclination $\alpha$ as shown. Then the angle of inclination $ \theta $ of free surface is :
Two immiscible liquid are filled in conical flask as shown in figure. The area of cross section is shown, a small hole of area a is made in lower end of cone. Find speed of liquid flow from hole
According to Bernoulli's equation $\frac{P}{{\rho g}} + h + \frac{1}{2}\,\frac{{{v^2}}}{g} = {\rm{constant}}$ The terms $A, B$ and $ C$ are generally called respectively:
A cubical block of wood of edge $10$ $cm$ and mass $0.92$ $kg$ floats on a tank of water with oil of rel. density $0.6$ to a depth of $4$ $cm$ above water. When the block attains equilibrium with four of its sides edges vertical
A tank $5 \,m$ high is half filled with water and then is filled to the top with oil of density $0.85 \,g/cm^3$. The pressure at the bottom of the tank, due to these liquids is ........ $g/cm^2$
Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be
The pressure acting on a submarine is $3 \times 10^{5}\;Pa$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is $1 \times 10^{5} \;Pa$ density of water is $10^{3}\, kg \,m ^{-3}, g =10 \,ms ^{-2}$ )