$A$ cylindrical vessel of $90 \ cm$ height is kept filled up to the brim. It has four holes $1, 2, 3, 4$ which are respectively at heights of $20 \ cm, 30 \ cm, 45 \ cm,$ and $50 \ cm$ from the horizontal floor $PQ.$ The water falling at the maximum horizontal distance from the vessel comes from:

  • A
    Hole number $4$
  • B
    Hole number $3$
  • C
    Hole number $2$
  • D
    Hole number $1$

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$A$ cylindrical vessel filled with water up to a height of $H$ stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of the vessel and the plane is $\mu$ and the total mass of the water plus the vessel is $M$. What should be the minimum diameter of the hole so that the vessel begins to move on the floor if the plug is removed? (Here,the density of water is $\rho$)

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$A$ cylindrical tank has a hole of $1 \ cm^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \ cm^3/s$,then the maximum height up to which water can rise in the tank is: $(g = 9.8 \ m/s^2)$ (in $cm$)

$A$ cylinder of height $h$ is filled with water and is kept on a block of height $h/2$. The level of water in the cylinder is kept constant. Four holes numbered $1, 2, 3$ and $4$ are at the side of the cylinder and at heights $0, h/4, h/2$ and $3h/4$ from the base of the cylinder, respectively. When all four holes are opened together, the hole from which water will reach the farthest distance on the plane $PQ$ is the hole number:

$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water,the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

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