
A Water barrel having water up to depth ‘d’ is placed on a table of height ‘h’. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance ‘R’ from the barrel, then the value of‘d’ is:
A. \[\dfrac{{4h}}{{{R^2}}}\]
B. \[4h{R^2}\]
C. \[\dfrac{{{R^2}}}{{4h}}\]
D. \[\dfrac{h}{{4{R^2}}}\]
Answer
578.7k+ views
Hint: Use Torricelli’s theorem and the maximum height and range by the projectile motion. Torricelli’s theorem states that the velocity with which a liquid flows out through a narrow hole is equal to that which a freely falling body would acquire in falling through a vertical distance equal to the depth of the hole below the surface of the liquid.
Complete step by step answer:Stepwise solution:
If a liquid contained in a tank flows out from a hole, then by Torricelli’s equation, we have
$v = \sqrt {2gd} $ ………………(1)
Where, $v$is the velocity with which the liquid is flowing.
As the water is flowing through the hole, it has zero initial velocity in vertical direction. It flows out in the form of a parabolic jet and strikes the ground at a distance $R$ from the tank after the time $t$.
Using the relation $S = ut + \dfrac{1}{2}a{t^2}$
In vertical direction, $h = \dfrac{1}{2}g{t^2}$ $ \Rightarrow t = \sqrt {\dfrac{{2h}}{g}} $ ………………(2)
Therefore, the horizontal range is given by $R = vt$
Substituting the values of $v$ and $t$ from equations (1) and (2), we get
$R = \sqrt {2gd} \sqrt {\dfrac{{2h}}{g}} $ $ \Rightarrow R = \sqrt {4hd} \Rightarrow d = \dfrac{{{R^2}}}{{4h}}$
Hence the correct option is (C)
Note:Bernoulli’s theorem states the principle of conservation of energy applied to a liquid in motion. This theorem states that for the streamline flow of an ideal liquid, the sum of the pressure energy, potential energy and the kinetic energy per unit mass remains constant at every cross-section throughout the liquid. Torricelli extends the concept of Bernoulli’s to find the velocity of efflux, that is, the velocity with which a liquid flows out through a narrow hole.
Complete step by step answer:Stepwise solution:
If a liquid contained in a tank flows out from a hole, then by Torricelli’s equation, we have
$v = \sqrt {2gd} $ ………………(1)
Where, $v$is the velocity with which the liquid is flowing.
As the water is flowing through the hole, it has zero initial velocity in vertical direction. It flows out in the form of a parabolic jet and strikes the ground at a distance $R$ from the tank after the time $t$.
Using the relation $S = ut + \dfrac{1}{2}a{t^2}$
In vertical direction, $h = \dfrac{1}{2}g{t^2}$ $ \Rightarrow t = \sqrt {\dfrac{{2h}}{g}} $ ………………(2)
Therefore, the horizontal range is given by $R = vt$
Substituting the values of $v$ and $t$ from equations (1) and (2), we get
$R = \sqrt {2gd} \sqrt {\dfrac{{2h}}{g}} $ $ \Rightarrow R = \sqrt {4hd} \Rightarrow d = \dfrac{{{R^2}}}{{4h}}$
Hence the correct option is (C)
Note:Bernoulli’s theorem states the principle of conservation of energy applied to a liquid in motion. This theorem states that for the streamline flow of an ideal liquid, the sum of the pressure energy, potential energy and the kinetic energy per unit mass remains constant at every cross-section throughout the liquid. Torricelli extends the concept of Bernoulli’s to find the velocity of efflux, that is, the velocity with which a liquid flows out through a narrow hole.
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