Velocity of Efflux and Torricelli's law Questions in English

(B) The rate of flow of water through a hole at the bottom of a bucket is given by Torricelli's Law,$v = \sqrt{2gh_{eff}}$,where $h_{eff}$ is the effective acceleration due to gravity.
In a stationary lift,the effective acceleration is $g_{eff} = g$. Thus,$R_0 \propto \sqrt{g}$.
When the lift moves upward with acceleration $a$,the effective gravity becomes $g_{eff} = g + a$. Since $g + a > g$,the rate of flow $R_u$ increases,so $R_u > R_0$.
When the lift moves downward with acceleration $a$,the effective gravity becomes $g_{eff} = g - a$. Since $g - a < g$,the rate of flow $R_d$ decreases,so $R_d < R_0$.
Combining these,we get $R_u > R_0 > R_d$.

(B) According to Torricelli's law,the velocity of efflux $v$ of a liquid from a small hole at a depth $h$ below the free surface is given by the formula:
$v = \sqrt{2gh}$
Given:
Height of the cylinder $h = 20\;m$
Acceleration due to gravity $g = 10\;m/s^2$
Substituting the values into the formula:
$v = \sqrt{2 \times 10 \times 20}$
$v = \sqrt{400}$
$v = 20\;m/s$
Therefore,the velocity of efflux is $20\;m/s$.

(A) The total pressure at the bottom of the tank is $P = P_{\text{atm}} + h\rho g = 3 \times 10^5 \text{ N/m}^2$.
Since the atmospheric pressure $P_{\text{atm}} = 1 \times 10^5 \text{ N/m}^2$,the gauge pressure (pressure due to the liquid column) is $P_l = h\rho g = P - P_{\text{atm}} = 3 \times 10^5 - 1 \times 10^5 = 2 \times 10^5 \text{ N/m}^2$.
The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gh}$.
Since $P_l = h\rho g$,we have $h = \frac{P_l}{\rho g}$.
Substituting this into the velocity formula: $v = \sqrt{2g \left( \frac{P_l}{\rho g} \right)} = \sqrt{\frac{2P_l}{\rho}}$.
Given the density of water $\rho = 10^3 \text{ kg/m}^3$,we get $v = \sqrt{\frac{2 \times 2 \times 10^5}{10^3}} = \sqrt{4 \times 10^2} = \sqrt{400} \text{ m/s}$.

(C) The time $t$ required to empty a cylindrical vessel of cross-sectional area $A$ through a hole of area $A_0$ at the bottom,when filled to a height $H$,is given by the formula:
$t = \frac{A}{A_0} \sqrt{\frac{2H}{g}}$
From this relation,we can see that the time $t$ is directly proportional to the square root of the height $H$:
$t \propto \sqrt{H}$
Therefore,the ratio of the times $t_2$ and $t_1$ for heights $H_2 = 4h$ and $H_1 = h$ is:
$\frac{t_2}{t_1} = \sqrt{\frac{H_2}{H_1}} = \sqrt{\frac{4h}{h}} = \sqrt{4} = 2$
Thus,$t_2 = 2t_1 = 2t$.

(A) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume of water flowing out per second.
Volume of water flowing out per second is given by $Q_{out} = A \sqrt{2gh}$, where $A = 1 \, cm^2$ is the area of the hole, $g = 980 \, cm/s^2$ is the acceleration due to gravity, and $h$ is the height of the water.
Volume of water flowing in per second is $Q_{in} = 70 \, cm^3/s$.
At maximum height, $Q_{in} = Q_{out}$.
$70 = 1 \times \sqrt{2 \times 980 \times h}$
$70 = \sqrt{1960 \times h}$
Squaring both sides: $4900 = 1960 \times h$
$h = \frac{4900}{1960} = 2.5 \, cm$.

(C) The velocity of water coming out of the hole (velocity of efflux) is given by Torricelli's law as $v = \sqrt{2gD}$.
The height of the hole from the bottom of the tank is $(H - D)$.
The time $t$ taken by the water to reach the ground is calculated using the equation of motion for a freely falling body: $(H - D) = \frac{1}{2}gt^2$,which gives $t = \sqrt{\frac{2(H - D)}{g}}$.
The horizontal distance $x$ covered by the water is the product of the horizontal velocity and the time taken to reach the ground:
$x = v \times t$
$x = \sqrt{2gD} \times \sqrt{\frac{2(H - D)}{g}}$
$x = \sqrt{2gD \times \frac{2(H - D)}{g}}$
$x = \sqrt{4D(H - D)}$
$x = 2\sqrt{D(H - D)}$

(B) The horizontal range $R$ of water issuing from a small hole at a depth $y$ below the free surface of a liquid in a vessel of total height $H$ is given by $R = 2\sqrt{y(H-y)}$.
Here,the total height of the water column is $H = 90 \ cm$.
The holes are at heights $h_1 = 20 \ cm, h_2 = 30 \ cm, h_3 = 45 \ cm,$ and $h_4 = 50 \ cm$ from the floor.
The depth of each hole from the free surface is $y = H - h$.
For hole $1$: $y_1 = 90 - 20 = 70 \ cm$.
For hole $2$: $y_2 = 90 - 30 = 60 \ cm$.
For hole $3$: $y_3 = 90 - 45 = 45 \ cm$.
For hole $4$: $y_4 = 90 - 50 = 40 \ cm$.
The horizontal range is maximum when the depth $y$ is equal to half the total height,i.e.,$y = H/2 = 90/2 = 45 \ cm$.
Comparing the depths,hole $3$ is at a depth of $45 \ cm$ from the free surface.
Therefore,the water falling at the maximum horizontal distance comes from hole $3$.

(B) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
The rate of flow is $\frac{dV}{dt} = A_0 v = A_0 \sqrt{2gh}$,where $A_0$ is the area of the orifice.
Since $V = Ah$ (where $A$ is the cross-sectional area of the vessel),we have $A \frac{dh}{dt} = -A_0 \sqrt{2gh}$.
Integrating this,the time $t$ taken to empty a vessel of height $H$ is $t = \int_{H}^{0} \frac{A}{-A_0 \sqrt{2g}} h^{-1/2} dh = \frac{A}{A_0} \sqrt{\frac{2H}{g}}$.
Thus,$t \propto \sqrt{H}$.
Given $t_1 = 10 \text{ minutes}$ for height $H$,for height $H/2$,the time $t_2$ is:
$t_2 = t_1 \sqrt{\frac{H/2}{H}} = \frac{t_1}{\sqrt{2}}$.
$t_2 = \frac{10}{\sqrt{2}} = 5\sqrt{2} \approx 5 \times 1.414 = 7.07 \text{ minutes}$.
Rounding to the nearest integer,the time is $7 \text{ minutes}$.

(C) Let $A$ be the area of cross-section of the hole,$v$ be the initial velocity of efflux,and $d$ be the density of water.
According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
The volume of water flowing out per second is $V = Av$.
The mass of water flowing out per second is $m = Avd$.
The rate of change of momentum (thrust force) is $F = \frac{dp}{dt} = (Avd)v = Adv^2$.
This thrust force acts upwards on the cylinder due to the reaction of the water jet.
Substituting $v^2 = 2gh$,the upward reaction force is $F = Ad(2gh) = 2Adgh$.
Given $A = \frac{1}{4} \ cm^2$,$d = 1 \ g/cm^3$,$g = 980 \ cm/s^2$,and $h = 25 \ cm$.
$F = 2 \times \frac{1}{4} \times 1 \times 980 \times 25 = 12250 \ dynes$.
Since $1 \ gm-wt = 980 \ dynes$,the force in $gm-wt$ is $F = \frac{12250}{980} = 12.5 \ gm-wt$.
As this force acts upwards,it causes a decrease in the effective weight of the cylinder on the balance.

(C) Let the velocities of the liquid exiting from the upper hole $A$ and the lower hole $B$ be $v_A$ and $v_B$ respectively.
According to the principle of conservation of momentum,the force exerted by the liquid jet is equal to the rate of change of momentum,$F = \frac{dp}{dt} = \rho a v^2$.
Since the holes are on opposite sides,the forces act in opposite directions. The net force $F$ required to keep the tank in equilibrium is:
$F = F_B - F_A = \rho a v_B^2 - \rho a v_A^2 = \rho a (v_B^2 - v_A^2)$
Applying Bernoulli's theorem between the two holes,where $h$ is the vertical distance between them:
$P_A + \frac{1}{2}\rho v_A^2 + \rho g h_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g h_B$
Taking the pressure at both holes to be atmospheric pressure $(P_A = P_B = P_{atm})$,we get:
$\frac{1}{2}\rho v_A^2 + \rho g h = \frac{1}{2}\rho v_B^2 + 0$ (taking the lower hole as reference level $0$)
$v_B^2 - v_A^2 = 2gh$
Substituting this into the force equation:
$F = \rho a (2gh) = 2\rho agh$.

(A) According to Torricelli's law,the velocity of efflux at height $h$ is $v = \sqrt{2gh}$.
The rate of volume flow out of the hole is $dV/dt = a \cdot v = a \sqrt{2gh}$.
Since the volume of liquid in the vessel decreases as the level drops,we have $dV = -A \cdot dh$ (where $dh$ is the change in height).
Equating the two expressions for $dV/dt$: $-A \cdot dh = a \sqrt{2gh} \cdot dt$.
Rearranging for $dt$: $dt = -\frac{A}{a \sqrt{2g}} \cdot h^{-1/2} \cdot dh$.
Integrating from $H_1$ to $H_2$ to find the total time $t$: $t = \int_{0}^{t} dt = -\frac{A}{a \sqrt{2g}} \int_{H_1}^{H_2} h^{-1/2} \cdot dh$.
$t = \frac{A}{a \sqrt{2g}} \int_{H_2}^{H_1} h^{-1/2} \cdot dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{H_2}^{H_1}$.
$t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_1} - \sqrt{H_2}]$.

(C) The time $t$ taken for the water level to drop from height $H_1$ to $H_2$ is given by the formula: $t = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{H_1} - \sqrt{H_2})$,where $A$ is the cross-sectional area of the container and $A_0$ is the area of the hole.
For the first interval (from $h$ to $\frac{h}{2}$): $t_1 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{\frac{h}{2}})$.
For the second interval (from $\frac{h}{2}$ to $0$): $t_2 = \frac{A}{A_0} \sqrt{\frac{2}{g}} (\sqrt{\frac{h}{2}} - \sqrt{0})$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{\sqrt{h} - \sqrt{\frac{h}{2}}}{\sqrt{\frac{h}{2}}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$.

(B) According to Torricelli's Law,the velocity of efflux $v$ of a liquid from a hole at a depth $h$ below the free surface is given by the formula: $v = \sqrt{2gh}$.
Given:
Height $h = 20 \ m$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting the values into the formula:
$v = \sqrt{2 \times 10 \times 20}$
$v = \sqrt{400}$
$v = 20 \ m/s$.
Therefore,the velocity of the water coming out is $20 \ m/s$.

(B) The pressure at the bottom of the tank is given by $P = h \rho g = 3 \times 10^{5} \, N/m^{2}$.
According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
Since $P = h \rho g$,we can write $h = P / (\rho g)$.
Substituting this into the velocity formula: $v = \sqrt{2g(P / \rho g)} = \sqrt{2P / \rho}$.
Given density of water $\rho = 10^{3} \, kg/m^{3}$ and $P = 3 \times 10^{5} \, N/m^{2}$.
$v = \sqrt{(2 \times 3 \times 10^{5}) / 10^{3}} = \sqrt{6 \times 10^{2}} = \sqrt{600} \, m/s$.

(C) The time $t$ taken for a vessel of cross-sectional area $A$ to drain through an orifice of area $A_0$ from an initial height $H$ is given by the formula: $t = \frac{A}{A_0} \sqrt{\frac{2H}{g}}$.
From this relation,we can see that the time $t$ is directly proportional to the square root of the height,i.e.,$t \propto \sqrt{H}$.
Given the initial height $H_1 = h$ and time $t_1 = t$.
For the new height $H_2 = 4h$,let the time be $t_2$.
Using the ratio: $\frac{t_2}{t_1} = \sqrt{\frac{H_2}{H_1}} = \sqrt{\frac{4h}{h}} = \sqrt{4} = 2$.
Therefore,$t_2 = 2t$.

(A) The rate of flow of water out of the hole is given by Torricelli's law: $Q_{out} = A \cdot v = A \sqrt{2gh}$.
Given the rate of water being poured in is $Q_{in} = 70 \, cm^3/s$.
For the water level to be at a maximum height, the rate of water flowing in must equal the rate of water flowing out: $Q_{in} = Q_{out}$.
Substituting the values: $70 = 1 \times \sqrt{2 \times 980 \times h}$.
Squaring both sides: $70^2 = 2 \times 980 \times h$.
$4900 = 1960 \times h$.
$h = \frac{4900}{1960} = 2.5 \, cm$.

(D) The velocity of the liquid through the orifice is given by Torricelli's law: $v = \sqrt{2gy}$.
The height of the orifice from the ground is $(h - y) + h = 2h - y$. The time taken by the liquid to reach the ground is:
$t = \sqrt{\frac{2(2h - y)}{g}}$.
The horizontal range $x$ is given by the product of horizontal velocity and time:
$x = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{2(2h - y)}{g}} = \sqrt{4y(2h - y)}$.
To find the maximum range,we maximize $x^2$ with respect to $y$:
$x^2 = 4y(2h - y) = 8hy - 4y^2$.
Differentiating with respect to $y$ and setting to zero:
$\frac{d(x^2)}{dy} = 8h - 8y = 0 \implies y = h$.
Substituting $y = h$ back into the expression for $x$:
$x_m = \sqrt{4h(2h - h)} = \sqrt{4h^2} = 2h$.
Since both $y = h$ and $x_m = 2h$ are correct,the correct option is $(d)$.

(A) The height of the water level above the hole is $D$. The height of the hole above the ground is $(H-D)$.
Using Torricelli's law,the velocity of the water jet emerging from the hole is $v = \sqrt{2gD}$.
The time $t$ taken by the water jet to reach the ground from the height $(H-D)$ is given by the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = H-D$,$u = 0$,and $a = g$.
$H-D = 0 + \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2(H-D)}{g}}$.
The horizontal distance $x$ (range) is the product of horizontal velocity and time:
$x = v \times t = \sqrt{2gD} \times \sqrt{\frac{2(H-D)}{g}}$.
$x = \sqrt{2gD \times \frac{2(H-D)}{g}} = \sqrt{4D(H-D)} = 2\sqrt{D(H-D)}$.

(D) According to Torricelli's Theorem,the velocity of efflux,i.e.,the velocity with which the liquid flows out of a hole,is equal to $\sqrt{2gh}$,where $h$ is the depth of the hole below the liquid surface.
Let the side of the cube be $a$. When the box is full,the depth of the spout below the liquid surface is $h = a$. Therefore,the initial speed is:
$v_0 = \sqrt{2ga}$
When the cubical box is half empty and tilted at $45^{\circ}$ such that the spout is at the lowest point,the liquid surface forms a plane passing through the diagonal of the cube. The height of the liquid surface above the spout is equal to the distance from the bottom corner to the diagonal plane,which is half the length of the face diagonal of the cube:
$h' = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$
Now,the speed of the wine from the spout is:
$v' = \sqrt{2gh'} = \sqrt{2g \left( \frac{a}{\sqrt{2}} \right)} = \sqrt{\frac{2ga}{\sqrt{2}}} = \frac{\sqrt{2ga}}{\sqrt[4]{2}} = \frac{v_0}{\sqrt[4]{2}}$

(B) According to the diagram,the tube is open to the atmosphere at the top and its lower end is submerged in water at a depth $h = 20 \ cm = 0.2 \ m$ above the hole.
Applying Bernoulli's theorem between the surface of the water in the tube (which is at atmospheric pressure $P_0$) and the hole (also at atmospheric pressure $P_0$):
$P_0 + \rho g h + 0 = P_0 + 0 + \frac{1}{2} \rho v^2$
Here,$h$ is the vertical distance between the water level in the tube and the hole.
$v^2 = 2gh$
$v = \sqrt{2gh}$
Given $g = 10 \ m/s^2$ and $h = 0.2 \ m$:
$v = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \ m/s$.

(C) The time taken to decrease the water level from $H_1$ to $H_2$ in a tank of cross-sectional area $A$ with an orifice of area $a$ is given by $t = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H_1} - \sqrt{H_2})$.
For the first interval,the height decreases from $H$ to $H/\eta$. Thus,$T_1 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H} - \sqrt{H/\eta})$.
For the second interval,the height decreases from $H/\eta$ to $0$. Thus,$T_2 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H/\eta} - \sqrt{0}) = \frac{A}{a} \sqrt{\frac{2}{g}} \sqrt{H/\eta}$.
Given $T_1 = T_2$,we equate the expressions:
$\sqrt{H} - \sqrt{H/\eta} = \sqrt{H/\eta}$.
$\sqrt{H} = 2\sqrt{H/\eta}$.
Squaring both sides,$H = 4(H/\eta)$.
Therefore,$\eta = 4$.

(D) According to Torricelli's Law,the velocity of efflux is $v = \sqrt{2gh}$.
The rate of change of volume of water in the vessel is given by $\frac{dV}{dt} = -A_h v$,where $A_h$ is the area of the hole and $A$ is the cross-sectional area of the vessel.
$A \frac{dh}{dt} = -A_h \sqrt{2gh}$
Integrating this expression from height $H$ to $0$ in time $t_0$:
$\int_{H}^{0} \frac{dh}{\sqrt{h}} = -\frac{A_h}{A} \sqrt{2g} \int_{0}^{t_0} dt$
$2\sqrt{H} = \frac{A_h}{A} \sqrt{2g} \cdot t_0$
This implies $t_0 \propto \sqrt{H}$.
For a new height $H' = 4H$,the new time $t'$ is:
$\frac{t'}{t_0} = \sqrt{\frac{H'}{H}} = \sqrt{\frac{4H}{H}} = 2$
Therefore,$t' = 2t_0$.

(C) Let $H$ be the height of the liquid in the tank and $A_h$ be the area of the hole. The area of cross-section of the tank is $A_t = N A_h$.
The velocity of efflux $v$ of the liquid from the hole at the center (at depth $h = H/2$) is given by Torricelli's law: $v = \sqrt{2gh} = \sqrt{2g(H/2)} = \sqrt{gH}$.
The force $F$ exerted on the tank due to the efflux of the liquid is given by $F = \rho A_h v^2$,where $\rho$ is the density of the liquid.
Substituting $v^2 = gH$,we get $F = \rho A_h (gH)$.
The mass of the liquid in the tank is $M = \rho V = \rho (A_t H) = \rho (N A_h H)$.
The initial acceleration $a$ of the tank is given by Newton's second law: $a = \frac{F}{M}$.
$a = \frac{\rho A_h g H}{\rho N A_h H} = \frac{g}{N}$.

(B) The rate of inflow is given by $Q_{in} = 10^{-4} \; m^3/s$.
The rate of outflow through the hole is given by $Q_{out} = A_{hole} \times v$, where $v$ is the velocity of efflux.
According to Torricelli's law, the velocity of efflux is $v = \sqrt{2gh}$.
For the water level to remain constant (steady state), the rate of inflow must equal the rate of outflow: $Q_{in} = Q_{out}$.
$10^{-4} = (10^{-4}) \times \sqrt{2gh}$.
Dividing both sides by $10^{-4}$, we get $1 = \sqrt{2gh}$.
Squaring both sides, $1 = 2gh$.
Solving for $h$, $h = \frac{1}{2g}$.
Using $g = 9.8 \; m/s^2$, $h = \frac{1}{2 \times 9.8} = \frac{1}{19.6} \approx 0.051 \; m$.

(C) Let the total height of the liquid surface from the ground be $h_{total} = 2H + H = 3H$.
Let the hole be at a height $y$ from the ground. The depth of the hole from the liquid surface is $h = 3H - y$.
The velocity of efflux is $v = \sqrt{2gh} = \sqrt{2g(3H - y)}$.
The time taken for the liquid to reach the ground is $t = \sqrt{\frac{2y}{g}}$.
The horizontal range is $R = v \cdot t = \sqrt{2g(3H - y)} \cdot \sqrt{\frac{2y}{g}} = 2\sqrt{y(3H - y)} = 2\sqrt{3Hy - y^2}$.
For maximum range,the term inside the square root $f(y) = 3Hy - y^2$ must be maximum.
Taking the derivative with respect to $y$ and setting it to zero: $\frac{df}{dy} = 3H - 2y = 0$.
Thus,$y = 1.5H$.
Therefore,the distance from the ground is $1.5H$.

(D) The range of efflux $x$ for a hole at height $h$ from the bottom in a vessel filled with liquid to a total height $H$ is given by $x = 2 \sqrt{(H - h)h}$.
Given that the range of efflux is the same for two holes at heights $h_1$ and $h_2$,we have:
$2 \sqrt{(H - h_1)h_1} = 2 \sqrt{(H - h_2)h_2}$
$(H - h_1)h_1 = (H - h_2)h_2$
$Hh_1 - h_1^2 = Hh_2 - h_2^2$
$H(h_1 - h_2) = h_1^2 - h_2^2$
$H(h_1 - h_2) = (h_1 - h_2)(h_1 + h_2)$
$H = h_1 + h_2$
The range of efflux $x = 2 \sqrt{Hh - h^2}$ is maximum when the term inside the square root is maximum. Let $f(h) = Hh - h^2$. For maximum,$f'(h) = 0$.
$H - 2h = 0 \Rightarrow h = \frac{H}{2}$.
Substituting $H = h_1 + h_2$,we get the height for maximum range as $h = \frac{h_1 + h_2}{2}$.

(D) The range of water issuing from a hole at depth $h$ is given by $R = 2\sqrt{h(H-h)}$,where $H$ is the total height of the water column. Assuming the hole is at the bottom $(h=H)$,the range is $R = 2\sqrt{h^2} = 2h$.
For a hole at depth $h=10 \ m$,the velocity of efflux is $v = \sqrt{2gh}$. The range is $R = v \cdot t$,where $t = \sqrt{2(H-h)/g}$.
If we apply an extra pressure $P_{ext}$ on the surface,the effective pressure at the hole becomes $P_{eff} = P_{atm} + P_{ext} + \rho gh$. The velocity of efflux becomes $v' = \sqrt{2(P_{ext} + \rho gh)/\rho} = \sqrt{2gh + 2P_{ext}/\rho}$.
To make the range $2R$,the velocity of efflux must become $2v$ (since time of flight remains constant if the hole position is fixed).
Thus,$v' = 2v \implies v'^2 = 4v^2$.
$2gh + 2P_{ext}/\rho = 4(2gh) = 8gh$.
$2P_{ext}/\rho = 6gh \implies P_{ext} = 3\rho gh$.
Given $\rho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$,and $h = 10 \ m$:
$P_{ext} = 3 \times 10^3 \times 10 \times 10 = 3 \times 10^5 \ Pa$.
Since $1 \ atm = 10^5 \ Pa$,the extra pressure required is $3 \ atm$.

(B) Let $d$ be the depth of water in the barrel.
According to Torricelli's law,the velocity of efflux $v$ at the base is $v = \sqrt{2gd}$.
The time $t$ taken by the water to reach the ground from the table of height $h$ is given by $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.
The horizontal distance $R$ covered by the water stream is $R = v \times t$.
Substituting the values,$R = \sqrt{2gd} \times \sqrt{\frac{2h}{g}} = \sqrt{4dh} = 2\sqrt{dh}$.
Squaring both sides,$R^2 = 4dh$.
Therefore,the depth of water $d = \frac{R^2}{4h}$.

(B) The pressure at the hole is due to the atmospheric pressure,the weight of the piston,and the height of the water column.
Let $A = 1000 \ cm^2 = 0.1 \ m^2$ be the area of the piston and $m = 10 \ kg$ be its mass.
The pressure exerted by the piston is $P_p = \dfrac{mg}{A} = \dfrac{10 \times 9.8}{0.1} = 980 \ Pa$.
The equivalent height of water column $h_p$ that produces this pressure is $h_p = \dfrac{P_p}{\rho g} = \dfrac{980}{1000 \times 9.8} = 0.1 \ m = 10 \ cm$.
The total effective height of the water column above the hole is $H_{eff} = h_p + h = 10 \ cm + 50 \ cm = 60 \ cm = 0.6 \ m$.
Using Torricelli's law,the velocity of efflux is $v = \sqrt{2gH_{eff}}$.
$v = \sqrt{2 \times 9.8 \times 0.6} = \sqrt{11.76} \approx 3.43 \ m/s$.
Thus,the velocity is approximately $3.4 \ m/s$.

(D) For the first hole at depth $h/2$ in the liquid of density $\rho$:
Using Torricelli's law,$v_1 = \sqrt{2g(h/2)} = \sqrt{gh}$.
For the second hole at depth $3h/2$ from the surface,it is at a depth $h/2$ within the liquid of density $2\rho$:
Applying Bernoulli's theorem between the top surface and the second hole:
$P_0 + \rho gh + 2\rho g(h/2) = P_0 + \frac{1}{2}(2\rho)v_2^2$
$\rho gh + \rho gh = \rho v_2^2$
$2\rho gh = \rho v_2^2$
$v_2 = \sqrt{2gh}$.
Now,calculating the ratio:
$\frac{v_1}{v_2} = \frac{\sqrt{gh}}{\sqrt{2gh}} = \frac{1}{\sqrt{2}}$.

(D) In steady state,the volume flow rate of water entering the vessel must be equal to the volume flow rate of water leaving the vessel through the hole.
Let $Q$ be the rate of inflow,$a$ be the area of the hole,and $h$ be the height of the water level.
The rate of outflow is given by Torricelli's law: $Q_{out} = a v = a \sqrt{2gh}$.
Equating inflow and outflow: $Q = a \sqrt{2gh}$.
Rearranging for $h$: $h = \frac{Q^2}{2 g a^2}$.
Given $Q = 100 \ cm^3 s^{-1}$,$a = 1 \ cm^2$,and $g = 1000 \ cm s^{-2}$.
Substituting the values: $h = \frac{(100)^2}{2 \times 1000 \times (1)^2} = \frac{10000}{2000} = 5 \ cm$.

(A) According to Torricelli's law,the velocity of efflux $v$ from a small hole at a depth $h$ below the free surface is given by $v = \sqrt{2gh}$.
The rate of volume flow out of the hole is $\frac{dV}{dt} = a \cdot v = a \sqrt{2gh}$,where $a$ is the area of the hole.
Since the tanks have the same height $h$ and the holes are identical (same area $a$),the velocity of efflux and the initial rate of flow are the same for all three tanks.
However,the time taken to empty depends on the volume $V$ and the rate of flow. Since the total volume $V$ is the same for all tanks and the rate of flow is determined by the height $h$ (which is also the same),the time required to empty the tanks is independent of the shape of the container.
Therefore,all three tanks will take the same amount of time to empty.

(A) The force exerted by the water jet exiting the hole is given by $F = \rho A v^2$,where $A$ is the area of the hole and $v$ is the velocity of the water.
Using Torricelli's Law,the velocity of the water at the bottom is $v = \sqrt{2gH}$.
Thus,the force is $F = \rho A (2gH)$.
For the vessel to begin moving,this force must overcome the limiting friction force,$f_s = \mu Mg$.
Equating the two: $\rho A (2gH) = \mu Mg$.
Substituting $A = \frac{\pi D^2}{4}$,we get $\rho \left( \frac{\pi D^2}{4} \right) (2gH) = \mu Mg$.
Simplifying the equation: $\frac{\rho \pi D^2 g H}{2} = \mu Mg$.
Solving for $D$: $D^2 = \frac{2\mu M}{\pi \rho H}$.
Therefore,$D = \sqrt{\frac{2\mu M}{\pi \rho H}}$.

(B) According to Torricelli's law,the velocity of efflux $v$ is given by $v = \sqrt{2gy}$.
The time taken for the liquid to reach the ground from a height $(H-y)$ is $t = \sqrt{\frac{2(H-y)}{g}}$.
The horizontal range $x$ is given by $x = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{2(H-y)}{g}} = 2\sqrt{y(H-y)}$.
To find the maximum range,we differentiate $x^2 = 4(Hy - y^2)$ with respect to $y$ and set it to zero:
$\frac{d}{dy}(4Hy - 4y^2) = 4H - 8y = 0 \implies y = \frac{H}{2}$.
Substituting $y = \frac{H}{2}$ into the expression for $x$,we get $x_{max} = 2\sqrt{\frac{H}{2}(H - \frac{H}{2})} = 2\sqrt{\frac{H^2}{4}} = H$.
Therefore,$x$ is maximum at $y = \frac{H}{2}$ and the maximum value is $H$.
Thus,option $B$ is the correct answer.

(C) Applying Bernoulli's principle at the top surface of the liquid and at the hole:
$P_0 + h_1\rho_1g + h_2\rho_2g = P_0 + \frac{1}{2}\rho_1v^2$
Here,$P_0$ is the atmospheric pressure,and we assume the velocity of the top surface is negligible compared to the velocity $v$ at the hole.
Subtracting $P_0$ from both sides:
$h_1\rho_1g + h_2\rho_2g = \frac{1}{2}\rho_1v^2$
Dividing by $\rho_1$:
$h_1g + \frac{h_2\rho_2g}{\rho_1} = \frac{1}{2}v^2$
$v^2 = 2g(h_1 + \frac{h_2\rho_2}{\rho_1})$
$v = \sqrt{2g(h_1 + \frac{h_2\rho_2}{\rho_1})}$

(A) According to Bernoulli's principle for the top surface of the water and the hole:
$P_{top} + \rho gh + \frac{1}{2}\rho v_{top}^{2} = P_{atm} + \frac{1}{2}\rho v^{2}$
Since the vessel is closed and the top pressure is $0.9 P_{0}$,and assuming $v_{top} \approx 0$:
$0.9 P_{0} + \rho gh = P_{0} + \frac{1}{2} \rho v^{2}$
$\rho gh - 0.1 P_{0} = \frac{1}{2} \rho v^{2}$
Substituting the values: $10^{3} \times 10 \times (9/4) - 0.1 \times 10^{5} = \frac{1}{2} \times 10^{3} \times v^{2}$
$22500 - 10000 = 500 v^{2}$
$12500 = 500 v^{2}$
$v^{2} = 25$
$v = 5 \text{ m/s}$

(B) The total height of the water in the tank is $H = 1 \, m = 100 \, cm$.
The depth of the hole from the top surface is $h = 20 \, cm$.
The height of the hole from the bottom of the tank is $y = H - h = 100 - 20 = 80 \, cm$.
The velocity of efflux is $v = \sqrt{2gh}$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2(H-h)}{g}}$.
The horizontal range $R$ is given by $R = v \times t = \sqrt{2gh} \times \sqrt{\frac{2(H-h)}{g}} = 2 \sqrt{h(H-h)}$.
Substituting the values: $R = 2 \sqrt{20 \times (100 - 20)} = 2 \sqrt{20 \times 80} = 2 \sqrt{1600} = 2 \times 40 = 80 \, cm$.

(D) Let the top surface of the liquid be point $1$ and the hole be point $2$. Applying Bernoulli's theorem between the top surface (point $1$) and the hole (point $2$):
$P_0 + \rho_1 g \left( \frac{H}{2} \right) + \rho_2 g \left( \frac{H}{2} - h \right) = P_0 + \frac{1}{2} \rho_2 v^2$
Here,$\rho_1 = d$,$\rho_2 = 2d$,and $P_0$ is the atmospheric pressure.
Substituting the values:
$d g \left( \frac{H}{2} \right) + 2d g \left( \frac{H}{2} - h \right) = \frac{1}{2} (2d) v^2$
$g \left( \frac{H}{2} + H - 2h \right) = v^2$
$g \left( \frac{3H}{2} - 2h \right) = v^2$
$v^2 = g \left( \frac{3H - 4h}{2} \right)$
$v = \sqrt{\frac{(3H - 4h)g}{2}}$

(A) The area of the tank $A = \pi R^2$,where $R = 0.08\,m$.
The area of the hole $a = \pi r^2$,where $r = 5 \times 10^{-3}\,m$.
According to the equation of continuity and Torricelli's law,the rate of change of volume is $A \frac{dh}{dt} = -a \sqrt{2gh}$.
Rearranging for time $t$: $dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.
Integrating from $h = 0.16\,m$ to $h = 0$:
$t = \frac{A}{a \sqrt{2g}} \int_{0}^{0.16} h^{-1/2} dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{0}^{0.16}$.
$t = \frac{\pi (0.08)^2}{\pi (5 \times 10^{-3})^2 \sqrt{2 \times 9.8}} \times 2 \times \sqrt{0.16}$.
$t = \frac{0.0064}{25 \times 10^{-6} \times 4.427} \times 0.8 \approx 46.26\,s$.

(A) Let $v_1$ be the velocity of water at the hole (exit velocity) and $v_2$ be the velocity of water when it hits the ground.
Using Torricelli's Law,the velocity at the hole is $v_1 = \sqrt{2gH}$.
Using the equation of motion for a freely falling body,the velocity at the ground is $v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{2gH + 2gh} = \sqrt{2g(H + h)}$.
According to the equation of continuity,the volume flow rate remains constant:
$A_1 v_1 = A_2 v_2$
$\pi r^2 v_1 = \pi x^2 v_2$
$r^2 \sqrt{2gH} = x^2 \sqrt{2g(H + h)}$
Squaring both sides:
$r^4 (2gH) = x^4 (2g(H + h))$
$x^4 = r^4 \frac{H}{H + h}$
$x = r \left( \frac{H}{H + h} \right)^{\frac{1}{4}}$

(B) According to Bernoulli's principle,the pressure at the bottom of the tank is equal to the sum of the hydrostatic pressures of the two liquids.
$P = h_w \rho_w g + h_k \rho_k g$
Here,$h_w = 3\,m$,$\rho_w = 1000\,kg/m^3$,$h_k = 2\,m$,and $\rho_k = 0.8 \times 1000 = 800\,kg/m^3$.
$P = (3 \times 1000 \times 10) + (2 \times 800 \times 10) = 30000 + 16000 = 46000\,Pa$.
Using Torricelli's law for the exit velocity $v$ at the bottom,where the pressure energy is converted into kinetic energy:
$P = \frac{1}{2} \rho_w v^2$
$46000 = \frac{1}{2} \times 1000 \times v^2$
$v^2 = \frac{46000 \times 2}{1000} = 92$
$v = \sqrt{92} \approx 9.6\,ms^{-1}$.

(B) Let the rate of fall of the water level be $-\frac{dh}{dt}$.
According to the equation of continuity,the volume of water leaving the hole per unit time is equal to the volume of water lost by the vessel per unit time.
$A \left( -\frac{dh}{dt} \right) = a_{hole} \cdot v$
Where $a_{hole} = \pi a^2$ is the area of the hole and $v = \sqrt{2gh}$ is the velocity of efflux (Torricelli's Law).
So,$A \left( -\frac{dh}{dt} \right) = \pi a^2 \sqrt{2gh}$.
Rearranging the terms to integrate:
$dt = -\frac{A}{\pi a^2 \sqrt{2g}} h^{-1/2} dh$.
Integrating from $t = 0$ to $T$ (total time) and $h = h$ to $0$:
$\int_0^T dt = -\frac{A}{\pi a^2 \sqrt{2g}} \int_h^0 h^{-1/2} dh$.
$T = -\frac{A}{\pi a^2 \sqrt{2g}} \left[ \frac{h^{1/2}}{1/2} \right]_h^0$.
$T = -\frac{A}{\pi a^2 \sqrt{2g}} \cdot 2 [0 - \sqrt{h}] = \frac{2A}{\pi a^2 \sqrt{2g}} \sqrt{h}$.
$T = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}} = \frac{\sqrt{2}A}{\pi a^2} \sqrt{\frac{h}{g}}$.

(A) The rate of flow of water (volume flow rate) is given by $Q = A \times v$,where $A$ is the area of the hole and $v$ is the velocity of efflux.
According to Torricelli's law,the velocity of efflux at a depth $d$ is $v = \sqrt{2gd}$.
For the square hole at depth $h$: Area $A_1 = l^2$,velocity $v_1 = \sqrt{2gh}$. Thus,$Q_1 = l^2 \sqrt{2gh}$.
For the circular hole at depth $4h$: Area $A_2 = \pi r^2$,velocity $v_2 = \sqrt{2g(4h)} = 2\sqrt{2gh}$. Thus,$Q_2 = \pi r^2 (2\sqrt{2gh})$.
Given that the rates of flow are equal,$Q_1 = Q_2$:
$l^2 \sqrt{2gh} = 2\pi r^2 \sqrt{2gh}$
$l^2 = 2\pi r^2$
$r^2 = \frac{l^2}{2\pi}$
$r = \frac{l}{\sqrt{2\pi}}$

(A) The horizontal range $R$ of water issuing from a hole at depth $y$ from the surface in a tank of total height $H$ is given by $R = 2\sqrt{y(H-y)}$.
For hole $A$,the depth from the surface is $h_1$,so the range is $R_A = 2\sqrt{h_1(H - h_1)}$.
For hole $B$,the height from the bottom is $h_2$,so the depth from the surface is $(H - h_2)$. The range is $R_B = 2\sqrt{(H - h_2)h_2}$.
Since the ranges are equal,$R_A = R_B$:
$2\sqrt{h_1(H - h_1)} = 2\sqrt{(H - h_2)h_2}$
Squaring both sides:
$h_1(H - h_1) = h_2(H - h_2)$
$Hh_1 - h_1^2 = Hh_2 - h_2^2$
$H(h_1 - h_2) = h_1^2 - h_2^2$
$H(h_1 - h_2) = (h_1 - h_2)(h_1 + h_2)$
Assuming $h_1 \neq h_2$,we get $H = h_1 + h_2$.
This implies that the ratio $h_1/h_2$ is not a fixed constant but depends on the specific values of $h_1, h_2$ and $H$ such that their sum is $H$. Thus,the ratio depends on $H$.

(B) The volume flow rate is given by $Q = A \cdot v$,where $v = \sqrt{2gh}$ is the velocity of efflux.
Given: $Q = 0.74\,m^3/min = \frac{0.74}{60}\,m^3/s$,radius $r = 2\,cm = 0.02\,m$,and $g = 9.8\,m/s^2$.
The area of the opening is $A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 4 \times 10^{-4}\,m^2$.
Substituting these into the equation $Q = A\sqrt{2gh}$:
$\frac{0.74}{60} = (3.14 \times 4 \times 10^{-4}) \sqrt{2 \times 9.8 \times h}$.
$\frac{0.01233}{1.256 \times 10^{-3}} = \sqrt{19.6h}$.
$9.816 = \sqrt{19.6h}$.
Squaring both sides: $96.36 = 19.6h$.
$h = \frac{96.36}{19.6} \approx 4.92\,m$.
Thus,the depth is close to $4.8\,m$.

(A) The rate of change of volume of water in the tank is given by the difference between the inflow rate and the outflow rate.
$\frac{dV}{dt} = Q_{in} - Q_{out} = 0$ (since the height remains steady).
$Q_{in} = 10^{-4} \, m^3 s^{-1}$.
The outflow rate is given by Torricelli's law: $Q_{out} = a \sqrt{2gh}$, where $a = 1 \, cm^2 = 10^{-4} \, m^2$ and $g = 9.8 \, m s^{-2}$.
Equating inflow and outflow: $10^{-4} = 10^{-4} \sqrt{2 \times 9.8 \times h}$.
$1 = \sqrt{19.6 \times h}$.
Squaring both sides: $1 = 19.6 \times h$.
$h = \frac{1}{19.6} \approx 0.051 \, m$.
Converting to centimeters: $h = 0.051 \times 100 = 5.1 \, cm$.

(D) When a bucket is in free fall, it experiences an acceleration of $g$ downwards. The water inside the bucket also experiences the same acceleration $g$ downwards. Therefore, the effective gravitational force acting on the water relative to the bucket becomes zero. Since the pressure at any depth $h$ in a fluid is given by $P = P_{atm} + \rho g_{eff} h$, and here $g_{eff} = 0$, the pressure at the hole becomes equal to the atmospheric pressure. Consequently, there is no pressure difference to drive the water out of the hole, and the water stops flowing.

(B) According to Torricelli's law,the velocity of efflux is given by $v = \sqrt{2gh}$,where $h$ is the height of the water level above the hole.
Since $v$ depends only on the height $h$ and the acceleration due to gravity $g$,the velocity of outflow remains unchanged when the size of the hole is increased.
The volume of water flowing out per second (rate of flow) is given by $Q = A \times v$,where $A$ is the area of the hole.
Since the size (area $A$) of the hole increases,the volume of water flowing out per second will increase.

(C) Let $d_w$ and $d_o$ be the densities of water and oil respectively. The pressure at the bottom of the tank is given by $P = h_w d_w g + h_o d_o g$.
We can find the equivalent height $h$ of a water column that exerts the same pressure at the bottom: $h d_w g = h_w d_w g + h_o d_o g$.
Substituting the given values: $h = h_w + \frac{h_o d_o}{d_w} = 100 + \frac{400 \times 0.9}{1} = 100 + 360 = 460 \, cm$.
According to Torricelli's theorem,the velocity of efflux $v$ is given by $v = \sqrt{2gh}$.
Substituting the values: $v = \sqrt{2 \times 980 \times 460} = \sqrt{920 \times 980} \, cm/s$.

(A) The velocity of efflux $(v)$ of water from a hole at depth $h$ is given by Torricelli's law: $v = \sqrt{2gh}$.
Given: $h = 5\, m$,$g = 10\, m/s^2$,and area $A = 1\, cm^2 = 10^{-4}\, m^2$.
Calculating velocity: $v = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\, m/s$.
The volume flow rate $(Q)$ is given by $Q = A \times v$.
Substituting the values: $Q = 10^{-4}\, m^2 \times 10\, m/s = 10^{-3}\, m^3/s$.