Velocity of Efflux and Torricelli's law Questions in English

(D) The horizontal range $x$ of a liquid jet emerging from a hole at a depth $h$ below the free surface of a liquid in a tank of total height $H$ is given by the formula:
$x = 2\sqrt{h(H-h)}$
Given that the depth of the hole from the upper surface is $h = \frac{3H}{4}$.
Substituting this value into the formula:
$x = 2\sqrt{\frac{3H}{4} \left(H - \frac{3H}{4}\right)}$
$x = 2\sqrt{\frac{3H}{4} \times \frac{H}{4}}$
$x = 2\sqrt{\frac{3H^2}{16}}$
$x = 2 \times \frac{H}{4} \sqrt{3}$
$x = \frac{\sqrt{3}H}{2}$

(B) The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gh}$.
The total pressure at the bottom of the tank is the sum of atmospheric pressure $(P_{atm})$ and gauge pressure due to the water column $(h\rho g)$.
Given,$P_{total} = P_{atm} + h\rho g = 3 \, atm$.
Since $P_{atm} = 1 \, atm$,the gauge pressure is $h\rho g = 3 \, atm - 1 \, atm = 2 \, atm$.
Substituting the values: $h\rho g = 2 \times 10^5 \, N/m^2$.
$gh = \frac{2 \times 10^5}{\rho} = \frac{2 \times 10^5}{10^3} = 200 \, m^2/s^2$.
Now,substituting $gh$ into the velocity formula:
$v = \sqrt{2 \times (gh)} = \sqrt{2 \times 200} = \sqrt{400} \, m/s$.

(A) The velocity of efflux $v$ is given by the formula derived from Bernoulli's principle:
$v = \sqrt{\frac{2gh}{1 - (A_0/A)^2}}$
where $h$ is the height of the water column above the hole.
Given:
Total height of water = $3\,m$
Height of hole from bottom = $52.5\,cm = 0.525\,m$
$h = 3 - 0.525 = 2.475\,m$
$A_0/A = 0.1$
$g = 10\,m/s^2$
Substituting these values into the formula for $v^2$:
$v^2 = \frac{2gh}{1 - (A_0/A)^2}$
$v^2 = \frac{2 \times 10 \times 2.475}{1 - (0.1)^2}$
$v^2 = \frac{49.5}{1 - 0.01}$
$v^2 = \frac{49.5}{0.99} = 50\,m^2/s^2$

(A) Let the total height of the tank be $H = 10\,m$. The holes are at heights $h_1 = 3\,m$ and $h_2 = 7\,m$ from the base.
The depth of the holes from the top surface are $y_1 = H - h_1 = 10 - 3 = 7\,m$ and $y_2 = H - h_2 = 10 - 7 = 3\,m$.
The horizontal range $R$ of water issuing from a hole at depth $y$ is given by $R = 2\sqrt{y(H-y)}$.
For the first hole $(y_1 = 7\,m)$: $R_1 = 2\sqrt{7(10-7)} = 2\sqrt{7 \times 3} = 2\sqrt{21}\,m$.
For the second hole $(y_2 = 3\,m)$: $R_2 = 2\sqrt{3(10-3)} = 2\sqrt{3 \times 7} = 2\sqrt{21}\,m$.
Since $R_1 = R_2$,the water from both holes will fall at the same spot.

(A) The velocity of efflux $v$ is given by Torricelli's law: $v = \sqrt{2gh}$.
The volume flow rate $Q$ is given by $Q = A \times v = A \sqrt{2gh}$.
Given: Area $A = 2 \; mm^{2} = 2 \times 10^{-6} \; m^{2}$,height $h = 2 \; m$,and $g = 10 \; m/s^{2}$.
Substituting the values:
$Q = (2 \times 10^{-6}) \sqrt{2 \times 10 \times 2}$
$Q = (2 \times 10^{-6}) \sqrt{40}$
$Q = (2 \times 10^{-6}) \times 2\sqrt{10}$
$Q = 4 \sqrt{10} \times 10^{-6} \; m^{3}/s$.
Since $\sqrt{10} \approx 3.162$,
$Q \approx 4 \times 3.162 \times 10^{-6} \; m^{3}/s = 12.648 \times 10^{-6} \; m^{3}/s$.
Thus,the rate of flow is nearly $12.6 \times 10^{-6} \; m^{3}/s$.

(N/A) Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body.
Consider a tank containing a liquid of density $\rho$ with a small hole in its side at a height $y_{1}$ from the bottom. The air above the liquid,whose surface is at height $y_{2}$,is at pressure $P$.
The velocities $v_{1}$ and $v_{2}$ are at points $1$ and $2$ respectively. Using the equation of continuity for points $1$ and $2$:
$A_{1} v_{1} = A_{2} v_{2}$
$v_{2} = \frac{A_{1} v_{1}}{A_{2}}$
$A_{2}$ is the cross-sectional area of the tank and $A_{1}$ is the cross-sectional area of the hole.
Since $A_{2} \gg A_{1}$,it follows that $v_{2} \ll v_{1}$,so we can approximate $v_{2} \approx 0$.
Using Bernoulli's equation at points $1$ and $2$:
$P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho g y_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho g y_{2}$
Here,$P_{1} = P_{a}$ (atmospheric pressure),$P_{2} = P$,and $v_{2} = 0$. Substituting these:
$P_{a} + \frac{1}{2} \rho v_{1}^{2} + \rho g y_{1} = P + \rho g y_{2}$
$\frac{1}{2} \rho v_{1}^{2} = (P - P_{a}) + \rho g (y_{2} - y_{1})$
Let $h = y_{2} - y_{1}$ be the height of the liquid above the hole.
$\frac{1}{2} \rho v_{1}^{2} = (P - P_{a}) + \rho g h$
$v_{1} = \sqrt{2g h + \frac{2(P - P_{a})}{\rho}}$
If the tank is open to the atmosphere,$P = P_{a}$,then:
$v_{1} = \sqrt{2gh}$
This is Torricelli's law.

(N/A) Efflux refers to the process of a fluid flowing out of a container or vessel through an opening,such as a hole or a nozzle.
In the context of fluid mechanics,the velocity of efflux is the speed at which the fluid emerges from the opening.
According to Torricelli's law,for an ideal fluid,the velocity of efflux $v$ is given by the formula $v = \sqrt{2gh}$,where $g$ is the acceleration due to gravity and $h$ is the height of the fluid column above the opening.

(N/A) Torricelli's law states that the speed of efflux,$v$,of a fluid through an orifice at a depth $h$ below the free surface of the liquid is equal to the speed that a body would acquire in falling freely from rest through a vertical distance $h$.
The mathematical expression for the speed of efflux is given by $v = \sqrt{2gh}$,where $g$ is the acceleration due to gravity and $h$ is the height of the liquid column above the orifice.

(D) The height of the water level in the tank is $H = 1\, m$.
The height of the orifice from the ground is $h = 0.25\, m$.
The depth of the orifice from the top surface is $y = H - h = 1 - 0.25 = 0.75\, m$.
The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gy} = \sqrt{2 \times g \times 0.75} = \sqrt{1.5g}$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.25}{g}} = \sqrt{\frac{0.5}{g}}$.
The horizontal range $R$ is given by $R = v \times t = \sqrt{2g(H-h)} \times \sqrt{\frac{2h}{g}} = 2\sqrt{h(H-h)}$.
Substituting the values: $R = 2\sqrt{0.25 \times 0.75} = 2\sqrt{0.1875} = 2 \times 0.433 = 0.866\, m$.
Converting to centimeters: $0.866\, m = 86.6\, cm$.

(A) Let the height of the liquid in the vessel be $h$. The velocity of the emerging liquid is given by Torricelli's law as $v = \sqrt{2gh}$.
The force exerted by the emerging liquid on the vessel (thrust force) is $F_{thrust} = \rho a v^2 = \rho a (2gh) = 2 \rho agh$.
For the vessel to not slide,the frictional force $f$ must balance this thrust force. Thus,$f \geq F_{thrust}$.
The maximum frictional force is $f_{max} = \mu N$,where $N$ is the normal reaction from the surface. Since the vessel is light (mass is negligible),the normal reaction $N$ is equal to the weight of the liquid,$N = mg = (\rho A h) g$.
Therefore,$\mu (\rho A h g) \geq 2 \rho a g h$.
Simplifying this,we get $\mu \geq \frac{2a}{A}$.
Thus,the minimum coefficient of friction is $\frac{2a}{A}$.

(D) Let $H = 12\, \text{m}$ be the total height of the water column.
Let $h$ be the depth of the hole from the water surface.
The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gh}$.
The time taken for the water to reach the ground from a height $(H - h)$ is given by $t = \sqrt{\frac{2(H - h)}{g}}$.
The horizontal range $R$ is given by $R = v \times t = \sqrt{2gh} \times \sqrt{\frac{2(H - h)}{g}}$.
Simplifying the expression: $R = \sqrt{4h(H - h)} = 2\sqrt{hH - h^2}$.
To find the maximum range, we differentiate $R$ with respect to $h$ and set it to zero: $\frac{dR}{dh} = 0$.
$\frac{d}{dh}(2\sqrt{hH - h^2}) = 2 \cdot \frac{1}{2\sqrt{hH - h^2}} \cdot (H - 2h) = 0$.
This implies $H - 2h = 0$, so $h = \frac{H}{2}$.
Given $H = 12\, \text{m}$, we get $h = \frac{12}{2} = 6\, \text{m}$.

(C) The velocity of efflux from the hole is given by $v_{e} = \sqrt{2gh}$.
The horizontal range $x$ of the water jet is given by $x = v_{e} \times t$,where $t$ is the time of fall from height $H$.
Since $H = \frac{1}{2}gt^{2}$,we have $t = \sqrt{\frac{2H}{g}}$.
Thus,$x = \sqrt{2gh} \times \sqrt{\frac{2H}{g}} = 2\sqrt{hH}$.
The speed $v$ of the bucket must match the rate of change of the range $x$ to keep the water falling into it:
$v = \frac{dx}{dt} = \frac{d}{dt}(2\sqrt{hH}) = 2\sqrt{H} \frac{d}{dt}(\sqrt{h}) = 2\sqrt{H} \cdot \frac{1}{2\sqrt{h}} \cdot \frac{dh}{dt} = \sqrt{\frac{H}{h}} \cdot \frac{dh}{dt}$.
From the equation of continuity,$A_{0} \left(-\frac{dh}{dt}\right) = A v_{e} = A \sqrt{2gh}$.
Therefore,$\frac{dh}{dt} = -\frac{A}{A_{0}} \sqrt{2gh}$.
Substituting this into the expression for $v$ (taking magnitude): $v = \sqrt{\frac{H}{h}} \cdot \frac{A}{A_{0}} \sqrt{2gh} = \frac{A}{A_{0}} \sqrt{2gH}$.
Given $A = 5 \, cm^{2}$,$A_{0} = 500 \, cm^{2}$,$g = 10 \, m/s^{2}$,and $H = 5 \, m$:
$v = \left(\frac{5}{500}\right) \sqrt{2 \times 10 \times 5} = \frac{1}{100} \times \sqrt{100} = \frac{10}{100} = 0.1 \, m/s$.

(D) When a box with a hole is in free fall,both the water inside and the box experience the same acceleration due to gravity,$g$,directed downwards.
Since the box and the water are accelerating at the same rate,there is no relative acceleration between them.
Consequently,the pressure difference at the hole becomes zero,and no water comes out of the hole during the free fall of the box.

(A) According to Torricelli's Law,the velocity of efflux $(v)$ for a small hole at a depth $h$ below the free surface of a liquid is given by the formula:
$v = \sqrt{2gh}$
Here,$g$ is the acceleration due to gravity and $h$ is the height of the liquid column above the hole.
Since both vessels are identical and filled to the same height $h$,the velocity of efflux depends only on the height $h$ and the acceleration due to gravity $g$.
It is independent of the density $(\rho)$ of the liquid.
Therefore,$v_1 = \sqrt{2gh}$ and $v_2 = \sqrt{2gh}$.
Thus,$v_1 = v_2$.

(D) Let $H$ be the total height of the water column. Let $h^{\prime}$ be the depth of hole $B$ from the top surface and $h$ be the height of hole $A$ from the bottom.
For hole $B$ at depth $h^{\prime}$ from the top,the velocity of efflux is $v_B = \sqrt{2gh^{\prime}}$. The height of this hole from the ground is $H - h^{\prime}$. The time taken for water to reach the ground is $t_B = \sqrt{\frac{2(H - h^{\prime})}{g}}$.
The range $R_B = v_B \times t_B = \sqrt{2gh^{\prime}} \times \sqrt{\frac{2(H - h^{\prime})}{g}} = 2\sqrt{h^{\prime}(H - h^{\prime})}$.
For hole $A$ at height $h$ from the bottom,the depth from the top is $H - h$. The velocity of efflux is $v_A = \sqrt{2g(H - h)}$. The time taken to reach the ground is $t_A = \sqrt{\frac{2h}{g}}$.
The range $R_A = v_A \times t_A = \sqrt{2g(H - h)} \times \sqrt{\frac{2h}{g}} = 2\sqrt{h(H - h)}$.
For the same range,$R_A = R_B$:
$2\sqrt{h(H - h)} = 2\sqrt{h^{\prime}(H - h^{\prime})}$
$h(H - h) = h^{\prime}(H - h^{\prime})$
$hH - h^2 = h^{\prime}H - h^{\prime 2}$
$H(h - h^{\prime}) = h^2 - h^{\prime 2}$
$H(h - h^{\prime}) = (h - h^{\prime})(h + h^{\prime})$
Since $h \neq h^{\prime}$,we have $H = h + h^{\prime}$.
This implies that the two holes are at depths $h^{\prime}$ and $H - h^{\prime} = h$ from the top surface. For the same range,the sum of the depths of the two holes from the top surface must be equal to the total height of the water column. In the given configuration,if $h^{\prime}$ is the depth of $B$ and $h$ is the height of $A$ from the bottom,the condition for equal range is $h = h^{\prime}$,so the ratio $\frac{h^{\prime}}{h} = 1$.

(D) Let the total height of the water level from the ground be $H = 3 \,m + 1 \,m = 4 \,m$. Let the height of the hole from the ground be $y$. The depth of the hole from the free surface of the water is $h = H - y = 4 - y$.
The velocity of efflux is $v = \sqrt{2gh}$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2y}{g}}$.
The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2g(4-y)} \cdot \sqrt{\frac{2y}{g}} = 2\sqrt{y(4-y)}$.
For maximum range,the term inside the square root,$f(y) = 4y - y^2$,must be maximum.
Taking the derivative with respect to $y$ and setting it to zero: $\frac{df}{dy} = 4 - 2y = 0$,which gives $y = 2 \,m$.
Thus,the hole should be at a height of $2 \,m$ from the ground.

(D) Let the depth of water in the cylinder be $x$.
According to Torricelli's law,the velocity of efflux $(v)$ is given by $v = \sqrt{2 g x}$.
The water stream acts as a projectile launched horizontally from a height $H$.
The time taken $(t)$ by the water to reach the ground is given by the equation of motion $H = \frac{1}{2} g t^2$,which gives $t = \sqrt{\frac{2 H}{g}}$.
The horizontal range $(R)$ is the product of the horizontal velocity and the time of flight:
$R = v \times t$
$R = \sqrt{2 g x} \times \sqrt{\frac{2 H}{g}}$
$R = \sqrt{2 g x \cdot \frac{2 H}{g}}$
$R = \sqrt{4 x H}$
Squaring both sides,we get $R^2 = 4 x H$.
Therefore,the depth of water $x = \frac{R^2}{4 H}$.

(C) The volume flow rate (quantity of water per second) is given by the product of the area of the hole $(A)$ and the velocity of efflux $(v)$.
According to Torricelli's law,the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.
For the square hole: Area $A_1 = a^2$,depth $h_1 = x$. Velocity $v_1 = \sqrt{2gx}$.
Flow rate $Q_1 = A_1 v_1 = a^2 \sqrt{2gx}$.
For the circular hole: Area $A_2 = \pi r^2$,depth $h_2 = 4x$. Velocity $v_2 = \sqrt{2g(4x)} = 2\sqrt{2gx}$.
Flow rate $Q_2 = A_2 v_2 = \pi r^2 (2\sqrt{2gx})$.
Given $Q_1 = Q_2$,we have:
$a^2 \sqrt{2gx} = 2\pi r^2 \sqrt{2gx}$.
Dividing both sides by $\sqrt{2gx}$:
$a^2 = 2\pi r^2$.
$r^2 = \frac{a^2}{2\pi}$.
$r = \frac{a}{\sqrt{2\pi}}$.

(A) The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gD}$.
The water stream travels a vertical distance of $(H-D)$ to reach the ground.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $s = H-D$,$u = 0$ (initial vertical velocity),and $a = g$:
$H-D = 0 + \frac{1}{2}gt^2$
$t = \sqrt{\frac{2(H-D)}{g}}$
The horizontal distance $x$ is given by the product of horizontal velocity and time:
$x = v \times t$
$x = \sqrt{2gD} \times \sqrt{\frac{2(H-D)}{g}}$
$x = \sqrt{4D(H-D)}$
$x = 2[D(H-D)]^{1/2}$

(A) The horizontal range $d$ of a water jet from a hole at height $h_2$ from the bottom,with $h_1$ being the height of water above the hole,is given by $d = v \cdot t = \sqrt{2gh_1} \cdot \sqrt{\frac{2h_2}{g}} = 2\sqrt{h_1 h_2}$.
Here,$g$ is the effective acceleration due to gravity $(g_{eff})$ inside the lift.
$P$. When the lift accelerates vertically up,$g_{eff} = g + a > g$. Since $d \propto \frac{1}{\sqrt{g_{eff}}}$,$d$ decreases. Thus,$d < 1.2 \ m$.
$Q$. When the lift accelerates vertically down with $a < g$,$g_{eff} = g - a < g$. Thus,$d$ increases. So,$d > 1.2 \ m$.
$R$. When the lift moves with constant speed,$a = 0$,so $g_{eff} = g$. Thus,$d = 1.2 \ m$.
$S$. When the lift is falling freely,$a = g$,so $g_{eff} = g - g = 0$. Since there is no effective gravity,the pressure at the hole is zero,and no water leaks out.
Matching: $P-2, Q-3, R-1, S-4$.

(B) For tank $1$,the velocity of efflux at height $y$ is $v_1 = \sqrt{2gy}$.
By the equation of continuity,$A(-dy/dt) = a\sqrt{2gy}$,where $A$ is the cross-sectional area of the tank and $a$ is the area of the hole.
Integrating from $y=h$ to $y=0$:
$dt = (A/a\sqrt{2g}) \cdot (-dy/\sqrt{y})$
$t_1 = (A/a\sqrt{2g}) \int_0^h y^{-1/2} dy = (A/a\sqrt{2g}) \cdot 2\sqrt{h} = (A/a) \sqrt{2h/g}$.
For tank $2$,the velocity of efflux at height $y$ is $v_2 = \sqrt{2g(y+H)}$.
By the equation of continuity,$A(-dy/dt) = a\sqrt{2g(y+H)}$.
Integrating from $y=h$ to $y=0$:
$dt = (A/a\sqrt{2g}) \cdot (-dy/\sqrt{y+H})$
$t_2 = (A/a\sqrt{2g}) \int_0^h (y+H)^{-1/2} dy = (A/a\sqrt{2g}) \cdot 2[\sqrt{y+H}]_0^h = (A/a\sqrt{2g}) \cdot 2(\sqrt{h+H} - \sqrt{H})$.
Given $H = (16/9)h$,then $h+H = h + (16/9)h = (25/9)h$.
$t_2 = (A/a\sqrt{2g}) \cdot 2(\sqrt{25h/9} - \sqrt{16h/9}) = (A/a\sqrt{2g}) \cdot 2(5/3\sqrt{h} - 4/3\sqrt{h}) = (A/a\sqrt{2g}) \cdot 2(1/3\sqrt{h}) = (A/3a) \sqrt{2h/g}$.
Therefore,the ratio $t_1 / t_2 = [(A/a) \sqrt{2h/g}] / [(A/3a) \sqrt{2h/g}] = 3$.

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

(A) Let $H$ be the total height of the liquid column. The depths of the two holes from the free surface are $h_1 = 4 \ cm$ and $h_2 = 6 \ cm$.
The horizontal range $R$ of a water stream from a hole at depth $h$ is given by $R = 2\sqrt{h(H-h)}$.
Since the streams strike the ground at the same point,their ranges are equal:
$2\sqrt{h_1(H-h_1)} = 2\sqrt{h_2(H-h_2)}$
$h_1(H-h_1) = h_2(H-h_2)$
Substituting the given values:
$4(H-4) = 6(H-6)$
$4H - 16 = 6H - 36$
$2H = 20$
$H = 10 \ cm$.

(B) The horizontal range $R$ of a fluid jet issuing from a small hole at a depth $h$ below the free surface of a liquid in a tank of total height $H$ is given by $R = 2\sqrt{h(H-h)}$.
For hole $A$,the depth below the free surface is $h_1 = 5 \text{ m}$. The total height of the liquid is $H = 20 \text{ m}$.
Thus,$R_1 = 2\sqrt{5(20-5)} = 2\sqrt{5 \times 15} = 2\sqrt{75} = 10\sqrt{3} \text{ m}$.
For hole $B$,the depth below the free surface is $h_2 = 5 \text{ m} + 5 \text{ m} = 10 \text{ m}$.
Thus,$R_2 = 2\sqrt{10(20-10)} = 2\sqrt{10 \times 10} = 20 \text{ m}$.
The ratio is $\frac{R_1}{R_2} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$.

(A) The height of water in the tank becomes maximum when the rate of water flowing into the tank equals the rate of water flowing out of the hole.
Rate of water flowing in = $70 \ cm^3/s$.
Rate of water flowing out = $A \times v = A \sqrt{2gh}$,where $A = 1 \ cm^2$ and $g = 980 \ cm/s^2$.
Equating the rates: $1 \times \sqrt{2 \times 980 \times h} = 70$.
Squaring both sides: $2 \times 980 \times h = 70^2$.
$1960 \times h = 4900$.
$h = \frac{4900}{1960} = 2.5 \ cm$.

(D) Let the depths of the two holes from the free surface of the liquid be $y_1$ and $y_2$ respectively. Given that the difference in height is $h$,we have $y_2 - y_1 = h$.
The velocity of efflux for the top hole is $v_1 = \sqrt{2gy_1}$ and for the bottom hole is $v_2 = \sqrt{2gy_2}$.
The force exerted by the liquid jet on the tank is given by $F = \frac{dm}{dt} v = (A \rho v) v = A \rho v^2$,where $A$ is the area of the hole and $\rho$ is the density of the liquid.
Since the holes are on opposite sides,the forces act in opposite directions. The net horizontal force $F_{net}$ is $|F_2 - F_1| = |A \rho v_2^2 - A \rho v_1^2|$.
Substituting the velocities: $F_{net} = A \rho (2gy_2 - 2gy_1) = 2A \rho g (y_2 - y_1)$.
Since $y_2 - y_1 = h$,we get $F_{net} = 2A \rho g h$.
Therefore,the net horizontal force is proportional to $h$.

(B) The velocity of efflux from a hole at depth $h$ is given by Torricelli's law: $v = \sqrt{2gh}$.
The volume of water flowing out per second (flow rate) is $Q = A \cdot v$,where $A$ is the area of the hole.
For hole $P$ (square of side $a$ at depth $2h$): $A_P = a^2$ and $v_P = \sqrt{2g(2h)} = \sqrt{4gh} = 2\sqrt{gh}$.
Thus,flow rate $Q_P = a^2 \cdot 2\sqrt{gh}$.
For hole $Q$ (circle of radius $r$ at depth $8h$): $A_Q = \pi r^2$ and $v_Q = \sqrt{2g(8h)} = \sqrt{16gh} = 4\sqrt{gh}$.
Thus,flow rate $Q_Q = \pi r^2 \cdot 4\sqrt{gh}$.
Given $Q_P = Q_Q$,we have: $a^2 \cdot 2\sqrt{gh} = \pi r^2 \cdot 4\sqrt{gh}$.
$a^2 \cdot 2 = 4\pi r^2 \implies a^2 = 2\pi r^2$.
Taking the square root of both sides: $a = r\sqrt{2\pi}$.

(C) According to Torricelli's law,the velocity of efflux is given by $v = \sqrt{2gh}$.
For hole '$A$' at depth '$h$',the velocity is $V_A = \sqrt{2gh}$.
For hole '$B$' at depth '$4h$',the velocity is $V_B = \sqrt{2g(4h)} = 2\sqrt{2gh} = 2V_A$.
The volume flow rate $(Q)$ is given by $Q = A \cdot v$,where '$A$' is the area of the hole.
Given that the flow rate is the same for both holes: $Q_A = Q_B$.
$A_A \cdot V_A = A_B \cdot V_B$.
Since hole '$A$' is a square of side '$L$',$A_A = L^2$.
Since hole '$B$' is a circle of radius '$R$',$A_B = \pi R^2$.
Substituting the values: $L^2 \cdot V_A = (\pi R^2) \cdot (2V_A)$.
$L^2 = 2\pi R^2$.
$L = \sqrt{2\pi} \cdot R$.

(B) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
Let $A_1$ be the area of the square hole and $A_2$ be the area of the circular hole.
$A_1 = a^2$ and $A_2 = \pi r^2$.
The volume flow rate $Q$ is given by $Q = A \cdot v$.
Given that the volume flow rate is equal for both holes:
$A_1 \sqrt{2gy} = A_2 \sqrt{2g(16y)}$
$a^2 \sqrt{y} = (\pi r^2) \sqrt{16y}$
$a^2 \sqrt{y} = \pi r^2 (4 \sqrt{y})$
$a^2 = 4 \pi r^2$
$r^2 = \frac{a^2}{4 \pi}$
$r = \sqrt{\frac{a^2}{4 \pi}} = \frac{a}{2 \sqrt{\pi}}$

(C) According to Bernoulli's principle,the velocity of efflux $v$ is given by the formula:
$v = \sqrt{\frac{2(P - P_0)}{\rho}}$
where $P$ is the total pressure at the bottom,$P_0$ is the atmospheric pressure,and $\rho$ is the density of water.
Given:
$P = 4H$
$P_0 = H$
Substituting these values into the formula:
$v = \sqrt{\frac{2(4H - H)}{\rho}}$
$v = \sqrt{\frac{2(3H)}{\rho}}$
$v = \sqrt{\frac{6H}{\rho}}$

(B) According to Torricelli's law,the speed of efflux $(V)$ of water from an orifice at a depth '$h$' below the free surface is given by:
$V = \sqrt{2gh}$
From the figure,the depths of the orifices $O_1, O_2, O_3$ from the free surface are $h_1, h_2, h_3$ respectively.
It is clear from the figure that $h_1 < h_2 < h_3$.
Since $V$ is directly proportional to $\sqrt{h}$,we have:
$V_1 < V_2 < V_3$
Thus,the speed of efflux increases as the depth of the orifice increases.

(B) According to Bernoulli's principle for a fluid in motion,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant.
For the fluid inside the pipe,the initial pressure is $P_{1}$ when the velocity is zero.
When the valve is opened,the pressure drops to $P_{2}$ and the fluid gains a velocity $v$.
Applying Bernoulli's equation between the inside of the pipe and the outlet:
$P_{1} + 0 = P_{2} + \frac{1}{2} \rho v^{2}$
Rearranging the terms to solve for $v$:
$P_{1} - P_{2} = \frac{1}{2} \rho v^{2}$
$v^{2} = \frac{2(P_{1} - P_{2})}{\rho}$
$v = \sqrt{\frac{2(P_{1} - P_{2})}{\rho}}$
Thus,the correct option is $B$.

(C) The volume flow rate (or discharge) $Q$ from a hole at depth $y$ is given by $Q = A_v \cdot v$,where $A_v$ is the area of the hole and $v = \sqrt{2gy}$ is the velocity of efflux.
For hole $A$ at depth $h$: Area $A_A = L^2$,velocity $v_A = \sqrt{2gh}$. So,$Q_A = L^2 \sqrt{2gh}$.
For hole $B$ at depth $3h$: Area $A_B = \pi r^2$,velocity $v_B = \sqrt{2g(3h)} = \sqrt{6gh}$. So,$Q_B = \pi r^2 \sqrt{6gh}$.
Given that the flow rate is the same,$Q_A = Q_B$:
$L^2 \sqrt{2gh} = \pi r^2 \sqrt{6gh}$
$L^2 = \pi r^2 \frac{\sqrt{6gh}}{\sqrt{2gh}} = \pi r^2 \sqrt{3}$
Taking the square root of both sides:
$L = \sqrt{\pi r^2 \sqrt{3}} = r \sqrt{\pi} (3)^{\frac{1}{4}} = r (\pi)^{\frac{1}{2}} (3)^{\frac{1}{4}}$.

(C) According to the equation of continuity and Torricelli's law,the rate of flow of water is given by:
$A \left( -\frac{dh}{dt} \right) = a \sqrt{2gh}$
Rearranging the terms for integration:
$\int_{h}^{0} \frac{-dh}{\sqrt{h}} = \int_{0}^{t} \frac{a}{A} \sqrt{2g} dt$
Integrating both sides:
$[2\sqrt{h}]_{0}^{h} = \frac{a}{A} \sqrt{2g} t$
$2\sqrt{h} = \frac{a}{A} \sqrt{2g} t$
Thus,the time taken to empty the vessel is proportional to the square root of the height:
$t \propto \sqrt{h}$
Given that for height $h$,the time is $t_1 = t$,and for height $h_2 = 4h$,the time is $t_2$:
$\frac{t_2}{t_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{4h}{h}} = \sqrt{4} = 2$
Therefore,$t_2 = 2t_1 = 2t$.

(B) According to Torricelli's law,the velocity of efflux $v$ is given by $v = \sqrt{2gh}$,where $h$ is the depth of the hole from the free surface.
Given $h = 2 \text{ cm}$,so $v = \sqrt{2 \times g \times 2} = 2\sqrt{g}$.
The time $t$ taken for the water to reach the ground from a height $H = 8 \text{ cm}$ is given by $H = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 8}{g}} = \sqrt{\frac{16}{g}} = \frac{4}{\sqrt{g}}$.
The horizontal range $R$ is the product of the velocity of efflux and the time of flight:
$R = v \times t = (\sqrt{2gh}) \times \sqrt{\frac{2H}{g}} = 2\sqrt{hH}$.
Substituting the values $h = 2 \text{ cm}$ and $H = 8 \text{ cm}$:
$R = 2 \times \sqrt{2 \times 8} = 2 \times \sqrt{16} = 2 \times 4 = 8 \text{ cm}$.

(A) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$,where $A$ is the area of the tank and $a$ is the area of the hole.
This gives $dt = -\frac{A}{a\sqrt{2g}} h^{-1/2} dh$.
The time taken to fall from height $h_1$ to $h_2$ is $t = \int_{h_2}^{h_1} \frac{A}{a\sqrt{2g}} h^{-1/2} dh = \frac{A}{a\sqrt{2g}} [2\sqrt{h}]_{h_2}^{h_1} = \frac{2A}{a\sqrt{2g}} (\sqrt{h_1} - \sqrt{h_2})$.
Let $t_1$ be the time to fall from $h$ to $h/2$:
$t_1 = \frac{2A}{a\sqrt{2g}} (\sqrt{h} - \sqrt{h/2}) = \frac{2A}{a\sqrt{2g}} \sqrt{h} (1 - 1/\sqrt{2})$.
Let $t_2$ be the time to fall from $h/2$ to $0$:
$t_2 = \frac{2A}{a\sqrt{2g}} (\sqrt{h/2} - \sqrt{0}) = \frac{2A}{a\sqrt{2g}} \sqrt{h} (1/\sqrt{2})$.
The ratio is $\frac{t_1}{t_2} = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2} - 1$.

(C) The rate of flow of water through a hole is given by the product of the velocity of efflux and the area of the hole.
$\frac{\Delta V}{\Delta t} = v \times A$
Using Torricelli's law, the velocity of efflux is $v = \sqrt{2gh}$.
Thus, $\frac{\Delta V}{\Delta t} = \sqrt{2gh} \times A$ ... $(i)$
Given:
$h = 20 \,m$
$g = 10 \,m/s^2$
$\frac{\Delta V}{\Delta t} = 3 \times 10^{-3} \,m^3/min = \frac{3 \times 10^{-3}}{60} \,m^3/s = 0.5 \times 10^{-4} \,m^3/s = 5 \times 10^{-5} \,m^3/s$
Substituting the values into equation $(i)$:
$5 \times 10^{-5} = \sqrt{2 \times 10 \times 20} \times A$
$5 \times 10^{-5} = \sqrt{400} \times A$
$5 \times 10^{-5} = 20 \times A$
$A = \frac{5 \times 10^{-5}}{20} = 0.25 \times 10^{-5} \,m^2 = 2.5 \times 10^{-6} \,m^2$
Since $1 \,m^2 = 10^6 \,mm^2$, we have:
$A = 2.5 \times 10^{-6} \times 10^6 \,mm^2 = 2.5 \,mm^2$.

(A) The velocity of efflux $(v)$ through a hole at a depth $H$ below the free surface is given by Torricelli's law: $v = \sqrt{2gH}$.
The volume flow rate $(Q)$ is the product of the area of the hole $(a)$ and the velocity of efflux $(v)$:
$Q = a \times v = a \sqrt{2gH}$
Given:
$Q = 2 \times 10^{-3} \,m^3/s$
$a = 5 \,cm^2 = 5 \times 10^{-4} \,m^2$
$g = 10 \,ms^{-2}$
Substituting the values into the equation:
$2 \times 10^{-3} = 5 \times 10^{-4} \times \sqrt{2 \times 10 \times H}$
Divide both sides by $5 \times 10^{-4}$:
$4 = \sqrt{20H}$
Squaring both sides:
$16 = 20H$
$H = \frac{16}{20} \,m = 0.8 \,m = 80 \,cm$.

(A) Let $a$ be the area of the hole,$v_e$ be the velocity of efflux,and $h$ be the height of the liquid above the hole. Let $v$ be the speed with which the level decreases in the container.
From the equation of continuity,$a v_e = A v \Rightarrow v = \frac{a v_e}{A}$.
Using Bernoulli's theorem at the top surface and the orifice:
$p_0 + h \rho g + \frac{1}{2} \rho v^2 = p_0 + \frac{1}{2} \rho v_e^2$
$h \rho g + \frac{1}{2} \rho \left(\frac{a v_e}{A}\right)^2 = \frac{1}{2} \rho v_e^2$
$v_e^2 = \frac{2gh}{1 - (a/A)^2}$
Given $h = 3 \,m - 0.525 \,m = 2.475 \,m$,$a/A = 0.1$,and $g = 10 \,ms^{-2}$:
$v_e^2 = \frac{2 \times 10 \times 2.475}{1 - (0.1)^2} = \frac{49.5}{1 - 0.01} = \frac{49.5}{0.99} = 50 \,m^2 \,s^{-2}$.

(A) Let the height of the liquid in the tank be $H = 5 \,m$. Let the height of the bottom of the tank from the ground be $h_0 = 5 \,m$. Let the hole be made at a depth $y$ from the free surface of the liquid. The height of the hole from the ground is $h = h_0 + (H - y) = 5 + 5 - y = 10 - y$. The velocity of efflux is $v = \sqrt{2gy}$. The time taken for the liquid to reach the ground is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(10-y)}{g}}$. The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{2(10-y)}{g}} = 2\sqrt{y(10-y)}$. To maximize $R$, we maximize $f(y) = y(10-y) = 10y - y^2$. Taking the derivative with respect to $y$ and setting it to zero: $\frac{df}{dy} = 10 - 2y = 0$, which gives $y = 5 \,m$. Substituting $y = 5 \,m$ into the range formula: $R_{max} = 2\sqrt{5(10-5)} = 2\sqrt{25} = 2 \times 5 = 10 \,m$.

(C) The velocity of efflux is given by Torricelli's law as $v = \sqrt{2gh}$.
Let $A$ be the area of the tank and $a$ be the area of the hole. The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$.
Rearranging the terms,we get $dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.
Integrating from $h_1$ to $h_2$,the time taken is $t = \int_{h_2}^{h_1} \frac{A}{a \sqrt{2g}} h^{-1/2} dh = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h_1} - \sqrt{h_2})$.
For the first interval ($h$ to $h/2$): $t_1 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{h/2}) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1 - \frac{1}{\sqrt{2}})$.
For the second interval ($h/2$ to $0$): $t_2 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h/2} - 0) = \frac{A}{a} \sqrt{\frac{2h}{g}} (\frac{1}{\sqrt{2}})$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \frac{(\sqrt{2}-1)/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}-1$.

(B) Let the total height of the water surface from the ground be $Y = H + h$.
The velocity of efflux at a depth $x$ below the water surface is $v = \sqrt{2gx}$.
The height of the hole from the ground is $y = H + (h - x)$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2(H + h - x)}{g}}$.
The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2gx} \cdot \sqrt{\frac{2(H + h - x)}{g}} = 2\sqrt{x(H + h - x)}$.
To maximize $R$,we maximize the term $f(x) = x(H + h - x) = (H + h)x - x^2$.
Taking the derivative with respect to $x$ and setting it to zero: $f'(x) = (H + h) - 2x = 0$.
Thus,$x = \frac{H + h}{2}$.

(C) According to Torricelli's law derived from Bernoulli's equation, the velocity of efflux $v$ is given by $v = \sqrt{2gh}$.
Given $g = 10 \,m/s^2$ and $h = 10 \,m$, we have $v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 14.14 \,m/s$.
The diameter of the hole is $d = 4 \,mm = 4 \times 10^{-3} \,m$, so the radius $r = 2 \times 10^{-3} \,m$.
The area of the hole is $A = \pi r^2 = 3.14 \times (2 \times 10^{-3})^2 = 3.14 \times 4 \times 10^{-6} = 12.56 \times 10^{-6} \,m^2$.
The flow rate $Q$ is given by $Q = A \times v$.
$Q = (12.56 \times 10^{-6} \,m^2) \times (14.14 \,m/s) \approx 177.6 \times 10^{-6} \,m^3/s = 1.776 \times 10^{-4} \,m^3/s$.

(B) Given: Area of the hole, $A = 2 \,cm^2 = 2 \times 10^{-4} \,m^2$.
Volume flow rate, $Q = 100 \,cm^3/s = 100 \times 10^{-6} \,m^3/s = 10^{-4} \,m^3/s$.
At maximum height $h$, the rate of water entering the tank equals the rate of water exiting through the hole.
By Torricelli's Law, the velocity of efflux is $v = \sqrt{2gh}$.
The volume flow rate through the hole is $Q = A \times v$.
Substituting the values: $10^{-4} = (2 \times 10^{-4}) \times \sqrt{2gh}$.
Dividing both sides by $2 \times 10^{-4}$: $0.5 = \sqrt{2gh}$.
Squaring both sides: $0.25 = 2gh$.
Given $g = 10 \,m/s^2$, we have $0.25 = 2 \times 10 \times h$.
$0.25 = 20h$.
$h = \frac{0.25}{20} = 0.0125 \,m$.
Converting to cm: $h = 0.0125 \times 100 \,cm = 1.25 \,cm$.

(A) According to Torricelli's law,the velocity of efflux $v$ at a depth $h$ is given by $v = \sqrt{2gh}$.
Since the depth $h$ is the same for both tanks,the velocity of efflux for both tanks is the same: $v_A = v_B = v = \sqrt{2gh}$.
The mass flux (mass flow rate) is given by $\dot{m} = A \cdot v \cdot \rho$,where $A$ is the area of the hole,$v$ is the velocity,and $\rho$ is the density.
Given that the mass flux is equal for both tanks: $\dot{m}_A = \dot{m}_B$.
$A_A \cdot v_A \cdot \rho_A = A_B \cdot v_B \cdot \rho_B$.
Since $v_A = v_B$,we can cancel them from both sides:
$A_A \cdot \rho_A = A_B \cdot \rho_B$.
Given that the area of the hole in $B$ is twice that of $A$,$A_B = 2 A_A$.
Substituting this into the equation:
$A_A \cdot \rho_A = (2 A_A) \cdot \rho_B$.
Dividing both sides by $A_A \cdot \rho_B$:
$\frac{\rho_A}{\rho_B} = 2$.
Wait,re-evaluating the provided options and the logic: If $A_B = 2 A_A$,then $\rho_A = 2 \rho_B$,so $\rho_A / \rho_B = 2$. However,looking at the provided solution logic in the prompt,it assumes $v_2 = v_1/2$ which is incorrect for Torricelli's law. Given the standard physics,the ratio is $2$. Since $2$ is not an option,$I$ will re-verify the prompt's provided solution logic. The prompt's solution claims $A_1 v_1 = A_2 v_2$ which is the continuity equation for a pipe,not for efflux from a tank. For efflux,$v = \sqrt{2gh}$. Thus $\rho_A / \rho_B = A_B / A_A = 2$. If the intended answer is $1$,the area must be equal. Given the constraints,$I$ will provide the mathematically correct derivation.

(C) Let $A_v$ be the cross-sectional area of the vessel and $A$ be the area of the hole. According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
The rate of flow is given by $A_v \frac{dh}{dt} = -A \sqrt{2gh}$.
Separating the variables,we get $\frac{dh}{\sqrt{h}} = -\frac{A}{A_v} \sqrt{2g} dt$.
Integrating both sides from height $h$ to $0$ for time $t$:
$\int_{h}^{0} h^{-1/2} dh = -\int_{0}^{t} \frac{A}{A_v} \sqrt{2g} dt$
$[2\sqrt{h}]_{h}^{0} = -\frac{A}{A_v} \sqrt{2g} t$
$2\sqrt{h} = \frac{A}{A_v} \sqrt{2g} t$
Thus,$t \propto \sqrt{h}$.
Given that for height $h$,the time is $t$,so $t = k\sqrt{h}$.
For height $4h$,the new time $t'$ is $t' = k\sqrt{4h} = 2k\sqrt{h} = 2t$.