**1.** The surface of water in a water tank on the top of a
house is $4$ m above the tap level. Find the pressure of
water at the tap when the tap is closed. Is it necessary
to specify that the tap is closed ? Take g = $10$ $m/s^2$.

p = h $\rho$ g $\\$ it is necessary to specify that the tab is closed.Otherwise pressure will gradually decrease, as h decrease, because, of the tap is open, the pressure at the tap is atmospheric.

**2.** The heights of mercury surfaces in the two arms of the
manometer shown in figure (13-E1) are $2$ cm and $8$ cm.
Atmospheric pressure = $1.01$ x $10$ S $N/m^{2}.$ Find (a) the
pressure of the gas in the cylinder and (b) the pressure
of mercury at the bottom of the U tube

None

Pressure at the bottom of the tab should be same when considered for both limbs.$\\$ From the figure are shown, $\\$ $p_g$ + $\rho_{Hg}$ $\times$ $h_2$ $\times$ g = $p_a$ + $\rho_{Hg}$ $\times$ $\times$ $h_1$ $\times$ g $\\$ $p_g$ = $p_a$ + $\rho_{Hg}$ $\times$ $g(h_1 - h_2)$ $\\$ b) Pressure of mercury at the bottom of u tube $\\$ $p$ = $p_a$ +$\rho_{Hg}h_1$ $\times$$g$

**3.** A glass full of water has a bottom of area $20$ $cm^2$, top
of area $20$ $cm^2$, height $20$ cm and volume half a litre.
(a) Find the force exerted by the water on the bottom.
(b) Considering the equilibrium of the water, find the
resultant force exerted by the sides of the glass on the
water. Atmospheric pressure = $1.0$ $\times$ 10 $N/m^2$. Density
of water = 1000 kg/$m^3$ and g = $10$ $m/s^2$. Take all numbers
to be exact.

a)Force exerted at the bottom.$\\$ =Force due to the cylindrical water colum + atm.Force $\\$ = A $\times$ h $\times$ $\rho_w$ $\times$g + $p_a$ $\times$A $\\$ = A(h$\rho_w$g + $p_a$)$\\$ b)To find out the resultant force exerted by the sides of the glass,from the freebody, diagram of water inside the glass $\\$ $p_a$ $\times$A + mg = A $\times$ h $\times$$\rho_w$$\times$g+$F_s$+$p_a$$\times$A $\\$ $\Rightarrow$ mg = A $\times$ h $\times$$\rho_w$$\times$g+$F_s$ $\\$ This force is provided by the sides of the glass

**4.** The area of cross-section of the wider tube shown in
figure (13-E2) is $900$ $cm^{2}$. If the boy standing on the piston weighs $45$ kg, find the difference in the levels of
water in the two tubes.

From the figure shown $\\$ $p_a$ +h$\rho$g = $p_a$ +mg\A $\\$ $\Rightarrow$ h$\rho$g = mg\A $\\$ $\Rightarrow$ h = $\frac{m}{A\rho}$

**5.** Suppose the glass of the previous problem is covered by
a jar and the air inside the jar is completely pumped
out. (a) What will be the answers to the problem ?
(b) Show that the answers do not change if a glass of
different shape is used provided the height, the bottom
area and the volume are unchanged.

5 None

SolutionsIf the glass will be covered by a jar and the air is pumped out,the atmospheric pressure has no effect.$\\$ So,$\\$ a)Force exerted on the bottom.$\\$ =(h$\rho_w$g)$\times$A $\\$ b)mg = h $\times$ $\rho_w$ $\times$g $\times$ + $F_s$ $\\$ c)It glass of different shape is used provided the volume,height and area remain same,no change in answer wil occur

**6.** If water be used to construct a barometer, what would
be the height of water column at standard atmospheric
pressure ($76$ cm of mercury) ?

6 None

SolutionsStandard atmospheric pressure is always pressure exerted by $76$cm Hg column $\\$ = ($76$ $\times$ $13.6$ $\times$ g)Dyne/$cm^2$.$\\$ If water is used in the barometer.$\\$ Let h $\longrightarrow$height of water column. $\\$ $\therefore$ h $\times$ $\rho_w$ $\times$g

**7.** Find the force exerted by the water on a $2$$m^2$ plane
surface of a large stone placed at the bottom of a sea
$500$ m deep. Does the force depend on the orientation of
the surface ?

7 None

Solutionsa) F = P $\times$ A = (h$\rho_w$ $\times$ g)A $\\$ b)The force does not depend on the orientation of the rock as long as the surface area remains same

**8.** Water is filled in a rectangular tank of size
$3$m x $2$m x $1$ m. (a) Find the total force exerted by the
water on the bottom surface of the tank. (b) Consider a
vertical side of area $2$ m x $1$ m. Take a horizontal strip
of width ox metre in this side, situated at a depth of x
metre from the surface of water. Find the force by the
water on this strip. (c) Find the torque of the force
calculated in part (b) about the bottom edge of this side.
(d) Find the total force by the water on this side.
(e) Find the total torque by the water on the side about
the bottom edge. Neglect the atmospheric pressure and
take g = $10$ $m/s^2$.

a)F = A h $\rho$ g $\\$ b)The force exerted by water on the strip of width $\delta$X as shown,$\\$ dF = p $\times$ A $\\$ = (x$\rho$g) $\times$ A $\\$ c) Inside the liquid force act in every direction due to adhesion. $\\$ di = F $\times$ r $\\$ d)The total force by the water on that side is given by F = $\int_{0}^{1}$ 20000 X$\delta$x $\Rightarrow$ F= $20,000$[$\frac{x^2}{2}]^{1}_{0}$ $\\$ e)The torque by the water on that side is given by $\\$ i= $\int_{0}^{1}$ 20000 X$\delta$x (1-x)$\Rightarrow$ $20,000$[$\frac{x^2}{2}$ - $\frac{x^3}{3}]^{1}_{0}$

**9.** An ornament weighing $36$ g in air, weighs only $34$ g in
water. Assuming that some copper is mixed with gold
to prepare the ornament, find the amount of copper in
it. Specific gravity of gold is $19.3$ and that of copper is
$8.9$.

Here, $m_0$ = $m_{au}$ + $m_{cu}$ = 36 g $\\$ Let V be the volume of the ornament in $cm^3$ $\\$ So, V $\times$ $\rho_w$ x g = $2$ x g $\\$ $\Rightarrow$ ($V_{au}$ + $V_{cu}$) $\times$ $\rho_w$ x g = $2$ x g $\\$ $\Rightarrow$ $\big(\frac{m}{\rho_{au}} + \frac{m}{p_{au}}\big) $ $\rho_w$ x g = 2 x g $\\$ $\Rightarrow$ $\big(\frac{m_{Au}}{19.3} + \frac{m_{Au}}{8.9}\big)$ $\times$ 1 = $2$ $\\$ 8.9 $m_{Au}$ + 19.3 $m_{cu}$ = 2 $\times$ 19.3 $\times$ 8.9 = 343.54 .....(2) $\\$ From equation (1) and (2), 8.9 $m_{Au}$ + 19.3 $m_{cu}$ =343.54 $\\$ $\Rightarrow$ $\frac{8.9(m_{Au} + m_{cu}) = 8.9 \times 36}{m_{cu} = 2.225g}$ $\\$ So, the amount of copper in the ornament is 2.2 g.

**10.** Refer to the previous problem. Suppose, the goldsmith
argues that he has not mixed copper or any other
material with gold, rather some cavities might have been
left inside the ornament. Calculate the volume of the
cavities left that will allow the weights given in that
problem.

10 None

Solutions$\big(\frac{m_{Au}}{p_{Au}} + V_c \big)$ $\rho_w$ $\times$ g = 2 $\times$ g (where $V_c$ = Volume Of Cavity)

**11.** A metal piece of mass $160$ g lies in equilibrium inside
a glass of water (figure 13•E4). The piece touches the
bottom of the glass at a small number of points. If the
density of the metal is $8000$ $kg/m^3$, find the normal force
exerted by the bottom of the glass on the metal piece.

11 None

Solutionsmg = U + R (Where U = Upward thrust) $\\$ $\Rightarrow$ mg -U = R $\\$ $\Rightarrow$ R = mg - v $\rho_w$ g (because, U=v$\rho_w$g) $\\$ = mg - $\frac{m}{p}$ $\times$ $\rho_w$ $\times$g

**12.** A ferry boat has internal volume $1$ $m^3$ and weight $50$ kg.
(a) Neglecting the thickness of the wood, find the fraction
of the volume of the boat immersed in water. (b) If a
leak develops in the bottom and water starts coming in,
what fraction of the boat's volume will be filled with
water before water starts coming in from the sides ?

a)Let $V_i$ $\rightarrow$ volume of boat inside water = volume of water displace in $m^3$. $\\$ Since, Weight of the boat is balanced by the buoyant force.$\\$ $\Rightarrow$ mg = $V_i$ $\times$ $\rho_w$ $\times$ g $\\$ b)Let, $v^1$ $\rightarrow$ volume of boat filled with water before water starts coming in from the sides. $\\$ mg + $v^1$ $\rho_w$ $\times$ g = V $\times$ $\rho_w$ $\times$ g

**13.** A cubical block of ice floating in water has to support a
metal piece weighing $0.5$ kg. What can be the minimum
edge of the block so that it does not sink in water ?
Specific gravity of ice = $0.9$.

13 None

SolutionsLet X $\rightarrow$ minimum edge of the ice block in cm. $\\$ So, mg +$W_{ice}$ = U.(Where U = Upward thrust) $\\$ $\Rightarrow$ 0.5 $\times$ g + $X^3$ $\times$ $\rho_{ice}$ $\times$ g = $X^3$ $\times$ $\rho_w$ $\times$ g

**14.** A cube of ice floats partly in water and partly in K.oil
(figure $13$-$E5$). Find the ratio of the volume of ice
immersed in water to that in K.oil. Specific gravity of
K.oil is $0.8$ and that of ice is $0.9$.

14 None

Solutions$V_ice$ = $V_k$ + $V_w$ $\\$ $V_ice$ $\times$ $\rho_ice$ $\times$ g = $V_k$ $\times$ $\rho_k$ $\times$ g + $V_w$ $\times$ $\rho_w$ $\times$ g $\\$ $\Rightarrow$($V_k$ + $V_w$) $\times$ $\rho_{ice}$ = $V_k$ $\times$ $\rho_k$ + $V_w$ $\times$ $\rho_w$ $\\$ $\Rightarrow$ $\frac{V_w}{V_k}=1$.

**15.** A cubical box is to be constructed with iron sheets $1$ mm
in thickness. What can be the minimum value of the
external edge so that the cube does not sink in water ?
Density of iron = $8000$ $kg/m^3$ and density of water
= $1000$ $kg/m^3$.

$V_ig$ =V $\rho_w$ g

**16.** A cubical block of wood weighing $200$ g has a lead piece
fastened underneath. Find the mass of the lead piece
which will just allow the block to float in water. Specific
gravity of wood is $0.8$ and that of lead is $11.3$.

16 None

Solutions($m_w + m_{pb}$)g = ($V_w + V_{pb}$)$\rho$ $\times$ g $\\$ $\Rightarrow$ ($m_w + m_{pb}$) = $\big(\frac{m_{w}}{\rho_{w}} + \frac{m_{pb}}{\rho_{pb}}\big)$ $\rho$

**17.** A cubical metal block of edge $12$ cm floats in mercury
with one fifth of the height inside the mercury. Water
is poured till the surface of the block is just immersed
in it. Find the height of the water column to be poured.
Specific gravity of mercury = $13.6.$

Given, x = 12 cm $\\$ Length of the edge of the block $\rho_{Hg}$ = 13.6 gm/cc $\\$ Given that , initially $\frac{1}{5}$ of block is inside mercury $\\$ Let $\rho_b \rightarrow$ density of block in gm/cc.$\\$ $\therefore (X)^3 \times \rho_b \times g = (X)^2 \times (\frac{X}{5}) \times \rho_{Hg} \times g$ $\\$ $\Rightarrow 12^3 \times \rho_b = 12^2 \times \frac{12}{5} \times 13.6$ $\\$ $\rho = \frac{13.6}{5} gm/cc$ $\\$ After water poured , let x = hight of water column.$\\$ $V_b= V_{Hg}+V_w = 12^3$ $\\$ Where $V_{Hg} and V_w$ are volume of block inside mercury and water respectively $\\$ $\therefore (V_b \times \rho_b \times g ) = (V_{Hg} \times \rho_{Hg} \times g ) + (V_w \times \rho_w \times g )$ $\\$ $\Rightarrow ( V_{Hg} + V_w) \rho_b = V_{Hg} \times \rho_{Hg} + V_w \times \rho_w $ $\\$ $\Rightarrow ( V_{Hg} + V_w) \times \frac{13.6}{5} = V_{Hg} \times 13.6 + V_w \times 1 $ $\\$ $\Rightarrow (12)^3 \times \frac{13.6}{5} = (12-x) \times (12)^2 \times 13.6 +(x) \times (12)^2 \times 1$ $\\$ $\Rightarrow x = 10.4 cm$

**18.** Solve the previous problem if the lead piece is fastened
on the top surface of the block and the block is to float
with its upper surface just dipping into water.

18 None

SolutionsMg = w $\Rightarrow$ ($m_{w} + m_{pb}$)g = $V_w$ $\times$ $\rho$ $\times$ g

**19.** A hollow spherical body of inner and outer radii $6$ cm
and $8$ cm respectively floats half submerged in water.
Find the density of the material of the sphere.

Here, Mg = Upward thrust $\\$ $\Rightarrow$ $V_{pg}$ = $\frac{V}{2}$($\rho_{w}$) $\times$ g (where $\rho_{w}$ = density of water) $\\$ $\Rightarrow$ $\big(\frac{4}{3}\pi r^{3}_{2} - \frac{4}{3}\pi r^{3}_{1} \big)$ $\rho$ $\\$ $\Rightarrow (r^{3}_{2} - r^{3}_{1}) \times \rho = \frac{1}{2}r^{3}_{2} \times 1 = 865 kg/ m^3$

**20.** A solid sphere of radius $5$ cm floats in water. If a
maximum load of $0.1$ kg can be put on it without wetting
the load, find the specific gravity of the material of the
sphere.

20 None

SolutionsW1 +W2 = U $\\$ $\Rightarrow mg+ V \times \rho_s \times g = V \times \rho_w \times g$ (where $\rho_s$ = density of sphere in gm/cc) $\\$ $\Rightarrow 1 - \rho_s = 0.19$ $\\$ $\Rightarrow \rho_s = 1-(0.19) = 0.8gm/cc$ $\\$ So,specific gravity of the material is $0.8$

**21.** Find the ratio of the weights, as measured by a spring
balance, of a $1$ kg block of iron and a $1$ kg block of wood.
Density of iron = $7800$ $kg/m^3$, density of wood
= $800$ $kg/m^3$ and density of air = $1.293$ $kg/m^3$.

21 None

Solutions$W_i = mg - V_i \rho_{air} \times g = \big(m - \frac{m}{p_i}\rho_{air}\big)g$ $\\$ $W_w = mg - V_w \rho_{air} \times g = \big(m - \frac{m}{p_w}\rho_{air}\big)g$

**22.** A cylindrical object of outer diameter $20$ cm and mass
$2$ kg floats in water with its axis vertical. If it is slightly
depressed and then released, find the time period of the
resulting simple harmonic motion of the object.

22 None

SolutionsDriving force U = V$\rho_{w}$g $\\$ $\Rightarrow a = \pi r^2 (X) \times \rho_{w}g \Rightarrow T = 2 \pi \sqrt \frac{displacement}{Acceleration}$

**23.** A cylindrical object of outer diameter $10$ cm, height $20$
cm and density $8000$ $kg/m^3$ is supported by a vertical
spring and is half dipped in water as shown in
figure$(13-E6)$. (a) Find the elongation of the spring in
equilibrium condition. (b) If the object is slightly
depressed and released, find the time period of resulting
oscillations of the object. The spring constant
= $500$ N/m.

1 None

23 None

Solutionsa) F + U = mg(Where F= kx) $\\$ $\Rightarrow$ =$kx$ + V$\rho_{w}g = mg$ $\\$ b) F= $kx$ + $ V\rho_{w}$ $\times$ g $\\$ $\Rightarrow ma = KX + \pi r^2 \times (X) \times \rho_{w} \times g = (K + \pi r^2 \times \rho_{w} \times g)$X $\\$ $\Rightarrow w^2 \times (X) = \frac{K + \pi r^2 \times \rho_w \times g}{m} \times (X)$ $\\$ $\Rightarrow$ T =2 $\pi$$\sqrt \frac{m}{K + \pi r^2 \times \rho_{w} \times g}$

**24.** A wooden block of mass $0.5$ kg and density $800$ $kg/m^3$
is fastened to the free end of a vertical spring of spring
constant $50$ N/m fixed at the bottom. If the entire system
is completely immersed in water, find (a) the elongation
(or compression) of the spring in equilibrium and (b) the
time-period of vertical oscillations of the block when it
is slightly depressed and released.

24 None

SolutionsA) $mg =kX + V\rho_{w}g$ $\\$ b) $a = kX/m$ $\\$ $w^2$X = $kX/m$ $\\$ $T$ = 2 $\pi$ $\sqrt \frac{m}{k}$

**25.** A cube of ice of edge $4$ cm is placed in an empty
cylindrical glass of inner diameter $6$ cm. Assume that
the ice melts uniformly from each side so that it always
retains its cubical shape. Remembering that ice is lighter
than water, find the length of the edge of the ice cube
at the instant it just leaves contact with the bottom of
the glass.

1 None

25 None

SolutionsLet x $\rightarrow$ edge of ice block $\\$ When it just leaves contact with bottom of the glass .$\\$ h $\rightarrow$ height of water melted from ice $\\$ W = U $\\$ $\Rightarrow x^3 \times \rho_{ice} \times g = x^2 \times h \times \rho_{w} \times g$ $\\$ Again, volume of water formed, from melting of ice is given by, $\\$ $4^3 - x^3 = \pi \times r^2 \times h -x^2h$ $\\$ $\Rightarrow 4^3 - x^3 = \pi \times 3^2 \times h -x^2h$ $\\$ Putting h =$0.9$ x $\Rightarrow$ x = $2.26$cm.

**26.** A U-tube containing a liquid is accelerated horizontally
with a constant acceleration $a_o$. If the separation between
the vertical limbs is $l$, find the difference in the
heights of the liquid in the two arms.

If $p_a \rightarrow$ atm.Pressure $\\$ A $\rightarrow$ area of cross secton $\\$ h $\rightarrow$ increase in hright $\\$ $p_aA + A \times L \times \rho \times a_0 = p a^A + h \rho g \times A$ $\\$ $\Rightarrow hg =a_0L \Rightarrow a_0 \frac{L}{g}$

**27.** At Deoprayag (Garhwal, UP) river Alaknanda mixes
with the river Bhagirathi and becomes river Ganga.
Suppose Alaknanda has a width of $12$ m, Bhagirathi has
a width of $8$ m and Ganga has a width of $16$ m. Assume
that the depth of water is same in the three rivers. Let
the average speed of water in Alaknanda be $20$ km/h
and in Bhagirathi be $16$ km/h. Find the average speed
of water in the river Ganga.

27 None

SolutionsVolume of water, discharged from Alkananda + vol are of water discharged from bhagirathi = Volume of water flow in Ganga

**28.** Water flows through a horizontal tube of variable
cross-section (figure 13-E7). The area of cross-section at
A and B are 4 mm 2 and 2 mm 2 respectively. If 1 cc of
water enters per second through A, find (a) the speed of
water at A, (b) the speed of water at B and (c) the
pressure difference PA - Pi.

a) $a_A \times V_A = Q_A$ $\\$ b) $a_A \times V_A = a_B \times V_b$ $\\$ c) $\frac{1}{2} \rho V_A^{2} + P_A =\frac{1}{2} \rho V_B^{2} + P_B$ $\\$ $\Rightarrow (P_A -P_B) = \frac{1}{2} \rho (V_B^{2} - V_A^{2})$

**29.** Suppose the tube in the previous problem is kept vertical
with A upward but the other conditions remain the same.
The separation between the cross-sections at A
and B is $15/16$ cm. Repeat parts (a), (b) and (c) of the
previous problem. Take g = $10$ $m/s^2$.

29 None

SolutionsFrom Bernouli's equation, $\frac{1}{2} \rho V_A^{2} + \rho g h_A + p_A = \frac{1}{2} \rho V_B^{2} + \rho g h_B + p_B.$ $\\$ $\Rightarrow (P_A - P_B) = (\frac{1}{2}) \rho (V_B^{2} -V_A^{2})$

**30.** Suppose the tube in the previous problem is kept vertical
with B upward. Water enters through B at the rate of
$1$ $cm^3/s$ . Repeat parts (a), (b) and (c). Note that the
speed decreases as the water falls down.

$\frac{1}{2} \rho V_B^{2} +\rho g h_B + p_{B} = \frac{1}{2} \rho V_A^{2} + \rho g h_A +p_{A}$

**31.** Water flows through a tube shown in figure $(13-E8)$. The
areas of cross-section at A and B are $1$ cm $2$ and $0.5$ $cm^2$
respectively. The height difference between A and B is
$5cm$. If the speed of water at A is $10 cm/s$ find (a) the
speed at B and (b) the difference in pressures at A and B.

$\frac{1}{2} \rho V_A^{2} + \rho g h_A + P_A = \frac{1}{2} \rho V_B^{2} + \rho g h_B + P_B$ $\\$ $\Rightarrow P_B - P_A = \frac{1}{2} \rho(V_A^{2} - V_B^{2}) + \rho g (h_A - h_B)$

**32.** Water flows through a horizontal tube as shown in figure
$(13-E9$). If the difference of heights of water column in
the vertical tubes is $2 cm$, and the areas of cross-section
at A and B are $4 cm^2$ and $2 cm^2$ respectively, find the
rate of flow of water across any section.

$\overline{V_A}a_A = \overline{V_B} \times a_B$ $\\$ $\Rightarrow \frac{1}{2} \rho V_{A}^2 + \rho g h_A +p_A =\frac{1}{2} \rho V_{B}^2 + \rho g h_B +p_B$ $\\$ $\Rightarrow \frac{1}{2} \rho V_{A}^2 +p_A =\frac{1}{2} \rho V_{B}^2 +p_B$ $\\$ $\Rightarrow p_A -p_B = \frac{1}{2} \times \rho(V_{B}^2 - V_{A}^2)$ $\\$ Rate of flow = $V_a \times a_A$

**33.** Water flows through the tube shown in figure $(13-E10)$.
The areas of cross-section of the wide and the narrow
portions of the tube are $5 cm^2$ and $2 cm^2$ respectively.
The rate of flow of water through the tube is $500 cm^3/s$.
Find the difference of mercury levels in the U-tube.

33 None

Solutions$V_Aa_A = V_B a_B \times \frac{V_A}{B} = \frac{a_B}{a_A}$ $\\$ $5V_A =2V_B \Rightarrow V_B = \frac{5}{2}V_A$ $\\$ $\frac{1}{2} \rho V_{A}^2 + \rho g h_A +p_A =\frac{1}{2} \rho V_{B}^2 + \rho g h_B +p_B$ $\\$ $\Rightarrow p_A -P_B =\frac{1}{2} \rho (V_{B}^2 -V_{B}^2)$ (because $p_A - p_B = h \rho_m g$

**34.** Water leaks out from an open tank through a hole of
area $2 mm^2$ in the bottom. Suppose water is filled up to
a height of $80 cm$ and the area of cross-section of _the
tank is $0.4 m^2$. The pressure at the open surface and at
the hole are equal to the atmospheric pressure. Neglect
the small velocity of the water near the open surface in
the tank. (a) Find the initial speed of water coming out
of the hole. (b) Find the speed of water coming out
when half of water has leaked out. (c) Find the volume
of water leaked out during a time interval dt after the
height remained is h. Thus find the decrease in height
dh in terms of h and dt. (d) From the result of part
(c) find the time required for half of the water to leak
out.

34 None

Solutions$P_A + \frac{1}{2} \rho V_{A}^2 = P_B +\frac{1}{2} \rho V_{B}^2 \Rightarrow P_A -P_B =\frac{1}{2} \rho V_{B}^2$ {$V_A$ = 0} $\\$ $\Rightarrow \rho gh =\frac{1}{2} \rho V_{B}^2$ {$P_A = P_{atm} + \rho gh$} $\\$ $\Rightarrow V_B = \sqrt {2gh}$ $\\$ a) V = $\sqrt {(2gh)}$ $\\$ b) V = $\sqrt {2g \frac{(h}{2}} = \sqrt gh$ $\\$ c) V = $\sqrt {2gh}$ $\\$ V = aV $\times$ dt $\\$ AV = aV $\\$ $\Rightarrow A \times \frac{dh}{dt} = a \times \sqrt {2gh} \Rightarrow dh = \frac{a \times \sqrt {2gh} \times dt}{A}$ $\\$ d) dh = $\frac{a \times \sqrt 2gh \times dt}{A} \Rightarrow T = \frac{A}{a} \sqrt {\frac{2}{g}}[\sqrt H_1 - \sqrt H_2]$

**35.** Water level is maintained in a cylindrical vessel upto a
fixed height H. The vessel is kept on a horizontal plane.
At what height above the bottom should a hole be made
in the vessel so that the water stream coming out of the
hole strikes the horizontal plane at the greatest distance
from the vessel (figure $13-E11$).

V = $\sqrt {2g(H-h)}$ $\\$ t =$\sqrt {(\frac{2h}{g})}$ $\\$ x = v $\times$ t = $\sqrt{(2g(H-h) \times \frac{2h}{g}} = 4\sqrt {(Hh - h^2)}$ $\\$ So, $\Rightarrow \frac{d}{dh}(Hh -h^2) = 0 \Rightarrow 0 = H- 2h \Rightarrow h = \frac{H}{2}$.