Hc Verma I for Class 12 Science Physics Chapter 13 - Fluid Mechanics

Answer:

Density is defined as mass per unit volume. It tells how closely packed are the molecules inside an object. When molecules are tightly packed, we say that the object is dense; when molecules are far apart, we say that the object is light.
The density of a liquid depends on the mass of the molecules that make up the liquid and the closeness of the molecules of the liquid. It is not always true that the molecules of a dense liquid are heavier than the molecules of a lighter liquid.
For example, alcohol is less dense than oil. Alcohol molecules are mostly carbon and hydrogen atoms, so they are similar to oil. But they also contain an oxygen atom, which makes them a little heavy. For this reason, you might think that alcohol is denser than oil. But because of their shape and size, alcohol molecules do not pack as efficiently as oil molecules. This property of molecules makes alcohol less dense than oil.

Answer:

We know that pressure is the application of force over a particular area. In mathematical terms, pressure is equal to force divided by area; that is, with more force comes more pressure and with more area comes less pressure. An iron rod has more surface area than the pointed tip of a needle. That is why the pointed needle exerts more pressure than the iron rod (with the same force) when pressed against our skin.

Answer:

In case of an incompressible liquid, the density is independent of the variations in pressure and always remains constant. But it is not so in case of a compressible liquid. Thus, the given equation will not be strictly valid for a compressible liquid.

Answer:

Using the equation P_o − P = ρ_ogz, we get:
P = P_o − ρ_ogz
The pressure calculated by using this equation will be more than the actual pressure because density at a height of 10 km above Madras will be less than ρ_o.
Yes, the answer will change if we also consider the variation in g. Because g decreases with height, it will have the same effect on pressure as that of density.

Answer:

Yes, the normal to the free surface of the liquid passes through the centre of the Earth. Because of the gravitational force, the free surface of the liquid takes the shape of the surface of the Earth. Also, because the gravitational force is directed towards the centre of the Earth, the normal to the free surface also passes through the centre of the Earth (in all cases).

Answer:

The length of the mercury column will be more than 76 cm. The pressure depends on the height of the highest point of the mercury from the ground and not on the length of the liquid column. 

$$\begin{gathered} \mathrm{Let}: \\ l=\mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{mercury} \mathrm{column} \\ \theta =\mathrm{Angle} \mathrm{at} \mathrm{which} \mathrm{the} \mathrm{tube} \mathrm{is} \mathrm{inclined} \mathrm{with} \mathrm{the} \mathrm{vertical} \\ \mathrm{Given}: \\ h=76 \mathrm{cm} \\ l\cos \theta =h \\ \mathrm{or}, l=\frac{h}{\cos \theta } \\ \therefore l>h \end{gathered}$$
Or,
l > 76 cm

Answer:

No, mercury cannot be pulled up into the pump by this process. The level up to which mercury can rise is 76 cm (to maintain equal pressure at points A and B).

Answer:

Pressure at point A is 76 cm of mercury. Therefore, mercury will rise to full length of the tube, i.e., 1 m, to maintain equal pressure at points A and B. Inside the satellite, ${\mathrm{g}}_{\mathrm{effective}}=0$, so the pressure due to height of the mercury column will be zero.

Answer:

Two pressures are acting upon point P:
(1) Pressure due to mercury level above point P equals to P₁ (say)
(2) Atmospheric pressure = P₀ (inwards)
And,
P₀ > P₁
As the inward pressure is more, the mercury will not come out of the hole.

Answer:

Archimedes' principle is not valid in case of an elevator accelerating upwards, but it is valid for a car accelerating on a level road.
According to Archimedes' principle,
Buoyant force, B = Weight of the substituted liquid
Or,
B = mg

The above principle is satisfied in case of a car accelerating on a level road.
In case of an elevator, the buoyant force will be as below:
B = mg + ma  (If the elevator is going upwards with an acceleration a)
Thus, Archimedes' principle is not valid in this case.

Answer:

Whether an object sinks or floats in a liquid depends upon the density of the two. We know that sea water has dissolved salts in it, which increase its density. So, sea water exerts more buoyancy force (in the upward direction) on the swimmer than that exerted by fresh water. This helps the person to swim easily in sea water compared to fresh water.

Answer:

The water of mass equal to the mass of the ice cube will take less volume compared to the ice cube. The water will not overflow when the ice melts because the ice will displace the space it would take if it were in a liquid state.

Answer:

Some rocks should be added to increase the force acting in the downward direction. It will help the boat to pass under the bridge. If some rocks are removed, the upthrust of water on the boat will be greater than the weight of the boat. So, the boat will rise in water and will fail to pass under the bridge.

Answer:

Let a be the area of cross section and v be the velocity of water.
According to the equation of continuity,
av = Constant
or, $v \alpha \frac{1}{a}$
It means the larger the area of cross section, the smaller will be the velocity of liquid and vice versa.
Thus, as the water comes out of the vertical pipe, its velocity increases and area of cross section decreases.

Answer:

According to the equation of continuity, if the exit hole of the pipe is partially closed, the water stream comes out with more velocity due to decrease in area. This results in the water stream going to a larger distance.

Answer:

This can be explained through Bernoulli's principle, which states that the higher the air speed, the lower the pressure in that area. Because the air inside the car does not move, the pressure in the car is atmospheric. Because air moves outside the car (directly above it), the pressure is low. The canvas top of the Gipsy car is pushed upwards because the pressure inside the car is greater than the pressure directly above the car.

Answer:

(b) the forces between the molecules is stronger in solids than in liquids

The forces between the particles of a solid are stronger than those between the particles of a liquid, so the particles cannot move freely but can only vibrate. Thus, a solid has stable, definite shape and volume. A solid can only change its shape by force (when broken or cut), whereas a liquid can easily change its shape because of weak inter-particle forces.

Answer:

(b) the first is valid but not the second

For a point inside the elevator, pressure can be defined as $P=\underset{∆s\to 0}{\lim }\frac{F}{∆S}$. It is independent of the acceleration of the elevator.

The modified form of the second equation, which will be valid in the given case, is given by
$P_1-P_2=\rho (g+a_0) z$
Here, acceleration a₀ (say) due to elevator accelerating upwards is also taken into account.$$

Answer:

(c) maximum in vessel C

Here, the height of the liquid column is maximum in vessel C. Thus, the force on the base of vessel C, i.e., $F=P_0+h\mathrm{\rho }g$ where P₀ is atmospheric pressure, is maximum.

Answer:

(d) equal in all the vessels

The force on the base is given by
$$\begin{gathered} F=h\rho g\times A \\ \Rightarrow F=\left(hA\rho \right)g \\ \Rightarrow \mathrm{F}=(\mathrm{Volume}\times \mathrm{Density})\times \mathrm{g} \\ \Rightarrow F=\mathrm{mg} \end{gathered}$$
In the question, the masses are equal. So, the force on the base is the same in all cases.

Answer:

(d) zero

At both points A and B, pressure is equal to atmospheric pressure.  

$$\begin{gathered} \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ {P}_{\mathrm{A}}=P_{\mathrm{B}}=P_{\mathrm{atm}} \\ \Rightarrow P_{\mathrm{B}}-P_{\mathrm{A}}=0 \end{gathered}$$

Answer:

(b) decrease

As the air inside the jar is pumped out, the air pressure decreases. Thus, the pressure in the liquid near the bottom of the beaker decreases.

Answer:

(b) the elevator only

The two points in the same horizontal line will not have equal pressure if the liquid is accelerated horizontally. There should be vertical acceleration.

Answer:

(a) P₁ = 0, P₂ = atmospheric pressure

The upper part of the tube contains vacuum as the mercury goes down and no air is allowed in. Thus, the pressure at the upper end, i.e., at the surface of mercury in a barometer tube is zero (P₁ = 0). However, the pressure at the surface of mercury in the cup or any another point at the same horizontal plane is equal to the atmospheric pressure.

Answer:

(c) < 76 cm

If the elevator goes up at an increasing speed, then the effective value of g increases.
We know:
$P=\mathrm{\rho }gh$
So, h will have a lesser value for the same value of P if g increases.

Answer:

(c) > 76 cm

When the elevator is going upwards with acceleration a, the effective acceleration is a' = (g + a).
Thus, pressure is given by
$P=h\mathrm{\rho }\mathit{(}g\mathit{+}a\mathit{)}\mathit{ }$
Air pressure in the elevator = $P=h\text{'}\mathrm{\rho }g$
Because the pressure is the same, h' > h.
∴ Air pressure > 76 cm

Answer:

(d) < 76 cm

Because of the trapped air, the pressure at the upper end of the mercury column inside the tube is not zero.
In other words, ${\mathrm{P}}_0>0$.
Using this relation, we get:
$$\begin{gathered} P_{\mathrm{atm}}=P_0+\mathrm{\rho }gh \\ \mathrm{Here}, \\ \mathrm{\rho }=\mathrm{Density} \mathrm{of} \mathrm{mercury} \\ h=\mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{mercury} \mathrm{column} \\ \because P_0>0 \\ \mathrm{And}, \\ {P}_{\mathrm{atm}}>\mathrm{\rho }gh \\ \therefore 76 \mathrm{cm} \mathrm{of} \mathrm{Hg}>\mathrm{\rho }gh \\ \mathrm{or}, h<76 \mathrm{cm} \end{gathered}$$

Answer:

(c) 44 N

Upthrust exerted by the water on the block = Change in the reading of the spring balance
                                                                     = (20 − 16) N = 4 N
Downthrust = 4 N
Actual weight of the beaker containing water = 40 N
∴ Effective weight = (40 + 4) N = 44 N

Answer:

(c) same part in the water

When more air is pushed into the bottle from the pump, the pressure of air increases on the wood as well as on the water surface with the same amount. So, the level of water and wood does not change. Thus, the piece of wood floats with the same part in the water.

Answer:

(c) will remain the same

In the absence of water, the force acting on the bottom of the vessel is due to the air and the cube. Now, when water is filled in the vessel, the force due to the water and the cube is greater. The extra force is balanced by the buoyant force acting on the cube in the upward direction.

Answer:

(a) P₁ = P₂ = P₃

If the fluid is in equilibrium, then the pressure is the same at all points in the same horizontal level.

Answer:

(c) passes through a point to the left of the centre

When the box is accelerated towards right, the water in the box experiences a pseudo force (ma) towards left, where m is the mass of water. So, the resultant normal force by the water on the top of the box passes through a point to the left of the centre.

Answer:

(b) F₁ > F₂

When the box is accelerated towards right, the water in the box experiences a pseudo force (ma) towards left, where m is the mass of water. So, the force F₁ exerted by the water on the the left wall of the box is greater.

Answer:

(d) each case

This happens in accordance with the equation of continuity.
As the area of the cross section of cylindrical tube AB is constant, the velocity of water will also be the same. The equation is derived from the principle of conservation of mass and it is true for every case, i.e., when the tube is either horizontal or vertical.

Answer:

(c) energy

The principle behind the Bernoulli theorem is the law of conservation of energy. It states that energy can be neither created nor destroyed; it merely changes from one form to another. 

Answer:

(d) P_A = P_B  because  the cross-sectional areas at A and B are equal.

According to Bernoulli's theorem, pressures at points A and B of the horizontal tube will be equal if water has the same velocity at these points.
Also, according to the equation of continuity, velocity at points A and B will be equal only if the cross-sectional areas at A and B are equal.
So, P_A = P_B only if the cross-sectional areas at A and B are equal.

Answer:

(a) v₁ = v₂

The velocity of efflux does not depend on the density of the liquid. It only depends on the height h (given same in the question) and acceleration due to gravity g (constant value here).

$v_1=v_2=v=\sqrt{2gh}$

Answer:

(c) The water level will rise to a height v²/2g and then stop.

From the principle of continuity and Bernoulli's equation, ​we have:
$$\begin{gathered} v^2=2gh \\ \Rightarrow h=\frac{v^2}{2g} \end{gathered}$$

So, h is the maximum height up to which the water level will rise if the water is ejected at a speed v.

Answer:

(a) The solid exerts a force equal to its weight on the liquid.
(b) The liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid.
(c) The weight of the displaced liquid equals the weight of the solid.

Force exerted by any solid on a liquid = F = mg = W = Weight of the solid 
According to Archimedes' principle, any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
Also, any floating object displaces its own weight of fluid. Thus, we can say that the weight of the object is equal to the weight of the fluid displaced.

Answer:

Given:
Height of the water tank above the tap level, h = 4 m
Acceleration due to gravity, g = 10 m/s²
Density of water, ρ = 10³ kg/m³
When the tap is closed, the pressure of the water in the tap is
P = hρg
On substituting the respective values in the formula, we get:
P = 4 × 10³ × 10
   = 40,000 N/m²

It is necessary to specify that the tap is closed because if the tap is open, then the pressure gradually decreases as h decreases and also because the pressure in the tap is atmospheric.

Answer:

(a) Given:
Height of the first arm, h₁ = 8 cm
Height of the second arm, h₂ = 2 cm
Density of mercury, ρ_Hg = 13.6 gm/cc
Atmospheric pressure, p_a = 1.01 × 10⁵ N/m² = 1.01 × 10⁶ dyn/cm²
Now,
Let p_g be the pressure of the gas.
If we consider both limbs, then the pressure at the bottom of the tube will be the same.



According to the figure, we have:
$$\begin{gathered} p_g+{\rho }_{\mathrm{H}g}\times h_2\times g=p_a+{\mathrm{\rho }}_{\mathrm{Hg}}\times h_1\times g \\ \Rightarrow p_g=p_a+{\mathrm{\rho }}_{\mathrm{H}g}\times g(h_1-h_2) \\ =1.01\times {10}^6+13.6\times 980\times (8-2) \mathrm{dyn}/{\mathrm{cm}}^2 \\ =(1.01\times {10}^6+13.6\times 980\times 6) \mathrm{dyn}/{\mathrm{cm}}^2 \\ =1.09\times {10}^{5 }\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Pressure of the mercury at the bottom of the U-tube:
p_Hg = pa + ρ_Hg× h₁ × g
      $$\begin{gathered} =1.01\times {10}^6+13.6\times 8\times 980 \\ =1.12\times {10}^5 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Answer:

 Given:
Area of the wider tube, A = 900 cm²
Weight of the boy, m = 45 kg
Density of water, $\rho$ = 10³ kgm⁻³
Let h be the difference in the levels of water in the tubes and p_a be the atmospheric pressure.


As per the figure, we have:
$p_a\mathit{+}h\rho g\mathit{=}{p}_a\mathit{+}\frac{mg}{A}$
$$\begin{gathered} \Rightarrow h\rho g\mathit{=}\frac{mg}{A} \\ \Rightarrow h\mathit{=}\frac{m}{\rho A} \\ \Rightarrow h=\frac{m}{1000\times \mathrm{A}} \\ =\frac{45}{1000\times 900\times {10}^{-4}}=\frac{1}{2} \mathrm{m}=50 \mathrm{cm} \end{gathered}$$

Answer:

Given:
$$\begin{gathered} \mathrm{Atmospheric} \mathrm{pressure}, p_{\mathrm{a}}=1.0\times {10}^5 \mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Density} \mathrm{of} \mathrm{water}, {\rho }_W={10}^3 \mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{Acceleration} \mathrm{due} \mathrm{to} \mathrm{gravity}, g=10 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Volume} \mathrm{of} \mathrm{water}, V=500 \mathrm{mL}\approx 500 \mathrm{g}\approx 0.5 \mathrm{kg} \end{gathered}$$
Area of the top of the glass, A = 20 m²
Height of the glass, h = 20 cm​

(a) Force exerted on the bottom of the glass = Atmospheric force + Force due to cylindrical water column or glass
$$\begin{gathered} =p_{\mathrm{a}}\times A+A\times h\times {\rho }_{\mathrm{w}}\times g \\ =A\left({\rho }_a+h{\rho }_{w}g\right) \\ =20\times {10}^{-4}\left({10}^5+20\times {10}^{-2}\times {10}^3\times 10\right) \\ =204 \mathrm{N} \end{gathered}$$

(b) Let F_​s be the force exerted by the sides of the glass. Now, from the free body diagram of water inside the glass, we can find out the resultant force exerted by the sides of the glass.
Thus, we have:
$$\begin{gathered} {\mathrm{p}}_{\mathrm{a}}\times A+mg=A\times h\times {\rho }_{\mathrm{w}}\times g+F_{\mathrm{s}}+{\mathrm{p}}_{\mathrm{a}}\times A \\ \Rightarrow mg=A\times h\times {\rho }_{\mathrm{w}}\times g+F_{\mathrm{s}} \\ \Rightarrow 0.5\times 1=20\times {10}^{-4}\times 20\times {10}^{-2}\times {10}^{-3}\times 10+F_{\mathrm{s}} \\ \Rightarrow F_s=5-4=1 \mathrm{N} \left(\mathrm{upward}\right) \end{gathered}$$

Answer:

When the glass is covered by a jar and the air is pumped out of the jar, atmospheric pressure has no effect on the glass.

(a) Force exerted on the bottom:
$\left(h{\mathrm{\rho }}_{\mathrm{w}}g\right)\times A$
$$\begin{gathered} =\left(20\times {10}^{-2}\times {10}^3\times 10\right)20\times {10}^{-4} \\ =4 \mathrm{N} \\ \left(b\right) mg=h\times {\rho }_{\mathrm{w}}\times g\times A+F_{\mathrm{s}} \\ \Rightarrow F_{\mathrm{s}}=5-4=1 \mathrm{N} \end{gathered}$$
(c) If we use a glass of different shape with same volume, height and area, then there will not be any change in the answer.

Answer:

Case: When water is used in a barometer instead of mercury
Pressure exerted by 76 cm of mercury column, P = 76 × 13.6 × g dyn/cm²
Density of water, ${\rho }_w$ = 10³ kg/m³
Let h be the height of the water column
$$\begin{gathered} \because P=h{\rho }_{w}g \\ \therefore 76\times 13.6\times g= h\times {\rho }_{\mathrm{w}}\times g \\ \Rightarrow h=\frac{76\times 13.6}{1} \\ \Rightarrow h=1033.6 \mathrm{cm} \end{gathered}$$

Answer:

(a) W₂ = W₁
(c) W₂ < W₁ + w

According to the question, the density of air inside and outside the balloon is the same. So, the weight w of air inside the balloon is equal to the weight of displaced air. Thus, the spring balance will not register any difference because the balloon will experience buoyant force equal to w that cancels out the weight of the added air.

Answer:

(c) decrease if it is taken partially out of the liquid
(d) be in the vertically upward direction.

The force exerted by the liquid on the solid is the vertically upward force (buoyant force) that opposes the weight of the immersed solid.​ As more and more volume of the solid is immersed in the liquid, the buoyant force increases.
Buoyant force depends on the weight of the displaced liquid. So, maximum upward buoyant force acts on the solid when it is completely immersed in the liquid. It decreases if the solid is taken partially out of the liquid. Once the object is immersed in the liquid, then pushing it further in the liquid does not increase the buoyant force.

Answer:

(b) The pressure at the surface of the water will decrease.
(c) The force by the water on the bottom of the vessel will decrease.

As air is pumped out of the hole, there is a decrease in the atmospheric pressure above the water surface in the vessel. Due to this, pressure at the surface of the water decreases. Thus, the force exerted by the water on the bottom of the vessel also decreases.

Answer:

(c) the kinetic energies of all the particles arriving at a given point are the same
(d) the momenta of all the particles arriving at a given point are the same

In a streamline flow, every fluid particle arriving at a given point has the same velocity v. Thus, the kinetic energies ($\frac{1}{2}m{v}^2$ ) and momenta (mv) of all particles arriving at a given point are the same, as the mass of a particle is constant.

Answer:

(a) Flow in both the tubes are steady.
(b) Flow in both the tubes are turbulent.
(c) Flow is steady in A but turbulent in B.

In a steady flow, the velocity of liquid particles reaching a particular point is the same at all times, but if the liquid is pushed in the tube at a rapid rate, i.e., if the flow rate increases, then the flow may become turbulent. Here, the flow rate is the volume of fluid per unit time per unit area flowing past a point.
Large volume of water passes through tube B compared to tube A. Thus, the flow rate is greater in tube B than in tube A. So, if the flow is turbulent in A, then the flow in B cannot be steady. Therefore, the first three options are possible.

Answer:

(c) The pressures are equal if the tube has a uniform cross section.
(d) The pressures may be equal even if the tube has a nonuniform cross section.

In streamline flow in a tube, every particle of the liquid follows the path of its preceding particle and the velocity of all particles crossing a particular point is the same. However, the velocity of the particles at different points in their path may not necessarily be the same. Thus, by applying Bernoulli's theorem and equation of continuity, we can say that if the tube has a uniform cross section, the pressures will be equal; and if the tube has a non-uniform cross section, the pressures may or may not be equal.

Answer:

(a) area of the hole
(b) density of the liquid

The emergent speed v of the liquid flowing from the hole in the bottom of the tank is given by
$v=\sqrt{2gh}$
Here, g is acceleration due to gravity and h is height of the liquid from the hole.
Thus, it is clear from the above relation that the speed of the liquid depends on the height of the liquid from the hole and on the acceleration due to gravity. It does not depend on the area of the hole and the density of the liquid.

Answer:

Given:
Depth of the stone from the water surface, h = 500 m
Area of the plane surface of the large stone, A = 2 m² 
Density of water, ρ_w = 10³ kgm^−3
​Force (F) is given by
$$\begin{gathered} F=P\times A=\left(h{\rho }_w\times g\right)A \left(P=\mathrm{Pressure}\right) \\ \Rightarrow F=\left(500\times {10}^3\times 10\right)\times 2 \\ ={10}^7 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
The force does not depend on the orientation of the rock when the surface area of the stone remains the same.

Answer:

Dimensions of the rectangular tank:
Length, l = 3 m
Breadth, b = 2 m
Height, h = 1 m
Area of the bottom surface of the tank, A = $2\times 3=6 {\mathrm{m}}^2$
Density of water, ρ_w = 1000 kgm⁻³

(a) Total force exerted by water on the bottom surface of the tank:
$$\begin{gathered} f=Ah{\rho }_{\mathrm{w}}g \\ =6\times 1\times {10}^3\times 10 \\ =6\times {10}^4=60,000 \mathrm{N} \end{gathered}$$

(b) Force exerted by water on the strip of width δx:
     $$\begin{gathered} \mathrm{d}f=p\times A=\left(x{\rho }_{w}g\right)\times \mathrm{A} \\ =x\times {10}^3\times 10\times 2\times \delta x \\ =20,000x\delta x \mathrm{N} \end{gathered}$$

(c) Inside the tank, the water force acts in every direction due to adhesion. Therefore, torque is given by
     $$\begin{gathered} di=F\times r \\ =20,000\times \mathrm{\delta }x(1-x) \mathrm{N} \end{gathered}$$

(d) Total force exerted by water on the side about the bottom edge $\left(F\right)$:
         $$\begin{gathered} F={\int }_0^{1}20,000 x\delta x \\ \Rightarrow F=20,000{\left[\frac{x^2}{2}\right]}_0^1 \\ =10,000 \mathrm{N} \end{gathered}$$
(e) Torque by the water on the side $\left(\tau \right)$:
      $$\begin{gathered} \tau =20,000\times {\int }_0^{1}x\delta x\left(1-x\right) \\ =20,000{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1 \\ =20,000\times \left[\frac{1}{2}-\frac{1}{3}\right] \\ =\frac{20,000}{6} \mathrm{Nm} =\frac{10000}{3} \mathrm{Nm} \end{gathered}$$

Answer:

Given:
Weight of the ornament in air, m₁a = 36 gm
Weight of the ornament in water, m₂w = 34 gm
Specific gravity (density) of gold, ρ_Au = 19.3 gm/cc
Specific gravity (density) of copper, ρ_Cu = 8.9 gm/cc
Using Archimedes' principle, we get:
Loss of weight = Weight of displaced water
                        = 36 − 34
                        = 2 gm
Let m_Au and m_Cu be the masses of gold and copper, respectively.
Now, the mass of the ornament $\left(m_c\right)$ will be
m_c = m_Au + m_Cu= 36 gm       ...(i)
Now, let the volume of the ornament in cm be V.
$$\begin{gathered} \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ V\times {\rho }_{\mathrm{w}}\times g=2\times g \\ \Rightarrow (v_{\mathrm{Au}}+v_{\mathrm{Cu}})\times {\mathrm{\rho }}_w\times g=2\times g \left({\rho }_w=\mathrm{Density} \mathrm{of} \mathrm{water}\right) \\ \Rightarrow \left(\frac{m_{Au}}{{\rho }_{\mathrm{Au}}}+\frac{m_{Cu}}{{\rho }_{\mathrm{Cu}}}\right){\rho }_w\times g=2\times g \\ \Rightarrow \left(\frac{m_{Au}}{19.3}+\frac{m_{Cu}}{8.9}\right)\times 1=2 \\ \Rightarrow 8.9 m_{\mathrm{Au}}+19.3 m_{\mathrm{Cu}}=2\times 19.3\times 8.9 \\ =343.54 ...\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ 8.9 (m_{\mathrm{Au}}+m_{\mathrm{Cu}})=8.9\times 36 \\ \Rightarrow 8.9 m_{\mathrm{Au}}+8.9m_{\mathrm{Cu}}=320.40 ...\left(\mathrm{iii}\right) \\ \mathrm{From} \left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ 10.4 m_{\mathrm{Cu}}=23.14 \\ \Rightarrow m_{\mathrm{Cu}}=2.225 \mathrm{gm} \end{gathered}$$

Therefore, the amount of copper present in the ornament is 2.2 gm.

Answer:

Given:
Mass of copper, m_Au = 36 gm
$$\begin{gathered} \mathrm{Now}, \\ \left(\frac{m_{\mathrm{Au}}}{{\rho }_{\mathrm{Au}}}+V_c\right){\rho }_{\mathrm{w}}\times g=2\times g \\ \mathrm{Here}, \\ {V}_{\mathrm{c}}=\mathrm{Volume} \mathrm{of} \mathrm{cavity} \\ {\rho }_{\mathrm{Au}}= \mathrm{Density} \mathrm{of} \mathrm{gold} \\ {\rho }_w=\mathrm{Density} \mathrm{of} \mathrm{water} \\ \mathrm{On} \mathrm{substituting} \mathrm{the} \mathrm{respective} \mathrm{values}, \mathrm{we} \mathrm{get}: \\ \left(\frac{36}{19.3}+V_{\mathrm{c}}\right)\times 1=2 \\ \Rightarrow V_{\mathrm{c}}=2-\frac{36}{19.3} \\ =\frac{36-6}{19.3} \\ =0.112 {\mathrm{cm}}^3 \end{gathered}$$

Answer:

Given:
Mass of the metal piece, m =160 gm = 160 × 10⁻³ kg
Density of the metal piece, ρ_m = 8000 kg/m³
Density of the water, ρ_w = 1000 kg/m^3
​Let R be the normal reaction and U be the upward thrust.

From the diagram, we have:
mg = U + R
$\Rightarrow$R = mg −Vρ_wg            [U = Vρ_wg]
$\Rightarrow R=mg-\frac{m}{{\rho }_m}\times {\rho }_w\times g$
       $$\begin{gathered} =160\times {10}^{-3}\times \left(10-\frac{{10}^3\times 10}{8000}\right) \\ =160\times {10}^{-3}\times 10\left(1-\frac{1}{8}\right) \\ =1.4 \mathrm{N} \end{gathered}$$

Answer:

Answer:

Given:
Specific gravity of ice, ρ_ice = 0.9 gm/cc
Weight of the metal piece, m = 5 kg
Density of water, ${\rho }_w$ = 10³ kg/m^3​
Let x be the minimum edge of the ice block in cm.
We have:
mg + W_ice = U
Here,
U = Upward thrust
W_ice = Weight of the ice
$$\begin{gathered} \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ 0.5\times g+x^3\times {\rho }_{ice}\times g=x^3\times {\rho }_w\times g \left[\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{displaced}=x^3\right] \\ \Rightarrow 0.5\times {10}^3+x^3\times (0.9)=x^3\times 1 \\ \Rightarrow x^3\times (0.1)=(0.5)\times {10}^3 \\ \Rightarrow x^3=5\times {10}^3 \\ \Rightarrow x=17.09 \mathrm{cm} \\ \Rightarrow x=17 \mathrm{cm} \end{gathered}$$

Answer:

Given:
Specific gravity of water, ${\rho }_W$ = 1 gm/cc
Specific gravity of ice, ρ_ice = 0.9 gm/cc
Specific gravity of kerosene oil, ρ_k = 0.8 gm/cc
Now,
V_​ice = V_k + V_w
Here,
V_k = Volume of ice inside kerosene oil
V_w = Volume of ice inside water
V_ice = Volume of ice 
Thus, we have:
$$\begin{gathered} V_{\mathrm{ice}}\times {\mathrm{\rho }}_{\mathrm{ice}}\times g=V_{\mathrm{k}}\times {\rho }_{\mathrm{k}}\times g+V_{\mathrm{w}}\times {\rho }_{\mathrm{w}}\times g \\ \Rightarrow \left(V_k+V_w\right)\times {\rho }_{\mathrm{ice}}=V_{\mathrm{k}}\times {\rho }_{\mathrm{k}}+V_{\mathrm{w}}\times {\rho }_{\mathrm{w}} \\ \Rightarrow (0.9) V_{\mathrm{k}}+(0.9)V_{\mathrm{w}}=(0.8)V_{\mathrm{k}}+\left(1\right)\times V_{\mathrm{w}} \\ \Rightarrow (0.1)V_{\mathrm{w}}=0.1 V_{\mathrm{k}} \\ \Rightarrow \frac{V_{\mathrm{w}}}{V_{\mathrm{k}}}=1. \\ \Rightarrow V_{\mathrm{w}}:V_{\mathrm{k}}=1:1 \end{gathered}$$

Answer:

Given:
Density of iron, ρ_I = 8000 kg/m³ = 8 gm/u
Density of water, ρ_w = 1000 kg/m³ = 1 gm/u
Let x be the external edge of iron.
According to Archemedes' principle,
Weight displaced = Upward thrust
∴ w = u


For the given condition, we have:
Weight of the box = Buoyant force
⇒ V₁${\rho }_I$g = vρ_wg
⇒ (x² × (0.1) × 6) × 8 = x³ × 1            [Volume of iron = v₁ = 6  times the volume of each sheet]
⇒ x = 4.8 cm

Answer:

Given:
Density of wood, ρ_w = 0.8 gm/cc
Density of lead, ρ_pb = 11.3 gm/cc
Weight of the cubical wood block, m_w  = 200 g


The cubical block floats in water.
Now,
(m_w+ m_pb) × g = (V_w + V_pb)ρ × g
Here,
ρ = Density of water
V_w = Volume of wood
V_pb = Volume of lead
$$\begin{gathered} \Rightarrow (m_{\mathrm{w}}+m_{\mathrm{pb}})=\left(\frac{m_{\mathrm{w}}}{{\mathrm{\rho }}_{\mathrm{w}}}+\frac{m_{\mathrm{pb}}}{{\mathrm{\rho }}_{\mathrm{pb}}}\right)\mathrm{\rho } \\ \Rightarrow (200+m_{\mathrm{pb}})=\left(\frac{200}{0.8}+\frac{m_{\mathrm{pb}}}{11.3}\right)\times 1 \\ \Rightarrow m_{\mathrm{pb}}-\frac{m_{\mathrm{pb}}}{11.3}=250-200 \\ \Rightarrow \frac{10.3 m_{\mathrm{pb}}}{11.3}=50 \\ \Rightarrow m_{pb}=\frac{50\times 11.3}{10.3}=54.8 \mathrm{gm} \end{gathered}$$

Answer:

Given:
Mass of wood, m_w = 200 g
Specific gravity of wood, ${\rho }_W$ = 0.8 gm/cc
Specific gravity of lead, ${\rho }_{PB}$= 11.3 gm/cc
$$\begin{gathered} \mathrm{We} \mathrm{know}: \\ Mg=w \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ (m_w+m_{pb})g=V_w\times \rho \times g [\mathrm{\rho }=\mathrm{Density} \mathrm{of} \mathrm{water}] \\ \Rightarrow 200+m_{pb}=\frac{200}{0.8}\times 1 \\ \Rightarrow m_{pb}=250-200=50 \mathrm{gm} \end{gathered}$$

Answer:

Given:
Length of the edge of the metal block, x  = 12 cm
Specific gravity of mercury, ${\rho }_{Hg}$= 13.6 gm/cc
It is given that $\frac{1}{5}$th of the cubical block is inside mercury initially.
Let ${\rho }_b$ be the density of the block in gm/cc.
$$\begin{gathered} \therefore (x{)}^3\times {\rho }_b\times g=(x{)}^2\times \left(\frac{x}{5}\right)\times {\rho }_{\mathrm{Hg}}\times g \\ \Rightarrow (12{)}^3\times {\rho }_b\times g=(12{)}^2\times \frac{12}{5}\times 13.6 \\ \Rightarrow {\rho }_b=\frac{13.6}{5} \mathrm{gm}/\mathrm{cc} \end{gathered}$$
Let y be the height of the water column after the water is poured.
∴ V_b = V_Hg + V_w = (12)³
Here,
V_Hg = Volume of the block inside mercury
V_w  = Volume of the block inside water
$$\begin{gathered} \therefore (V_{\mathrm{b}}\times {\rho }_b\times g)=(V_{\mathrm{Hg}}\times {\rho }_{\mathrm{Hg}}\times g)+(V_{\mathrm{w}}\times {\rho }_{\mathrm{w}}\times g) \\ \Rightarrow (V_{\mathrm{Hg}}+V_w)\times \frac{13.6}{5}=V_{\mathrm{Hg}}\times 13.6+V_{\mathrm{w}}\times 1 \\ \Rightarrow (12{)}^3\times \frac{13.6}{5}=(12-y)\times (12{)}^2\times 13.6+\left(y\right)\times (12{)}^2\times 1 \\ \Rightarrow 12\times \frac{13.6}{5}=(12-y)\times 13.6+(y) \\ \Rightarrow 12.6y=13.6\left(12-\frac{12}{5}\right)=(13.6)\times (9.6) \\ \Rightarrow y=\frac{(9.6)\times (13.6)}{(12.6)}=10.4 \mathrm{cm} \end{gathered}$$

Answer:

Given:
Inner radius of the hollow spherical body, r₁ = 6 cm
Outer radius of the hollow spherical body, r₂ = 8 cm 
Let the density of the material of the sphere be $\rho$ and the volume of the water displaced by the hollow sphere be V.
If ${\rho }_w$ is the density of water, then:
$$\begin{gathered} \mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{displaced}=\left(\frac{V}{2}\right)\left({\rho }_{\mathrm{w}}\right)\times g \\ \mathrm{We} \mathrm{know}: \\ \mathrm{Upward} \mathrm{thrust}=\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{displaced} \\ \therefore \left(\frac{4}{3}\mathrm{\pi }{r}_{\mathit{3}}^{\mathit{2}}\mathit{-}\frac{\mathit{4}}{\mathit{3}}\mathrm{\pi }{r}_1^3\right)\rho =\left(\frac{1}{2}\right)\frac{4}{3}\mathrm{\pi }{r}_2^3\mathit{\times }{\rho }_{\mathrm{w}} \\ \Rightarrow \left(r_2^3-r_1^3\right)\times \rho =\left(\frac{1}{2}\right)r_2^3\times 1 \\ \Rightarrow (8{)}^3-(6{)}^3\times \rho =\left(\frac{1}{2}\right)(8{)}^3\times 1 \\ \Rightarrow \rho =\frac{512}{2\times (512-216)} \\ = \frac{512}{2\times 296}= 0.865 \mathrm{gm}/\mathrm{cc} =865 \mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$³

Answer:

Given:
Radius of the sphere, r = 5 cm
Mass of the maximum load, m = 0.1 kg
Let the weight of the sphere be W₁ and the weight of the load be W₂.
Now,
W₁ + W₂ = U
Here, U is the upward thrust.
Let V be the volume of the sphere.
$$\begin{gathered} \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ mg+V\times {\rho }_s\times g=v\times {\rho }_w\times g \\ \mathrm{Here}, \\ {\rho }_{\mathrm{s}}=\mathrm{Density} \mathrm{of} \mathrm{the} \mathrm{sphere} \mathrm{in} \mathrm{gm}/\mathrm{cc} \\ {\rho }_w=\mathrm{Density} \mathrm{of} \mathrm{water} \\ \mathrm{On} \mathrm{substituting} \mathrm{the} \mathrm{respective} \mathrm{values} \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ (0.1)\times {10}^3+\left(\frac{4}{3}\right)\times \mathrm{\pi }\times (5{)}^3\times {\rho }_{\mathrm{s}}=\left(\frac{4}{3}\right)\times \mathrm{\pi }\times (5{)}^3\times 1 \\ \Rightarrow 100=\left(\frac{4}{3}\right)\times \mathrm{\pi }\times 125\times (1-{\rho }_s) \\ \Rightarrow 1-{\rho }_s=\frac{3\times 100}{4\times \mathrm{\pi }\times 125}=0.19 \\ \Rightarrow {\rho }_s=1-(0.19) \\ =0.81 \mathrm{gm}/\mathrm{cc}=0.8 \mathrm{gm}/\mathrm{cc} \end{gathered}$$

Answer:

Given:
Density of iron, ρ_I= 7800 kgm⁻³
Density of wood, ρ_w = 800 kgm⁻³
Density of air, ρ_air = 1.293 kgm⁻³
$$\begin{gathered} \mathrm{Net} \mathrm{weight} \mathrm{of} \mathrm{Iron} \left(W_1\right): \\ {W}_I=m_{I}g-V_{\mathrm{I}}{\rho }_{air}\times g \\ \mathrm{Here}, m_{\mathrm{I}}\mathit{ }\mathrm{and} V_{\mathrm{I}} \mathrm{are} \mathrm{the} \mathrm{mass} \mathrm{and} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{iron}, \mathrm{respectively}. \\ \mathrm{Now}, \\ \left[m_{\mathrm{I}}-\left(\frac{m_{\mathrm{I}}}{{\rho }_I}\right){\mathrm{\rho }}_{air}\right]g \\ =\left[1-\left(\frac{1}{7800}\right)\times 1.293\right]\times (9.8) \\ \mathrm{Net} \mathrm{weight} \mathrm{of} \mathrm{wood}=W_w=m_{\mathrm{w}}g-V_w . {\mathrm{\rho }}_{air} g \\ \mathrm{Here}, m_{\mathrm{w}}\mathit{ }\mathrm{and} {\mathrm{V}}_{\mathrm{w}} \mathrm{are} \mathrm{the} \mathrm{mass} \mathrm{and} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wood}, \mathrm{respectively}. \\ \mathrm{Now}, \\ \left[m-\left(\frac{m}{{\rho }_w}\right){\mathrm{\rho }}_{air}\right]g \\ =\left(1-\frac{1}{800}\times 1.293\right)9.8 \\ \therefore \frac{W_{\mathrm{I}}}{W_w}=\frac{9.8\left({\displaystyle \frac{7800-1.293}{7800}}\right)}{9.8\left({\displaystyle \frac{800-1.293}{800}}\right)} \\ =\frac{7800-1.293}{800-1.293}\times \frac{8}{78}=1.0015 \end{gathered}$$

Answer:

Given:
Outer diameter of the cylindrical object, d = 20 cm
∴ Radius$, r=\frac{d}{2}=10 \mathrm{cm}$
Mass of the object, m = 2 kg
When the object is depressed into water, the net unbalanced force causes simple harmonic motion.
Let x be the displacement of the block from the equilibrium position.
Now,
$$\begin{gathered} \mathrm{Driving} \mathrm{force}=U=V{\rho }_{\mathrm{w}}g \\ \mathrm{Here}, \\ U \mathrm{is} \mathrm{the} \mathrm{upward} \mathrm{thrust}. \\ V\mathit{ }\mathrm{is} \mathrm{the} \mathrm{volume}. \\ {\rho }_{\mathrm{w}}\mathit{ }\mathrm{is} \mathrm{the} \mathrm{density} \mathrm{of} \mathrm{water}. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ ma=\mathrm{\pi }{r}^2\left(x\right)\times {\rho }_{\mathrm{w}}g \\ a\mathit{=}\frac{\pi r^{\mathit{2}}{\rho }_w\mathit{9}x}{\mathit{2}\mathit{\times }{\mathit{10}}^{\mathit{3}}} \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{Displacement}}{\mathrm{Acceleration}}} \\ =2\mathrm{\pi }\sqrt{\frac{\left(x\right)\times 2\times {10}^3}{\left(\mathrm{\pi }{r}^2{\mathrm{\rho }}_{w}g\left(x\right)\right)}} \\ =2\mathrm{\pi }\sqrt{\frac{2\times {10}^3}{\mathrm{\pi }\times (10{)}^2\times 1\times 10}} \\ =2\mathrm{\pi }\sqrt{\frac{2}{\mathrm{\pi }\times 10}}=0.5 \mathrm{s} \end{gathered}$$
Therefore, the required time period is 0.5 second.

Answer:

Given:
Diameter of the cylindrical object, d = 10 cm
∴ Radius, r = 5 cm
Height of the object, h = 20 cm
Density of the object, ρ_b = 8000 kg/m³ = 8 gm/cc
Density of water, ρ_w = 1000 kg/m³
Spring constant, k =  500 N/m = 500 × 10³ dyn/cm

(a) Now,
F + U = mg                   [F = kx]
Here, U is the upward thrust and x is the small displacement from the equilibrium position.
kx + Vρ_wg = mg
$$\begin{gathered} \Rightarrow 500\times {10}^3\times \left(x\right)+\left(\mathrm{\pi }{r}^2\right)\times \left(\frac{h}{2}\right)\times 1\times 1000=\mathrm{\pi }{r}^2\times h\times {\mathrm{\rho }}_b\times 1000 \\ \Rightarrow 500\times {10}^3\times \left(x\right)= \mathrm{\pi }{r}^2\times h\times 1000\left({\rho }_{\mathrm{b}}-\frac{1}{2}\right) \\ =\mathrm{\pi }\times (5{)}^2\times 20\times 1000\left({\rho }_{\mathrm{b}}-\frac{1}{2}\right) \\ \Rightarrow 50x=\mathrm{\pi }\times 25\times 2\times \left({\rho }_{\mathrm{b}}-\frac{1}{2}\right) \\ x=\mathrm{\pi }(8-0.5) \\ \mathrm{or}, x=\mathrm{\pi }\times 7.5=23.5 \mathrm{cm} \end{gathered}$$

(b) We know that X is the displacement of the block from the equilibrium position.
Now,
Driving force:
$$\begin{gathered} F=kX+V{\mathrm{\rho }}_w\times g \\ \Rightarrow ma=kx+\mathrm{\pi }{r}^2\times \left(X\right)\times {\rho }_{\mathrm{w}}\times g \\ =(k+\mathrm{\pi }{r}^2\times {\rho }_{\mathrm{w}}\times g)x \\ \Rightarrow {\mathrm{\omega }}^2\times \left(X\right)=\frac{\left(k+\mathrm{\pi }{r}^2\times {\rho }_{\mathrm{w}}\times g\right)}{m}\times \left(X\right) \\ \mathrm{In} \mathrm{SHM} \left(a={\omega }^{2}X\right), \mathrm{time} \mathrm{period} \left(T\right) \mathrm{is} \\ T=2\mathrm{\pi }\frac{m}{k+\mathrm{\pi }{r}^2\times {\rho }_w\times g} \\ =2\mathrm{\pi }\frac{\mathrm{\pi }\times 25\times 20\times 8}{500+{10}^3+\mathrm{\pi }\times 25\times 1\times 1000} \\ =0.935 \mathrm{s} \end{gathered}$$

Answer:

Given:
Mass of the wooden block, m = 5 kg
Density of the block, ρ = 800 kg/m³
Spring constant, k = 50 N/m
Density of water, ρ_w = 1000 kg/m³
(a) Using the free body diagram, we get:
mg = kx + Vρ_wg          (Here, Vρ_wg > mg. So, there will be some elongation.]
$$\begin{gathered} \Rightarrow (0.5)\times (10+50\times (x\left)\right)= \left(\frac{0.5}{800}\right)\times {10}^3\times \left(10\right) \\ \Rightarrow 50x=\left(0.5\right)\times \left(10\right)\times \left(\left(\frac{10}{8}\right)-1\right) \\ \Rightarrow 50x=\left(\frac{5}{4}\right) \\ \Rightarrow x=2.5 \mathrm{cm} \end{gathered}$$

(b) As the system is inside the water, the unbalance force will be the driving force, which is kx for SHM.
Hence, there will be no change in the buoyant force.
$$\begin{gathered} \therefore ma=kx \\ \Rightarrow a=kx/m \\ \Rightarrow w^{2}x=kx/m \\ \Rightarrow {\left(\frac{2p}{\mathrm{T}}\right)}^2=k/m \\ \Rightarrow T=2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ =2\mathrm{\pi }\times \sqrt{\left(\frac{0.5}{50}\right)}=\frac{\mathrm{\pi }}{5} \mathrm{s} \end{gathered}$$

Answer:

Let the length of the edge of the ice block when it just leaves contact with the bottom of the glass be x and the height of water after melting be h.
Given:
Inner diameter of the cylindrical glass = 6 cm
∴ Inner radius, r = 3 cm
Edge of the ice cube = 4 cm
$$\begin{gathered} \mathrm{Weight} = \mathrm{Upward} \mathrm{thurst} \\ \Rightarrow (x{)}^3\times {\rho }_{\mathrm{ice}}\times g=(x{)}^2\times h\times {\mathrm{\rho }}_{\mathrm{w}}\times g \\ \Rightarrow h=(0.9)x \end{gathered}$$
Again, volume of the water left from the melting of the ice is given by
$$\begin{gathered} (4{)}^3-(x{)}^3=\mathrm{\pi }\times (r{)}^2\times h-x^{2}h [\mathrm{Amount} \mathrm{of} \mathrm{water}=\mathrm{\rho }(r_2-x_2)h] \\ \Rightarrow (4{)}^3-(\mathrm{x}{)}^3=\mathrm{\pi }\times (3{)}^2\times h-x^{2}h \\ \mathrm{Putting} h=0.9x, \mathrm{we} \mathrm{get}: \\ (4{)}^3-(\mathrm{x}{)}^3=\mathrm{\pi }\times (3{)}^2\times (0.9)\mathrm{x} \end{gathered}$$
On solving the above equation, we get:
x = 2.26 cm

Answer:

Let a₀ be the acceleration with which U-tube is accelerated horizontally. So, the horizontal part will experience some inertial force.
Also,
p_a = Atmospheric pressure
A = Area of cross section
h = Increase in height of the liquid
According to the question,
$$\begin{gathered} p_{\mathrm{a}}A+A\times l\times \rho \times a_0=p_{\mathrm{a}}A+h\rho g\times A \\ \Rightarrow hg=a_{0}l \\ \Rightarrow h=\frac{a_{0}l}{g} \end{gathered}$$

Answer:

Given:
Average speed of water in Alaknanda = 20 kmh⁻¹
Average speed of water in Bhagirathi = 16 kmh⁻¹
Width of Bhagirathi = 8 m
Width of Alaknanda = 12 m
Width of Ganga = 16 m
Now,
Volume of water discharged from Alaknanda + Volume of water discharged from Bhagirathi = Volume of water flow in Ganga
Let:
d = Depth of the three rivers
⇒V_A × d × 12 + V_B × d × 8 = V_G × d × 16
Here, V_A, V_B and V_G be the average speeds of water in Alaknanda, Bhagirathi and Ganga, respectively.
Using the equation of continuity, we get:
$$\begin{gathered} 20\times 12+16\times 8=V_{\mathrm{G}}\times 16 \\ \Rightarrow V_{\mathrm{G}} \times 16= 368 \\ \Rightarrow V_{\mathrm{G}}=\frac{368}{16} \\ =23 \mathrm{km}/\mathrm{h} \end{gathered}$$

Answer:

(a) Given:
The areas of cross-sections of the tube at A and B are 4 mm² and 2 mm², respectively.
1 cc of water enters per second through A.
Let:
$\rho$ = Density of the liquid
P_A = Pressure of liquid at A
P_B = Pressure of liquid at B

(a)
$$\begin{gathered} \frac{Q_{\mathrm{A}}}{t}=\frac{1a}{\mathrm{s}}=\mathrm{Discharge} \\ \Rightarrow a_{\mathrm{A}}\times V_{\mathrm{A}}=Q_{\mathrm{A}} \\ \mathrm{Given}: \\ a_A=4 {\mathrm{mm}}^2=4\times {10}^{-2} {\mathrm{cm}}^2 \\ \therefore 4\times {10}^{-2}\times V_{\mathrm{A}}=1\mathrm{cc}/\mathrm{s} \\ \Rightarrow V_{\mathrm{A}}=25 \mathrm{m}/\mathrm{s} \\ \left(b\right) a_A\times V_{\mathrm{A}}=a_{\mathrm{B}}\times V_{\mathrm{B}} \\ \Rightarrow 4\times {10}^{-2}\times 25=2\times {10}^{-2}\times V_{\mathrm{B}} \\ \Rightarrow V_{\mathrm{B}}=50 \mathrm{cm}/\mathrm{s} \end{gathered}$$

(c) By Bernoulli's equation, we have:
$$\begin{gathered} \frac{1}{2}\rho v_{\mathrm{A}}^2+P_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+P_{\mathrm{B}} \\ \Rightarrow (P_{\mathrm{A}}-P_{\mathrm{B}})=\frac{1}{2}\rho \left(v_B^2-v_A^2\right) \\ =\frac{1}{2}\times 1\times (2500-625) \\ =\frac{1875}{2}=937.5 \mathrm{dyn}/{\mathrm{cm}}^2 \\ =93.75 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Answer:

Given:
Separation between the cross sections at A and B = 15/16 cm
Speed of water at A, v_A = 25 cm/s
Speed of water at B, v_B = 50 cm/s
(c) By Bernoulli's equation, we get:
$$\begin{gathered} \frac{1}{2}p{v}_{\mathrm{A}}^2+\mathrm{\rho }g{h}_{\mathrm{A}}+P_{\mathrm{A}}=\frac{1}{2}p{v}_{\mathrm{B}}^2+\mathrm{\rho }g{h}_{\mathrm{B}}+P_{\mathrm{B}} \\ \Rightarrow P_{\mathrm{A}}\mathit{-}{P}_{\mathit{B}}=\left(\frac{1}{2}\right)p\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right)+\rho g\left(h_{\mathrm{B}}-h_{\mathrm{A}}\right) [h_{\mathrm{A}}-h_{\mathrm{B}}=\left(\frac{15}{16}\right)\mathrm{cm}] \\ \Rightarrow P_{\mathit{A}}\mathit{-}{P}_{\mathrm{B}}=\frac{1}{2}(2500-625)-1000\times \frac{15}{16} \\ =0 \end{gathered}$$

Answer:

Water enters through B at the rate of 1 cm³s⁻¹.

(a) Speed of water at A, ${\overrightarrow{\mathrm{V}}}_{\mathrm{A}}=25 \mathrm{cm}/\mathrm{s}$
(b) Speed of water at B, ${\overrightarrow{\mathrm{V}}}_{\mathrm{B}}=50 \mathrm{cm}/\mathrm{s}$
(c) By Bernoulli's equation, we have:
$$\begin{gathered} \frac{1}{2}p{v}_A^2+\rho g{h}_A+P_A=\frac{1}{2}\rho {v_B}^2+\rho g{h}_B+P_B \\ \Rightarrow P_{\mathit{A}}\mathit{-}{P}_{\mathit{B}}=\frac{1}{2}\times 1\times 1875+\frac{1}{2}\times 1000\times \frac{15}{16} \\ =1875 \mathrm{dyn}/{\mathrm{cm}}^2=188 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Answer:

Given:
Difference in the heights of A and B = 5 cm
Area of cross section at A, a_a = 1 cm²
Area of cross section at B, a_b = 0.5 cm²
Speed of water at A, $v_A$ = 10 cms⁻¹

$$\begin{gathered} \left(a\right) \mathrm{From} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{continuity}, \mathrm{we} \mathrm{have}: \\ {\overrightarrow{V}}_{\mathrm{A}}\times a_{\mathrm{A}}={\overrightarrow{V}}_{\mathrm{B}}\times a_{\mathrm{B}} \\ \Rightarrow 10\times 1={\overrightarrow{V}}_{\mathrm{B}}\times 0.5 \\ \Rightarrow {\overrightarrow{V}}_{\mathrm{B}}=20 \mathrm{cm}/\mathrm{s} \end{gathered}$$
The required speed of water at cross section B is 20 cms⁻¹

$$\begin{gathered} \left(b\right) \mathrm{From} \mathrm{Bernoulli}\text{'}\mathrm{s} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \frac{1}{2}\rho v_{\mathrm{A}}^2+\rho g{h}_{\mathrm{A}}+P_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_{\mathrm{A}}+P_{\mathrm{B}} \\ \Rightarrow P_{\mathrm{B}}-P_{\mathrm{A}}=\frac{1}{2}\rho \left(v_{\mathrm{A}}^2-v_{\mathrm{B}}^2\right)+\rho g\left(h_{\mathrm{A}}-h_{\mathrm{B}}\right) \end{gathered}$$
Here,
P_A and P_B are the pressures at A and B, respectively.
h_A and h_B are the heights of points A and B, respectively.
ρ is the density of the liquid.
On substituting the values, we have:
$$\begin{gathered} P_B-P_A=\frac{1}{2}\times 1(100-400)+1\times 1000(5.0) \\ =-150+5000=4850 \mathrm{Dyne}/{\mathrm{cm}}^2 \\ =485 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
Therefore, the required pressure difference at A and B is 485 N/m².

Answer:

Given:
Difference in the heights of water columns in vertical tubes = 2 cm
Area of cross section at A, a_A = 4 cm²
Area of cross section at B, a_B = 2 cm²
Now, let v_A and v_B be the speeds of water at A and B, respectively.
From the equation of continuity, we have:
 $$\begin{gathered} v_A{a}_A=v_B\times a_B \\ \Rightarrow v_{\mathrm{A}}\times 4=v_{\mathrm{B}}\times 2 \\ \Rightarrow v_{\mathrm{B}}=2v_{\mathrm{A}} ...\left(\mathrm{i}\right) \end{gathered}$$
From Bernoulli's equation, we have:
$$\begin{gathered} \frac{1}{2}\rho v_A^2+\rho g{h}_A+p_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_B+p_{\mathrm{B}} \\ \Rightarrow \frac{1}{2}\mathrm{\rho }{v}_{\mathrm{A}}^2+p_{\mathrm{A}}=\frac{1}{2}\mathrm{\rho }{v}_{\mathrm{B}}^2+p_{\mathrm{B}} \\ \Rightarrow p_{\mathrm{A}}-p_{\mathrm{B}}=\frac{1}{2}\mathrm{\rho }\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right) \end{gathered}$$
Here,
p_A and p_B are the pressures at A and B, respectively.
h_A and h_B are the heights of water columns at point A and B, respectively.
ρ is the density of the liquid.
Thus, we have:
$$\begin{gathered} \frac{1}{2}\times 1\times \left(4v_{\mathrm{A}}^2-v_{\mathrm{A}}^2\right) \\ \Rightarrow 2\times 1\times 1000=\frac{1}{2}\times 1\times 3v_{\mathrm{A}}^2 \\ [p_{\mathrm{A}}-p_{\mathrm{B}}=2 \mathrm{cm}=2\times 1\times 1000 \mathrm{dyne}/{\mathrm{cm}}^2 (\mathrm{water} \mathrm{column}\left)\right] \\ \Rightarrow v_{\mathrm{A}}^2=\sqrt{\frac{4000}{3}}=36.51 \mathrm{cm}/\mathrm{s} \\ \therefore \mathrm{Rate} \mathrm{of} \mathrm{flow}=v_A{\mathrm{a}}_{\mathrm{A}}=36.51\times 4 \\ =146 {\mathrm{cm}}^3/\mathrm{s} \end{gathered}$$
Hence, the required rate of flow of water across any section is 146 cm³/s.

Answer:

Given:
Area of cross section of the wide portions of the tube, a_a = 5 cm²
Area of cross section of the narrow portions of the tube, a_b= 2 cm² 
Now, let v_a and v_b be the speeds of water at A and B, respectively.
Rate of flow of water through the tube = 500 cm³/s
$$\begin{gathered} \Rightarrow v_{\mathrm{A}}=\left(\frac{500}{5}\right)=100 \mathrm{cm}/\mathrm{s} \\ \mathrm{From} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{continuity}, \mathrm{we} \mathrm{have}: \\ {v}_{\mathrm{A}}{a}_{\mathrm{A}}=v_{\mathrm{B}}{a}_{\mathrm{B}} \\ \Rightarrow \frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{a_{\mathrm{B}}}{a_{\mathrm{A}}}=\frac{2}{5} \\ \Rightarrow 5v_{\mathrm{A}}=2v_{\mathrm{B}} \\ \Rightarrow v_{\mathrm{B}}=\left(\frac{5}{2}\right)v_{\mathrm{A}} ...\left(\mathrm{i}\right) \\ \mathrm{From} \mathrm{the} \mathrm{Bernoulli}\text{'}\mathrm{s} \mathrm{equation}, \mathrm{we} \mathrm{have}: \\ \frac{1}{2}\rho v_{\mathrm{A}}^2+\rho g{h}_{\mathrm{A}}+p_{\mathrm{A}}=\frac{1}{2}\rho v_{\mathrm{B}}^2+\rho g{h}_{\mathrm{B}}+p_{\mathrm{B}} \\ \Rightarrow p_{\mathrm{A}}-p_{\mathrm{B}}=\frac{1}{2}\left[p\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right)\right] \end{gathered}$$
Here,
ρ is the density of the fluid.
p_A and p_B are the pressures at A and B.
h is the difference of the mercury levels in the U-tube.
$$\begin{gathered} \mathrm{Now}, \\ h\times 13.6\times 980=\frac{1}{2}\times 1\times \frac{21}{4}(100{)}^2 [\mathrm{Using} \left(\mathrm{i}\right)] \\ \Rightarrow h=\frac{21\times (100{)}^2}{2\times 13.6\times 980\times 4} [\because p_{\mathrm{A}}-p_{\mathrm{B}}] \\ =1.969 \mathrm{cm} \end{gathered}$$

Therefore, the required difference is 1.969 cm.
      

Answer:

Given:
Area of the hole of the water tank = 2 mm²
Height of filled water, h = 80 cm
Area of the cross-section of the tanks, A = 0.4 m²
Acceleration due to gravity, g = 9.8 ms^$-2$
Pressure at the open surface and at the hole is equal to atmospheric pressure.

(a) Velocity of water:
v$=\sqrt{2gh}$
  $$\begin{gathered} = \sqrt{2\times 10\times 0.80} \\ = 4 \mathrm{m}/\sec \end{gathered}$$

(b) Velocity of water when the tank is half-filled and h $\mathrm{is} \frac{80}{2}, \mathrm{i}.\mathrm{e}., 40 \mathrm{cm}:$
$$\begin{gathered} = \sqrt{2\times 10\times 0.40} \\ = \sqrt{8} \mathrm{m}/\sec \end{gathered}$$

(c) Volume:
$$\begin{gathered} \mathrm{Volume}=Ah=A v \mathrm{dt} \\ =A\mathit{\times }\sqrt{2gh} dt \\ \mathit{=}\left(\mathit{2}\mathit{ }{\mathit{mm}}^{\mathit{2}}\right)\sqrt{2gh}dt \end{gathered}$$
Volume of the tank = Ah = V (say)
$\mathrm{i}.\mathrm{e}., \frac{d\mathrm{V}}{dt}=A\frac{dh}{dt}$
$$\begin{gathered} a_1{v}_1=A\frac{\mathrm{d}h}{\mathrm{d}t} \\ \Rightarrow 2\times {10}^{-6}\sqrt{2gh}=0.4\frac{\mathrm{d}h}{\mathrm{d}t} \\ \Rightarrow \mathrm{d}h=5\times {10}^{-6}\sqrt{2gh}\mathrm{d}t \end{gathered}$$
$$\begin{gathered} \left(d\right) \because dh=5\times {10}^{-6} \sqrt{2gh} dt \\ \therefore \frac{dh}{\sqrt{2gh}}=5\times {10}^{-6} dt \end{gathered}$$
On integrating, we get:
$$\begin{gathered} 5\times {10}^{-6}{\int }_0^{t}dt=\frac{1}{\sqrt{28}}{\int }_{0.8}^{0.4}\frac{d\mathrm{h}}{\sqrt{h}} \\ =5\times {10}^{-6}\times t=\frac{1}{\sqrt{28}}\times 2{\left[h^{\frac{1}{2}}\right]}_{0.8}^{0.4} \\ =5\times {10}^{-6}\times t=\frac{1}{\sqrt{28}}\times 2{\left[h^{\frac{1}{2}}\right]}_{0.8}^{0.4} \\ \Rightarrow t=\frac{1}{\sqrt{20}}\times 2\times \left[(0.4{)}^{1/2}-(0.8{)}^{1/4}\right]\times \frac{1}{5\times {10}^{-6}} \\ \Rightarrow t=\frac{1}{4.47}\times 2\times \frac{2}{3.16}\times \frac{1}{5\times {10}^{-6}}\times \frac{1}{3600} \mathrm{h} \\ = 6.51 \mathrm{h} \end{gathered}$$
Thus, the time required to leak half of the water out is 6.51 hours.

Answer:

It is given that H is the height of the cylindrical vessel.


Now, let h be the height of the hole from the surface of the tank.
The velocity of water $\left(v\right)$ is given by
$$\begin{gathered} v=\sqrt{2g(\mathrm{H}-h)} \\ \mathrm{Also}, \mathrm{let} t \mathrm{be} \mathrm{the} \mathrm{time} \mathrm{of} \mathrm{flight}. \\ \mathrm{Now}, \\ t=\sqrt{\frac{2h}{g}} \\ \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{maximum} \mathrm{horizontal} \mathrm{distance}. \\ \therefore x=v\times t \\ =\sqrt{2g(\mathrm{H}-h)}\times \sqrt{\frac{2h}{g}} \\ =\sqrt{4(\mathrm{H}h-h^2)} \end{gathered}$$
For x to be maximum,
$$\begin{gathered} \frac{d}{dh}\left(\mathrm{H}h-h^2\right)=0 \\ \Rightarrow 0=\mathrm{H}-2h \\ \Rightarrow h=\frac{\mathrm{H}}{2} \end{gathered}$$