Answer the following question.

Question 15 Marks

The mass of a tile is 500 g . If the density of the tile is $2.5 g / cm ^3$, what will be the weight of the tile when it is completely immersed in water?
$\left(g=9.8 m / s^2, \rho(\text { water })=1000 kg / m^3\right)$

Answer

Self

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Question 25 Marks

A body of mass 200 g and volume $400 cm^3$ is put in a bucket containing water. Will it float or sink? [ $\rho$ (water) $=1 g / cm ^3$ ]

Answer

Proceed as above.
$\rho$ (body) $=200 g 400 cm 3=0.5 g / cm ^3$
It is less than the density of water.
$\therefore$ The body will float in water.

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Question 35 Marks

A body of mass 200 g and volume $50 cm^3$ is put in a bucket containing water. Will it float or sink ?
$\left[\rho\right.$ (water) $=1 g / cm ^3$ ]

Answer

Data: $m =200 g, V =50 cm^3$,
$\rho$ (water) $=1 g / cm ^3$
Density $(\rho)=$ mass volume
$\therefore \rho \text { (body) }=20050 cm 3$
It is greater than the density of water.
Hence, the body will sink in water.

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Question 45 Marks

A body of volume loo $cm ^3$ is immersed completely in water. Find the weight of the water displaced by the body. 1 g $=9.8 m / s ^2 . p$ (water) $= kg / m ^3 ]$

Answer

Data: $V =100 cm^3=100 \times 10^{-6} m^3$
$=1 \times 10 m^3, p$ (water) $=10 kg / m ^3$
$g =9.8 m / s ^2$, weight of the displaced water ?
Density = mass volume
$\therefore$ Mass $=$ volume $\times$ density
Volume of the water displaced by the body $=1 \times 10^{-4} m^3$
$\therefore$ Mass of the water displaced,
$m=1 \times 10^{-4} m^3 \times 10 kg / m^3=0.1 kg$
$\therefore$ Weight of the water displaced
$=mg=0.1 kg \times 9.8 m / s^2=0.98 N .$
The weight of the water displaced by the body $=0.98 N$.

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Question 55 Marks

A metal block has thmensions $10 cm \times 5 cm \times 2 cm$ and the density of the metal is $8 \times 10^3 kg / m ^3$. It is kept on a table with the face $10 cm \times 5 cm$ in contact with the table. Find the force and pressure exerted by the block on the table. $\left(g=9.8 m / s ^2\right)$

Answer

Data : $1=10 cm, b =5 cm$,
$h =2 cm, p =8 \times 10 kg / m 3, g=9.8 m / s ^2$,
$A = lb =10 cm \times 5 cm=50 cm^2=50 \times 104 m^2$
$=5 \times 10^{-5} m^2$, force $=$ ?, pressure $=$ ?
Volume of the block $= lbh =$
$10 cm \times 5 cm \times 2 cm=100 cm^3$
$=100 \times 10^{-6} m^3=1 \times 10^{-4} m^3$
Mass of the block $=$ volume $\times$ density
( $\because$ density = mass/volume)
$\therefore$ Mass of the block,
$m =1 \times 10^{-4} m^3 \times 8 \times 10^3 kg / m ^3=0.8 kg$
Weight of the block $= mg =0.8 kg \times 9.8 m / s ^2=7.84 N$
$\therefore$ The force exerted by the block on the table $=7.84 N$.
force 7.84 NPressure $=\frac{\text { force }}{\text { area }}=\frac{7.84 N}{5 \times 10^{-3} m^2}$
$=1.568 \times 10^3 N / m^2 \text { or } 1.568 \times 10^3 Pa$
The pressure exerted by the block on the table $=1.568 \times 10 N / m 2$ or $1.568 \times 10^3$ Pa.

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Question 65 Marks

A force of 1000 N is applied over an area $50 cm \times 20 cm$. Find the corresponding pressure.

Answer

Data: $F =1000 N$,
$A =50 cm \times 20 cm=0.5 m \times 0.2 m=0.1 m^2$,
pressure $=? P=\frac{F}{A}=\frac{1000 N}{0.1 m^2}=10^4 N / m ^2$
The pressure $=10 N / m ^2$.

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Question 75 Marks

Measure the length, breadth, height and mass of a rectangular tiffin box. Find the weight of the box and calculate the pressure in two different positions as in Ex. (1) above.

Answer

Let $I =0.25 m, 6=0.1 m, h =0.05 m, F = W =0.5 N$
(i) $A = bh =0.1 m \times 0.05 m=0.005 m^2:$. Pressure $P=\frac{F}{A}=\frac{0 . N }{0.00 m m ^2}=100 Pa$.
(ii) $A =1 b=0.25 m \times 0.1 m=0.025 m 2$
$\therefore$ Pressure, $P=\frac{F}{A}=\frac{0.5 N}{0.02 mm^2}=20 Pa$.

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Question 85 Marks

(i) Calculate the pressure exerted by the wooden block when it is kept in the vertical position.
Given: The length of the wooden block is 80 cm, the breadth is 50 cm, the thickness is 20 cm and the weight is 500 N
(ii) Also calculate the pressure when the wooden block is kept in the horizontal position with its surface 80 cm × 50 cm touching the floor.

Answer

Data: $F = W =500 N, I =80 cm=0.8 m, b =50 cm=0.5 m . h =20 cm=0.2 m$
(i) $A = bh =0.5 m \times 0.2 m=0.1 m^2 P=\frac{F}{A}=\frac{500 N}{0.1 m^2}=5000 Maharash / m ^2$ or 5000 PaPa
The pressure exerted in the vertical position of the block $=5000 N / m ^2$ or 5000 Pa .
(ii) $A = lb =0.8 m \times 0.5 m=0.4 m^2$
$P=\frac{F}{A}=\frac{500 N}{0.4 m^2}=1250 N / Mah ^2$ or 1250 Pa
The pressure exerted in the horizontal position of the block $=1250 N / m 2$ or 1250 Pa .

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Question 95 Marks

A plastic ball is released underwater. State whether it will sink or come up to the surface of water. Give the reason.

Answer

A plastic ball released under water will come up to the surface of water. The density of water is greater than that of plastic. Hence, when a plastic ball is under water, the magnitude of the buoyant force exerted by water on the ball is greater than the magnitude of the weight of the ball.

Therefore, the ball will start moving upward. As it comes up with part of the ball above the water surface, the volume of the water displaced by the ball becomes less and hence at a certain stage, the buoyant force and the weight balance each other. Then the ball continues to remain in that state, as the net force on the ball becomes zero.

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Question 105 Marks

A piece of wood floats both in water and kerosine. In which liquid does it sink more during floating? Why?

Answer

The piece of wood sinks more in kerosine than in water during floating. The density of kerosine is less than that of water. The buoyant force on a body is proportional to the density of the liquid in which the body is immersed. When a body floats, the magnitude of the buoyant force acting on the body is equal to that of the weight of the body.

Hence, the volume of the liquid displaced by a floating body is inversely proportional to the density of the liquid. As a result, when a piece of wood floats in kerosine, it displaces greater volume of kerosine compared to the volume of water displaced when the piece of wood floats in water. Hence, it sinks more in kerosine than in water.

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Question 115 Marks

What is density of a substance? I Obtain its SI unit.

Answer

The density of a substance is the ratio of its mass to volume.
The SI unit of density = the SI unit of mass the SI unit of volume $= kg / m ^3$
[Note: Density is useful in determining the purity of a substance. The CGS unit of density is $g / cm ^3$.
$1 kg / m^3=10^3 g /(100 cm)^3=10^{-3} g / cm^3$
$\left.\therefore 1 g / cm^3=1000 kg / m^3\right]$

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Question 125 Marks

How does the doctor’s syringe work?

Answer

As shown in the figure, a syringe used by children when they play with coloured water consists of a cylinder made of plastic or metal fitted with a piston. One end of the cylinder is in the form of a narrow

tube. The snugly fitting piston can slide in , and out smoothly. The rod connected to the i piston passes through a hole in the centre of the lid and has a handle at the other end.

When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure. When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure.

Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out. To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

The tip of a syringe is fitted with a very fine and hollow needle. The required quantity of medicine can be taken in the syringe with the help of the piston. The medicine can then be injected into the body of a patient using the needle and the piston.

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Question 135 Marks

Explain the working of a syringe

Answer

tube. The snugly fitting piston can slide in , and out smoothly. The rod connected to the i piston passes through a hole in the centre of the lid and has a handle at the other end. When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure.

When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure. Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out.

To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

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Question 145 Marks

Explain the working of an ink dropper.

Answer

An ink dropper consists of a tube of glass or plastic, with one end tapering to a narrow opening and the other end fitted with a small rubber bulb. When the narrow open end is dipped into the ink and the rubber bulb is pressed, some air in the tube escapes through the open end. This reduces the air pressure inside the dropper.

On releasing the bulb, the atmospheric pressure on the ink pushes the ink into the dropper. The dropper is then taken out and its open end is held over the open barrel of the pen. The bulb is then pressed so that the ink in the dropper enters the pen.

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Question 155 Marks

Explain why a person may bleed from the nose when at a great height above the sea level.

Answer

The pressure exerted by the blood in blood capillaries is slightly more than the atmospheric pressure and acts in a direction opposite to that of the atmospheric pressure. Atmospheric pressure decreases with height and at a great height above the sea level, it is very low.

As a result, there arises a difference in the internal and external pressures on the walls of the cells and blood capillaries. If the difference is large, it may cause the cell wall and the blood capillaries to burst. Thus, the capillaries in the nose (and ear) may burst causing bleeding.

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Question 165 Marks

Think – Why?

1.Some people feel their ears popping at the top of a mountain.

2.Some people feel breathless as they climb higher and higher on a mountain.

3.We can enjoy a cold-drink or fruit juice with the help of a straw but can we imagine drinking a cold-drink or fruit juice on the moon using a straw?

4.People are often advised not to carry fountain pens while travelling by air.

Answer

1.Atmospheric pressure decreases with altitude. At the top of a mountain, it becomes less than the internal pressure in the ear. Hence, some people feel their ears popping at the top of a mountain.

2.Atmospheric pressure decreases with altitude. Hence, some people feel breathless as they climb higher and higher on a mountain.

3.When we suck air in the straw, the pressure of the air in the straw becomes less than that of the outside air on the cold drink or fruit juice in the bottle (or the glass). Hence, the cold drink or fruit juice rises in the straw and enters our mouth. We can then drink it. There is no atmosphere on the moon. Hence, it is not possible to enjoy a cold drink or fruituice on the moon by using a straw.

4.The ink in a fountain pen (filled at sea level at atmosphere pressure) may come out through its mouth while travelling by air if the outside pressure becomes less than the pressure in the ink holder of the pen. This can spoil the clothes/purse/bag. Hence, people are often advised not to carry fountain pens while travelling by air.

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Question 175 Marks

State the relation between 1 atmosphere and the pascal.

Answer

1 atmosphere $=$ (about) $101 \times 10^3 Pa$ (pascal)
[Note: The air pressure at the sea level is (about) 1 atmosphere.
1 atmosphere $=101325 Pa$.
$1 bar =10^3 mbar$ (millibar)
1 mbar w $10^3 Pa$ (hectopascal)
Atmospheric pressure is expressed in mbar or hectopascal (hPa).]

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Question 185 Marks

What is meant by atmospheric pressure?

Answer

The earth is surrounded by air from all sides. This layer of air is called the atmosphere. Its density is high up to about 16 km from the earth’s surface. Beyond that, up to about 400 km, its density is very low. Air, due to its weight, exerts pressure on the surface of the earth.

The pressure exerted by air or the atmosphere surrounding the earth is known as the atmospheric pressure. It is the ratio of the weight of the air to the area of the surface of the earth. It decreases with altitude as the density of air decreases with altitude and also the weight of the air column above a given place.

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Question 195 Marks

State the characteristics of the pressure due to a liquid (or a fluid in general).
OR
Write a short note on the pressure due to a liquid (a fluid in general).

Answer

The pressure at a point in a liquid (or a fluid) is due to the weight of the liquid (fluid) column above that point.
It acts on all sides of the container.
At a given depth it is the same in all directions.
It is independent of the size and shape of the container.
It is proportional to the height of the liquid (fluid) column above the given point.
It is proportional to the density of the liquid (fluid).
It is proportional to the acceleration due to gravity at the given place.

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Question 205 Marks

With neat diagrams, describe an experiment to show that pressure increases if the surface area is decreased, keeping the applied force the same.

Answer

Take a brick measuring $20 cm \times 10 cm \times 5 cm$. Take some clay in a glass trough. Add water to it and knead it into a soft dough.Place the brick on the dough with one of its faces measuring $20 cm \times 10 cm$ in contact with the dough. Observe how deep the brick penetrates into the dough..

Clean the brick and place it on the dough with one of its faces measuring $10 cm \times 5 cm$ in contact with the dough. Observe how deep the brick penetrates into the dough. You will find that the brick penetrates deeper in this case than that in the first case.

In the first case, the weight of the brick acts on a surface area of $200 cm^2$.
In the second case, the weight of the brick acts on a surface area of $50 cm^2$.
This shows that pressure increases if the surface area is decreased, keeping the applied force the same.

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Question 215 Marks

Explain with a suitable example that pressure varies inversely as the area of the surface on which the force is applied, if the force remains constant.

Answer

The tip of a pointed nail has an extremely small area, while that of a blunt nail has a comparatively large area. A given force creates a large pressure on the pointed nail and it can be easily hammered into the wood, while a very less pressure is created on the blunt nail and it cannot be easily hammered into the wood.

This shows that pressure varies inversely as the area of the surface on which the force is applied, if the force remains constant. If the same force is applied to surfaces having different areas, the pressure is more on the surface having a smaller area.

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Question 225 Marks

What is inertia of motion? Give two examples of inertia of motion.

Answer

The inherent property of a body by virtue of which it cannot change its state of motion is called the inertia of motion.
Examples:

When a fan is switched off, its blades continue to rotate for some time. Due to internal friction and friction with air, the blades of the fan stop rotating after some time.
Passengers in a bus experience a forwarderk when the bus stops suddenly due to application of brakes.

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Question 235 Marks

What is inertia of rest ? Give two examples of inertia of rest.

Answer

The inherent property of a body by virtue of which it cannot change its state of rest is called the inertia of rest.
Examples:

When we dust a carpet, the carpet moves but the dust particles in it remain at rest due to inertia and hence get separated from the carpet. Hence, the carpet becomes clean.
When a bus starts suddenly, the passengers experience a backwarderk due to inertia.

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Question 245 Marks

Explain: Force has magnitude as well as direction. OR Force is a vector quantity.

Answer

The effect of force applied to a body depends upon how much force we supply, i.e., the magnitude of the force, and the direction in which the force is applied. Consider a ball at rest on the ground. When ; we push it, it starts rolling. The greater the applied force, the greater is the speed acquired by the ball.

Consider a body moving in a straight line, If we apply a force in the direction of motion of the body, the speed of the body increases. On the contrary, if we apply a force in the direction opposite to that of motion of the body, the speed of the body decreases. These ( examples show that force has magnitude as well as direction, i.e., force is a vector quantity.

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Question 255 Marks

What is a non contact force? Give one example.

Answer

A force that acts between two objects even if the two objects are not in contact, is called a non contact force.
Example: The earth revolves around the Sun.

Make a list of some more examples in which contact and non contact forces are applied. Write the types of force.

1. Some examples in which contact forces are applied:

to cut an apple with a knife (muscular force, frictional force)
to lift a ball lying on the ground (muscular force, frictional force)

2. Some examples in which non contact forces are applied:

the motion of the earth around the Sun (gravitational force)
the motion of an electron around the nucleus of an atom (mainly the electric force).

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Science Answer the following question.

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