Force and Pressure Test

Result

Force and Pressure Test - 8

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Question 1
1 / -0

The density of a mixture of 1.5 m³ of water and 0.50 m³of alcohol of density 800 kg/m³ is

A
350 kg/m³

B
650 kg /m³

C
950 kg/m³

D
1250 kg/m³

Solution

Density of the mixture = Total mass of the mixture/Total volume of the mixture
Mass of the mixture (M) = Mass of the water + Mass of the alcohol
= (Density of the water x Volume of the water) + (Density of the alcohol x Volume of the alcohol)
Question 2
1 / -0

Two vessels A and B have the same base area and contain water up to the same height, but the mass of water in vessel A is four times that of water in vessel B. The ratio of pressure on the base of vessel A to that on the base of vessel B is

A
4 : 1

B
2 : 1

C
1 : 1

D
16 : 1

Solution

Pressure due to liquid column is given by:
P = ρgh, where h is the height and ρ is the density of the liquid.
Hence, pressure depends only of the height of the liquid column and is independent of the shape of the vessel and the mass of the liquid.
So, pressure on the base of both the vessels (A and B) is the same. So, the required ratio is 1 : 1.
Question 3
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The upthrust on a body immersed in a fluid is equal to the

A
mass of the liquid displaced

B
weight of the fluid displaced

C
volume of the fluid displaced

D
density of the fluid displaced

Solution

The upthrust on a body immersed in a fluid is equal to the weight of the fluid displaced.
Question 4
1 / -0

A force of 10 kg wt is applied on the head of a nail of area of cross-section (of the tip) 0.0001 cm². What is the pressure exerted?

A
10¹⁰ N m^-2

B
10⁶ N m^-2

C
10^-6 N m^-2

D
10^-10 N m^-2

Solution

Here, force = 10 kg wt = 100 N
Area = 0.0001 cm² = 10^-8m²Hence, pressure
Question 5
1 / -0

Find the pressure exerted on a surface having area equal to 25 m² by a thrust of 200 N.

A
18 Pa

B
8 Pa

C
16 Pa

D
9 Pa

Solution

If a thrust F acts on an area A, then:
Pressure =
Or, P = = 8 Pa
Therefore, the pressure exerted on the surface is 8 Pa.
Question 6
1 / -0

When you press a rubber sucker against a surface, most of the air between its cup and the surface escapes out. To pull the rubber sucker off the surface, the applied force should be

A
equal to the atmospheric pressure

B
large enough to overcome the atmospheric pressure

C
less than the atmospheric pressure

D
None of the above

Solution

The rubber sucker sticks to the surface because the pressure of atmosphere acts on it. To pull the sucker off the surface, the applied force should be large enough to overcome the atmospheric pressure.
Question 7
1 / -0

A body floats in a liquid contained in a beaker. The whole system, as shown in the figure, falls freely under gravity. The upthrust on the body due to the liquid is

A
zero

B
equal to the weight of the liquid displaced

C
equal to the weight of the body

D
equal to the weight of the immersed portion of the body

Solution

A body floats in a liquid contained in a beaker. The whole system, as shown in the figure, falls freely under gravity. The upthrust on the body due to the liquid is zero.
Question 8
1 / -0

Water is flowing through a pipe under constant pressure. At some place, the pipe becomes narrow. The pressure of water at this place

A
increases

B
decreases

C
remains the same

D
depends on several factors

Solution

As the area of cross section decreases velocity increases. At the narrow end as the area of cross section decreases, velocity increases.
As the velocity increases, pressure decreases.
Question 9
1 / -0

What happens when a stream of air is blown through a tube under one of the pans of a physical balance in equilibrium?

A
The pan will go up.

B
The pan will not be affected.

C
The pan will go down.

D
The pan will oscillate.

Solution

As the stream of air is blown through a tube under one of the pans, due to the high air speed, the pressure below the pan decreases as compared to the pressure above the pan. Hence, there will be a net downward force acting on the pan and the pan will go down.
Question 10
1 / -0

For a steel boat floating on a lake, the weight of the water displaced by the boat is

A
less than the weight of the boat

B
more than the weight of the boat

C
equal to the weight of the boat

D
equal to the weight of the part of the boat which is below the water level of the lake

Solution

As the boat is floating , the resultant force acting on the boat is zero
Weight of the body acting downward should be equal to the up thrust acting on the boat.
Question 11
1 / -0

A deep sea diver may hurt his eardrum during diving because of

A
lack of oxygen

B
high atmospheric pressure

C
high water pressure

D
All of the above

Solution

When a diver dives into the deep sea, there is water on the outside of the ear drum, but air on the inside. The pressure inside will still be equal to atmospheric pressure. However, on the outside, the pressure will be greater. This means that the pressure of air inside the ear will not cancel the pressure of the water outside. There will be net inward force due to this pressure difference, as a result of which the eardrum will stretch. This stretching of the eardrum hurts.
Question 12
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What happens to a solid object with density less than that of water when it is put in water?

A
The object dissolves in the water.

B
The object displaces a quantity of water greater than its volume.

C
The object settles to the bottom of the water.

D
The object floats on top of the water.

Solution

The object floats on top of the water as the density of water is more than the density of object.
Question 13
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If you have 5 mL of a liquid that has a mass of 20 g, then what is the density of the liquid?

A
4 x 10¹ kg/m³

B
4 x 10³ kg/m³

C
40 x 10³ kg/m³

D
4 x 10²kg/m³

Solution

We know:
1 L = 10^-3 m³Hence, 5 mL = 5 x 10^-6 m³Mass = 20 g = 20 x 10^-3 kg
Now, Density = Mass/Volume
Density =
Question 14
1 / -0

A raft is 2.35 m wide, 1.25 m long, and 0.365 m deep. What is the buoyant force on the raft, if only 0.125 m of the raft remains above water? (Take g = 10 m/s²)

A
10,500 N

B
3590 N

C
7050 N

D
6910 N

Solution

Upthrust acting on the object = Weight of the water displaced
Weight of the water displaced = Volume of water displaced (Volume of the object in the water) × Density × g
Volume of water displaced = 2.35 × 1.25 × 0.24 = 0.705 m³Hence, upthrust = 0.705 × 10³× 10 = 7050 N
Question 15
1 / -0

A cube of side 5 cm having density 1600 kg/m³ is submerged in the water with help of a rope. What is the weight of the liquid displaced? (Take g = 10 m/s²)

A
125 N

B
12.5 N

C
1.25 N

D
0.125 N

Solution

Weight of the water displaced = Volume of the water displaced (Volume of the object in the water) x Density of water x g
= (5 x 10^-2)³x 10³x 10 = 1.25 N