Chapter 1 Physical Environment Force And Pressure Solved Numerical Problems
Problem 1. Equal forces act on two masses ‘m’ and ‘2m’. If the acceleration acquired by ‘m’ is ‘f’, what is the acceleration acquired by 2m
Solution:
Given:
Equal forces act on two masses ‘m’ and ‘2m’. If the acceleration acquired by ‘m’ is ‘f’.
From the relation, force = mass x acceleration, the force acting on the mass ‘m’ is mf. Now the same force mf acts on the mass 2m, and since, acceleration = force/mass, acceleration of the second body = mass
mf/2m =f/2
Problem 2. A force of 4 kg wt acts on a body of mass 9.8 kg. Calculate the acceleration. [ Take g = 9.8 m/s²]
Solution:
Given:
A force of 4 kg wt acts on a body of mass 9.8 kg.
4 kg wt = 4 kg x 9.8 m/s² = 39.2
N. Now, force = mass x acceleration, 39.2
N = 9.8 kg x acceleration
Acceleration = 39.2/9.8kg=m/s²
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Problem 3. Two bodies A and B of equal mass are at rest. A is acted upon with 1 kg wt force and B with IN. Will the acceleration generated in them be equal?
Solution:
Given:
Two bodies A and B of equal mass are at rest.
A is acted upon with 1 kg wt force and B with IN.
Using the relation,acceleration=force/mass
acceleration of A =,1kg wt/m=9.8n/m, Where
‘m’ is the mass of A or B. Acceleration of B=1 n/m
The acceleration of A is 9.8 times greater than that of B.
Class 8 Force And Pressure Numerical WBBSE
Problem 4. A body of mass 200 g is moving with a velocity of 10 cm/s. Calculate the force required to stop it in 10 seconds.
Solution:
Given:
A body of mass 200 g is moving with a velocity of 10 cm/s.
Retardation (or negative acceleration),
a =Final velocity-initial velocity/time
a=0-10/10=-1cm/s²
Since this negative acceleration is produced by the force which retards the motion of the body,
Force = 200 g x l cm/s² = 200 dynes.
Problem 5. An object of mass 5 g is moving with an acceleration of 10 cm/s². Calculate the magnitude of the force acting on it.
Solution:
Given:
An object of mass 5 g is moving with an acceleration of 10 cm/s².
Mass of the object = 5 g Acceleration of the object = 10 cm/s²
Hence, the force acting on the object
= mass x acceleration
= 5 g x 10 cm/s²
= 50 dynes.
Problem 6. A force of magnitude 15 N is acting on an object of mass 10 kg. What will be the magnitude of the acceleration of the object produced by the force?
Solution:
Given:
A force of magnitude 15 N is acting on an object of mass 10 kg.
Mass of the object = 10 kg
Force applied on the body = 15 N
Force = mass x acceleration or, 15 N = 10 kg x acceleration
or, Acceleration of the object =15/10 m/ s² = 1.5 m/s².
Class 8 Force And Pressure Numerical WBBSE
Problem 7. A force of 4kg wt is acting on an object of mass 9.8 kg. Calculate the acceleration of the object produced by the force.
Solution:
Given:
A force of 4kg wt is acting on an object of mass 9.8 kg.
Force applied on the body = 4 kg wt = 4 kg x 9.8 m/s² = 39.2 N
Mass of the object = 9.8 kg
Force = mass x acceleration
or, 39.2 N = 9.8 kg x acceleration
or, Acceleration of the object = 39.2/9.8 m/ s²= 4 m/s².
Chapter 1 Physical Environment Force And Pressure Solved Numerical Problems
Problem 1. The density of water is 1 kg/liter. Express it in g/cc (gram per cubic centimeter) unit.
Solution :
Given:
The density of water is 1 kg/liter.
1 litre of water = 1000 cc of water
and, 1 kg of water = 1000 g of water
So, the density of water
=1kg /litre= 1kg/1 litre =1000g/1000cc=1g/cc.
Problem 2. The mass of 2 liters of a liquid is 4 kg. Calculate its density.
Solution:
Mass of the liquid = 4 kg
The volume of the liquid = 2 liter
So, the density of the liquid mass of the liquid/volume of the liquid = 4kg/2 liter =2kg/liter.
Problem 3. We have all seen mercury, a liquid metal, within the bulb at the bottom of a thermometer. Its density is 13.6 g/cc Calculate the mass of 1 liter of mercury.
Solution:
Density of mercury = mass of mercury/volume of mercury
1 liter = 1000 cc
Mass of mercury = (13.6 g/cc x 1000 cc) = 13,600 g.
For two different liquids of the same volume, the density of the liquid, the heavier it will be. For example, an equal volume of glycerine is heavier than an equal volume of water since its density is more than that of water.
For two immiscible liquids taken together in a container, the one with less density will float on the other. For example, oil (having lower density) floats on water (having higher density).
Class 8 Force And Pressure Numerical WBBSE
Densities of some common substances are given in
| Substances | Density g/cc | Density Kg/m³ |
| Water | 1 | 1000 |
| Glycerine | 1.26 | 1260 |
| Mercury | 13.6 | 13,600 |
| Air | 0.0012 | 1.2 |
| Oil | 0.78 | 780 |
| kerosene | 0.08 | 800 |
Problem 4. Determine the volume of 1000 kg kerosene, (density of kerosene is 0.8 g/cc)
Solution :
Mass of kerosene = 1000 kg = 1000 x 1000 g
Density of kerosene = 0.8 g/cc
We know, the density of kerosene = mass of kerosene/volume of kerosene or, volume of kerosene
= mass of kerosene/density of kerosene
= (1000 x 1000)/0.8
or, volume of kerosene = 12,50,000 cc.
Pressure of Liquid
When a solid block of 1 kg is kept on a table, the earth pulls it with a force of W = m x g = 1kg x 9.8m/s² = 9.8 Newton. The block is therefore applying pressure on the table on account of its weight.
Pressure is the force per unit area, applied to the surface of the object on which it is placed.
Pressure=Force/Area
Class 8 Force And Pressure Numerical WBBSE
SI unit of pressure is Newton per square meter (N/m²).
Let us consider a solid, wooden block of mass 5 kg, which is placed on a table. The total area of contact between the block and the table is 0.2 m².
We can calculate the pressure of the block exerted on the table.
Mass of the block = 5 kg
The magnitude of the force exerted by the block on the table
= 5 kg X 9.8 m/s² (D g = 9.8m/s²)
= 49 Newton
Force And Pressure Class 8 Numerical Questions WBBSE
Area of contact of the block and the table =
0.2 m²
So, the pressure of the block exerted on the table = Force/Area
= 49 Newton / 0.2 m² = 245 N/m²
Since all liquids have weight, so when we pour a liquid into a vessel or tumbler, then the weight of the liquid pushes down on the base of the vessel producing pressure.
Therefore, to calculate the pressure exerted by a liquid on the base of a vessel using the formula, pressure =, we are to substitute ‘Force’ with the weight of the liquid and ‘Area’ by the area of the base of the vessel in which the liquid is placed.
Let us consider a tumbler filled with water of mass 10 kg. The area of the floor of the tumbler is 0.5 m². We can measure the pressure exerted by the water on the base of the tumbler.
Mass of the water in the tumbler = 10 kg Magnitude of the force exerted by water on the base of the tumbler
= 10 kg x 9.8 m/s²
= 98 Newton
Area of the base of the tumbler = 0.5 m²
So, Pressure exerted by water on the base of the tumbler
Force And Pressure Class 8 Numerical Questions WBBSE
= Force / Area = 98 N / 0.5 m²
= 196 N/m²
Chapter 1 Physical Environment Force And Pressure Solved Numerical Problems
Problem 1. A block weighing 2 kg is placed over a table. The block occupies an area of 200 cm² on the table. Calculate the pressure applied by the block on the table.
Solution:
Given:
A block weighing 2 kg is placed over a table.
The block occupies an area of 200 cm² on the table.
Weight of the block = 2 kg
The force with which the earth pulls a mass of 1 kg towards its center = 9.8 newton
So, the force with which the earth pulls a mass of 2 kg towards its center = (9.8 x 2) newton = 19.6 newton.
So the force applied by the block on the table = 19.6 newton.
Area of the block = 200 cm² = 200/(100 x 100) m² = 0.02 m²
The pressure applied by the block on the table = force/area = (19.6 / 0.02) N/m² = 980 N/m².
Force And Pressure Class 8 Numerical Questions WBBSE
Problem 2. A tumbler contains 5 kg water. The area of the base of the tumbler is 0.2 square meters. What is the pressure applied by the water on the base of the tumbler?
Solution:
Given:
A tumbler contains 5 kg water.
The area of the base of the tumbler is 0.2 square meters.
Mass of water = 5 kg
The force with which the earth pulls a mass of 1 kg towards its center = 9.8 newton.
Hence, the force exerted by the water = (9.8 x 5) newton = 49 newton.
Area of the base of the tumbler = 0.2 m²
So, the pressure applied by the water on the base of the tumbler = Force / Area = 49 / 0.2 N/m³= 245 N/m².
In the above example, we have calculated the pressure exerted by water on the base of the tumbler.
But water and, in general, any liquid exert pressure not only on the base but also on the walls of the tumbler. The pressure exerted by a liquid on the sides of a container is called lateral pressure.
This is the reason that when the wall of the tumbler is pierced (that is a tiny hole is made in the wall of the tumbler), the water flows out from this hole. Several experiments can be carried out to understand various aspects of the pressure of liquids.