MOECDC Class 8 Force and Motion
MOECDC Class 8 Science and Technology | Force and Motion | Nepal Class 8 | Exercise | Question Answer
MOECDC Class 8 English Medium Science Chapter 6 Force and Motion Exercise Question Answer Solution.
| Board | Government of Nepal. |
| Class | Eight (8). |
| Subject | Science and Technology. |
| Chapter | Force and Motion |
| Topic | Exercise Solve / Question Answer / Notes. |
Exercise 1 :
(1) Choose the best alternative :
(a) What is the condition of an object called if doesn’t change its position with respect to the reference frame or surrounding?
(i) motion
(ii) velocity
(iii) rest
(iv) speed
Answer :
(iii) Rest – is the condition of an object called if does not change its position with respect to the reference frame or surrounding.
(b) What is the formula to calculate the relative velocity of two bodies moving in the opposite direction?
(i) VAB = VA I V B
(iii) V AB = VAX VB
(ii) VAB = VA-VB
(iv) V AB = VA+ VB
Answer :
(iv) (VAB) = VA + VB – is the formula to calculate the relative velocity of two bodies moving in the opposite direction.
(c) What is the shortest distance between any two points in a partcular direction?
(i) speed
(ii) displacement
(iii) average velocity
(iv) relative velocity
Answer :
(ii) Displacement – is the shortest distance between any two points in a particular direction.
(d) In which condition the average velocity of an object is calculated?
(i) non-unifonnniotion
(ii) uniform motion
(iii) zero acceleration
(iv) object at rest
Answer :
In (i) non-uniform motion – the average velocity of an object is calculated.
(e) What is the S.I. unit of acceleration?
(i) m/s
(ii) m/s
(iii) Nm/kg
(iv) m²
Answer :
The S.I. unit of acceleration is – (ii) m/s2.
(f) Which one of the given statements is true?
(i) negative acceleration is produced if an object moves with uniform motion.
(ii) acceleration is not produced if an object is moving with non-uniform motion.
(iii) positive acceleration is called retardation.
(iv) rate of change in the velocity of a body is called acceleration.
Answer :
(iv) Rate of change in the velocity of a body is called acceleration – is true.
(2) Differentiate between :
(a) distance and displace1nent
Answer :
Distance and displacement : Distance is the total movement of an object from its initial position to its final position. On the other hand, the shortest distance between any two points in a particular direction is called displacement.
(b) average velocity and relative velocity
Answer :
Average velocity and relative velocity: The total distance travelled by the body in unit time in a particular direction is called average velocity. On the other hand, the velocity of a body relative to the reference point or another body is called relative velocity.
(c) acceleration and retardation
Answer :
Acceleration and retardation: The rate of change in velocity of a body in a unit time is called acceleration. On the other hand, the rate of decrease velocity or negative acceleration of a body is called retardation.
(3) Answer the following questions :
(a) Define the reference point.
Answer :
A reference point is a point, place or object used to determine whether something is at rest or in motion.
(b) What is acceleration?
Answer :
The change in velocity of a body in a unit time is called acceleration.
(c) What do you understand by retardation?
Answer :
The rate of decrease velocity or negative acceleration of a body is called retardation.
(d) What is meant by relative velocity?
Answer :
The velocity of a body relative to the reference point or another body is called relative velocity.
(e) Write down the formula to calculate the relative velocity of a body moving in the same and opposite directions with examples.
Answer :
Two buses A and B are moving in the same direction with the velocity 15 m/s and 10 m/s respectively.
VA = 15 m/s VB = 10 m/s
Relative velocity of the bus A with respect to bus B, (VAB) = VA – VB
VAB = 15 – 10 = 5 m/s
If they are moving in the opposite direction, (VAB) = VA + VB
VAB = 15 + 10 = 25 m/s
(f) What is average velocity?
Answer :
The total distance travelled by the body in unit time in a particular direction is called average velocity.
(g) What is the condition in which the average velocity of a body is calculated? Write down formulae to calculate average velocity.
Answer :
In the case of non-uniform motion, the velocity of the body increases or decreases so we need to calculate its average velocity.
Average velocity = Total distance travelled in particular direction (S) / Total time taken (t)
(4) Solve these numerical problems :
(a) If a bus covers 4.2 km of distance in 7 minutes, then calculate its average velocity.
Answer :
Distance covered (s) = 4.2 km = 4200 m
Time taken (t) = 7 minutes = 420 seconds
Average Velocity of the bus (a) = s / t = 4200 / 420 = 10 m/s
(b) If a car starts to move from rest and its velocity becomes 30 m/s in 12 seconds then find its acceleration.
Answer :
Initial velocity (u) = 0 m/s
Final velocity (v) = 30 m/s
Time taken (t) = 12 seconds
Acceleration of the car (a) = v – u / t = (30 – 0) / 12 = 2.5 m/s2
(c) Find the final velocity of the car if it starts to move from rest with the acceleration of 5 m/s² in 6 seconds. Calculate the average velocity of the car.
Answer :
Initial velocity (u) = 0 m/s
Acceleration of the car (a) = 5 m/s2
Time taken (t) = 6 seconds
Final velocity (v) = a * t + u = 5 * 6 + 0 = 30 m/s [a = v – u / t]
Average velocity of the car (V) = v + u / 2 = 30 + 0 / 2 = 15 m/s
(d) If the brake is applied in a car moving with uniform velocity then the retardation of 2 m/s² is produced and the car stops after 4 seconds. Calculate the initial velocity of the car.
Answer :
Final velocity (v) = 0 m/s
Retardation (a) = 2 m/s2
Time taken (t) = 4 seconds
Initial velocity (u) = at – v = (2 * 4) – 0 = 8 m/s [a = v – u / t]
(e) If two motorcycles are moving at the speed of 50 km/hr and 60 km/hr in the same direction. Calculate the relative velocity between these motorcycles. If they were moving in opposite directions what would be the relative velocity between them?
Answer :
Speed of motorcycle 1 (S1) = 50 km/hr.
Speed of motorcycle 2 (S2) = 60 km/hr.
Relative velocity between these motorcycles when they are moving in the same direction = S2 – S1 = 60 – 50 = 10 km/hr.
Relative velocity between these motorcycles when they are moving in the opposite direction = S1 + S2 = 50 + 60 = 110 km/hr.
Exercise 2 :
(1) Choose the best alternative :
(a) Which one of the following simple machines is a second- class lever?
(i) shovel
(ii) dhiki
(iii) nut-cracker
(iv) hammer
Answer :
(iii) Nut-cracker – is a second-class lever.
(b) What is the principle of the lever in a balanced condition?
(i) Effort x Effort distance = Load x Load distance
(ii) Effort Load
(iii) Effort distance = Load distance
(iv) Mechanical advantage = Efficiency
Answer :
(i) Effort * Effort distance = Load * Load distance – is the principle of the lever in a balanced condition.
(c) What is the work done by effort called?
(i) input work
(ii) output work
(iii) friction
(iv) gravitational work
Answer :
The work done by effort is called – (i) input work.
(d) Which one of the following is not affected by friction?
(i) mechanical advantage
(ii) velocity ratio
(iii) efficiency
(iv) input work
Answer :
(ii) Velocity ratio – is not affected by friction.
(e) Which of the following given statements is true?
(i) The product of load and effort in the balanced condition of the lever is velocity ratio.
(ii) Effort covers less distance than load while lifting a load heavier than effort.
(iii) Both the mechanical advantage and velocity ratio have their units.
(iv) The ratio of output work to the input work expressed in percentage is called efficiency.
Answer :
(iv) The ratio of output work to the input work expressed in percentage is called efficiency – is true.
(2) Differentiate between :
mechanical advantage and velocity ratio input work and output work
Answer :
Mechanical advantage and velocity ratio input work and output work: Mechanical advantage is the ratio of two similar physical quantities of force. On the other hand, work done by effort in a simple machine is called input work. But work done by the machine when the effort is applied to it is called output work.
(3) Give reasons:
(i) The efficiency of the lever is never 100 %.
Answer :
The efficiency of the lever is never 100% because mechanical advantage is always less than the velocity ratio due to friction.
(ii) It is easier to push the load in a wheelbarrow while shifting
Answer :
It is easier to push the load in a wheelbarrow while shifting because in a wheelbarrow, we need less effort to lift the load as the load distance is less.
(iii) The load towards its wheel.
Answer :
The load towards its wheel because it is easier to lift the load if we put the load towards the wheel.
Mechanical advantage is always less than the velocity ratio
Answer :
Mechanical advantage is always less than the velocity ratio because mechanical advantage is affected by friction but not the velocity ratio.
(4) Answer the following questions :
(i) How can we lift the heavy load by using a lever?
Answer :
We have to put less effort to lift the heavy load by using a lever.
(ii) Which class lever is a scissor?
Answer :
Scissor is a first-class lever.
(iii) What can be done with a lever whose mechanical advantage is more than 1?
Answer :
If the mechanical advantage of a lever is more than 1, then such a lever multiplies the applied effort.
(iv) What are the other advantages of lever besides making our work easier?
Answer :
Besides making our work easier, a lever also helps us to complete our work quickly and correctly with less effort.
(v) What is called the work done by effort in a machine?
Answer :
The work done by effort in a machine is called input work.
(5) Solve these numerical problems :
(i) An effort of 75 N is required to lift the load of 300 N. Calculate the effort distance if the distance of the load from the fulcrum is 25 cm.
Answer :
Load (L) = 300 N
Load distance (L.D.) = 25 cm
Effort (E) = 75 N
Effort distance (E.D.) = L * L.D. / E = 300 * 25 / 75 = 100 cm [E * E.D. = L * L.D.]
(ii) The weight of Sarad is 550 N and Nirmal is 300 N. Sarad and Nirmal are playing see-saw. How far does the Nirmal sit from the centre to balance the Sarad who sits at a distance of 1.5 m from the center.
Answer :
Weight of Sarad = 550 N
Distance of Sarad from center = 1.5 m
Weight of Nirmal = 300 N
Distance of Nirmal from center = 550 * 1.5 / 300 = 2.75 m [E * E.D. = L * L.D.]
(iii) If 300 N effort is applied to lift a load of 900 N by using 2 meters long lever. Calculate the mechanical advantage, velocity ratio and efficiency of the lever if the load lies at a distance of 50 cm from the fulcrum.
Answer :
Effort (E) = 300 N
Load (L) = 900 N
Distance of load from the fulcrum (L.D.) = 50 cm = 50 / 100 = 0.5 m
Length of Lever = 2 m
Effort distance (E.D.) = (2 – 0.5) = 1.5 m
Mechanical advantage (MA) = L/E = 900 / 300 = 3
Velocity ratio (VR) = E.D. / L.D. = 1.5 / 0.5 = 3
Efficiency of the lever = MA / VR * 100% = 3 / 3 * 100% = 100%
Exercise 3 :
(1) Choose the best alternative :
(a) What is the S.I. unit of pressure?
(i) Nm
(ii) N/m²
(iii) m/s²
(iv) Nm²/kg
Answer :
The S.I. unit of pressure is – (ii) N/m2.
(b) Which of the following statement is true?
(i) More pressure is exerted if the area is less.
(ii) Pressure of an object is not affected by area.
(iii) Pressure increases with the increase in the area of the body.
(iv) Pressure remains the same although the force varies.
Answer :
(iii) Pressure increases with the increase in the area of the body – is true.
(c) What is the value of atmospheric pressure at sea level?
(i) 760 mmHg (iii) 1000 mmHg
(ii) 700 mmHg
(iv) 750 mmHg
Answer :
The value of atmospheric pressure at sea level is – (i) 760 mmHg.
(d) What is the pressure exerted by the water at the depth of 60 m where the value of acceleration due to gravity is 9.8 m/s? (density of water is 1000 kg/m³)
(i) 588 Pa.
(ii) 5880 Pa.
(iii) 58800 Pa.
(ii) 5880 Pa.
(iv) 588000 Pa.
Answer :
(iv) 588000 Pa.
(e) What is the name of the instrument used to measure atmospheric pressure?
(i) thermometer
(ii) barometer
(iii) manometer
(iv) lactometer
Answer :
The name of the instrument used to measure the atmospheric pressure is – (ii) barometer.
(f) Which of the following given instrument is used to measure the pressure exerted by compressed air?
(i) pressure gauge
(iii) manometer
(ii) rain gauge
(iv) barometer
Answer :
(i) Pressure gauge – is used to measure the pressure exerted by compressed air.
(g) What is the principle on which the air braking system is based?
(i) atmospheric pressure
(iii) liquid pressure
(ii) compressed air
(iv) pressure
Answer :
(ii) Compressed air – is the principle on which the air braking system is based.
(2) Give reasons :
(a) Football shoes have studs on their soles.
Answer :
Football shoes have studs on their soles because these help to grip the playground surface and prevent players from sliding while playing.
(b) Ploughshare is made sharper to plough the field.
Answer :
Ploughshare is made sharper to plough the field to make it easier to plough because more pressure will be exerted even while applying less force due to less area of the ploughshare.
(c) The bottom of the dam is made wider than its walls.
Answer :
The bottom of the dam is made wider than its walls because the pressure exerted by liquid increases with the increase in depth of the liquid column.
(d) The bucket is filled faster on the ground floor than on the upstairs
Answer :
The bucket is filled faster on the ground floor than on the upstairs because liquid pressure is high at the ground floor as the pressure exerted by liquid increases with the increase in depth of the liquid column.
(3) Answer the following questions :
(a) What is pressure? Write down any three applications of pressure in daily life.
Answer :
The impact of force on an area is called pressure.
Three applications of pressure in daily life,
(i) A water tank is kept at the maximum height in the house.
(ii) Atmospheric pressure is used to fill injections in a syringe.
(iii) Air pressure is used to operate automatic air cylinders.
(b) How can we tell that liquid exerts pressure?
Answer :
Liquid also exerts pressure on the surface of an object like a solid due to its weight. Liquid exerts pressure in all directions. The pressure exerted by a liquid depends on the depth of the liquid column, density of the liquid and acceleration due to gravity. The pressure exerted by liquid increases with the increase in depth of the liquid column and density of the liquid.
(c) Write down any two applications of liquid pressure in daily life?
Answer :
Two applications of liquid pressure in daily life,
(i) A water tank is kept at the maximum height in the house.
(ii) Deep sea driver wears special costume to go deep into the sea.
(d) What is atmospheric pressure? Mention it’s any two importance.
Answer :
The pressure exerted by atmospheric air is called atmospheric pressure.
Two importance of atmospheric pressure,
(i) Atmospheric pressure influences all objects and living beings on Earth.
(ii) Air flows from one place to another due to changes in the atmospheric pressure on the earth’s surface.
(e) Prove that the pressure exerted by the liquid is P = hdg. Where the symbols have their usual meaning.
Answer :
Let us consider the liquid of density ‘d’ is filled in the container of height ‘h’ having cross-section area ‘A’. The acceleration due to gravity at this place is ‘g’. The volume of liquid in the container is given by, V = A * h.
The force exerted by the liquid at bottom of the container is called the weight of that liquid.
The pressure exerted by the liquid column at its bottom is the force acting normally per unit are of the body.
Therefore, Pressure (P) = weight of liquid column (W) / cross section area (A)
Or, P = W/A = mass (m) * acceleration due to gravity (g) / A [w = mg]
Density (d) * (Volume (V) * acceleration due to gravity (g)) / A m = d * V
= d * V * g / A
= d * A * h * g / A …[V = A * h]
(f) Draw a neat and labelled diagram of a manometer.
Answer :
(g) What is atmospheric pressure? Explains with examples.
Answer :
The atmospheric pressure is the air pressure exerted per unit area of the earth’s surface. The air pressure is maximum at the sea level and as we go higher the atmospheric pressure decreases. The atmospheric pressure at sea level is 760 mmHg. Air flows from one place to another due to changes in the atmospheric pressure on t he earth’s surface.
More Solutions :
👉 (2) Information and Communication Technology
👉 (3) Living Beings and Their Structure
👉 (4) Biodiversity and Environment
👉 (8) Electricity and Magnetism
