Inter 1st Year

Students get through AP Inter 1st Year Physics Important Questions 6th Lesson Work, Energy and Power which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 6th Lesson Work, Energy and Power

Very Short Answer Questions

Question 1.
State the conditions under which a force does no work.
Answer:
Work done W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\) = |F||S|cosθ
Hence, if θ = 90° then cos 90° = 0 ⇒ W = 0

1. When force (F) and displacement (S) are mutually perpendicular then work done is zero.
2. Even though when force is applied, if displacement is zero then also work done W = 0.

Question 2.
Define Work, power and Energy. State their Si units. [Imp.Q]
Answer:
Work :
Work is said to be done, when a force acting on a body moves it, through some distance in the direction of the force

SI unit of Work : Joule
Power : The rate of doing work is called power
SI unit of Power : Watt.
Energy : The capacity to do work is called Energy.
SI unit of Energy : Joule

Question 3.
State the relation between the kinetic energy and momentum of a body. [Imp.Q]
Answer:

Question 4.
State the sign of work done by a force in the following.
a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
b) work done by gravitational force in the above case.
Answer:
a) Force applied by the man in the displacement of the bucket are in the same direction.
So work done by the man is positive.

b) Gravitational force is acting in the downward direction and bucket is moving in the upward direction. So, work done by the gravity is negative.

Question 5.
State the sign of work done by a force in the following. [Imp.Q]
a) Work done by friction on a body sliding down an inclined plane.
b) Work done by gravitational force in the above case.
Answer:
a) Friction always acts in a direction, opposite to the direction of motion.
Hence the work done is negative.

b) The work done is positive, because the angle between the gravitational force and displacement is acute.

Question 6.
State the sign of work done by a force in the following
(a) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
(b) Work done by resistive force of air on a vibrating pendulum in bringing it to rest
Answer:
(a) The work done is positive because the applied force and displacement of the body are in the same direction.

(b) The work done is negative because resistive force of air on a vibrating pendulum always acts in a direction opposite to the direction of motion of pendulum.

Question 7.
State if each of the following statements are true (or) false. Give reasons to your answer.
(a) Total energy of a system is always conserved, no matter what internal and externa! forces on the body are present.
(b) The. work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.
Answer:
(a) True. Total energy of an isolated system will be conserved, if all the forms of energy are taken into account. Energy in one form may be converted into another form, but total energy of the system remains constant.

(b) True. Because the gravitational force is conservative.

Question 8.
Which physical quantity remains constant. [Imp.Q]
(i) in an elastic collision (ii) in an inelastic collision?
Answer:
The total momentum and total K.E of the system remain constant in elastic collision.

The total momentum of the system only remains constant in an inelastic collision.

Question 9.
A body freely falling from a certain height ‘h’ after striking a smooth floor rebounds to a height h/2. What is the coefficient of restitution between the floor and the body? [TS 18][Imp.Q]
Answer:

Question 10.
What is the total displacement of a freely falling body, after successive rebounds from the same place of ground, before it comes to stop? Assume that ‘e’ is the coefficient of restitution between the body and the ground?
Answer:
If h is the height from which the body is falling freely, then the total displacement is equal to ‘h’ only.

Short Answer Questions

Question 1.
What is potential energy? Derive an expression for the gravitational potential energy. [Imp.Q]
Answer:
Potential Energy: The energy possessed by a body by virtue of its position is called Potential energy.
Ex : Water in a Reservoir, Energy stored in a stretched rubber band.

Expression for P.E :
Consider a body of mass ‘m’, which is at rest on the surface of earth.
The Force acting on the body is equal to its weight mg.
Force required to lift the body upwards through a height ‘h’ is mg.
Now, Work done W = Force × displacement = mg × h = mgh
The work done is stored as P.E in the body.
∴ Potential Energy P.E = mgh

Question 2.
A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance? [Imp.Q]
Answer:
(i) Given that, momentum is same for both lorry and car. Also, retarding forces are equal. Work done to stop the body = Change in Kinetic energy
F = ma ⇒ F = m(\(\frac{v-u}{t}\))
⇒ F × t = mv – mu
Both lorry and car are having same initial momentum, mu.
Both are finally stopped. So their final momenta(mv) are also same i.e, zero.
Their retarding forces (F) are also same. So they will be stopped in the same time.

(ii) Given that, momentum is same for both lorry and car. Also, retarding forces are equal.
Work done to stop the body = Kinetic energy

⇒ distance S ∝ \(\frac{1}{m}\) (Since, momentum and force are equal)
The heavier body (lorry) will come to rest in less distance.

Question 3.
Distinguish between conservative and non-conservative forces with one example each. [Imp.Q]
Answer:

Conservative force	Non – Conservative force
1) If the work done by a force around a closed path is zero, then such a force is called conservative force.	1) If the work done by a force around a closed path is a non-zero then such a force is called non-conservative force
2) The work done is independent of the path followed.	2) The work done depends on the path followed by the body.
3) Ex: Gravitational force.	3) Ex: Frictional force.

Question 4.
Show that in case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision. [AP 20; TS 18; May 08, Mar’07, Jun’06, Mar’05]
Answer:
Consider two spheres A and B of masses m₁ and m₂ moving with initial velocities u₁ and u₂ undergo a collision and after collision move with velocities v₁ and v₂ respectively.

According to law of conservation of momentum,
Total momentum before collision = Total momentum after collision.
m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂ ⇒ m₁u₁ – m₁v₁ = m₂v₂ – m₂u₂.
m₁(u₁ -v₁) = m₂(v₂ – u₂) ………. (1)
From the law of conservation of K.E

Thus,
Relative velocity of approach before collision = Relative Velocity of Separation after collision.

Question 5.
Show that two equal masses undergoing oblique elastic collision will move at right angles after collision, if the second body is initially at rest.
Answer:

Let A and B be two particles which have an elastic oblique (non-head on) collision with each other, the particle B being at rest. The particle A is called the incident particle and the particle B is called target particle.
Let m be the mass of each particle. Let the velocity of A before collision is u₁.
∴ Total momentum of A and B before collision along X-axis = mu₁ + 0 = mu₁.

Total momentum of A and B before collision along Y-axis = 0 + 0 = 0 (because in Y-direction there are no velocity components)

Let v₁, v₂ be the velocities of A and B after collision.
Since the collision is oblique, let v₁ and v₂ make angles θ and Φ respectively with X-axis as shown in the figure.

From the law of conservation of momentum.
Total momentum before collision = Total momentum after collision.

Thus the two identical particles move at right angles after elastic oblique collision.

Question 6.
Derive an expression for the height attained by a freely falling body after ‘n’ number of rebounds from the floor.
Answer:
Consider a ball dropped onto the floor from a height ‘h’.
Suppose it hits the floor with a velocity u, and rebounds to a height ‘h₁’, with a velocity v₁
The velocity of the body before and after collision is zero (u₂ = v₂ = 0)

Question 7.
Explain the law of conservation id’ energy [Imp.Q]
Answer:
The total mechanical energy of a system is conserved if the forces doing work on it are conservative. If some of the forces involved are non-conservative, part of the mechanical energy may get transformed into other forms such as heat, light and sound. However, the total energy of an isolated system does not change, as long as one accounts for all forms of energy.

Energy may be transformed from one form to another form but total energy of an isolated system remains constant. Energy can neither be created nor destroyed. This law is called law of conservation of energy.

Long Answer Questions

Question 1.
Develop the notions of work and kinetic energy and show that it leads to work-energy theorem. [IPE’ 14; Imp.Q; TS 15, 19, 22; AP 17]
Answer:
Work :
The product of ‘force component along the displacement’ and magnitude of this displacement is called work.

Formula:
Work W= (F cosθ)S = FS cosθ
Kinetic Energy :
The energy possessed by a body by virtue of its motion is called Kinetic energy.

Formula:
Kinetic Energy, K = \(\frac{1}{2}\)mv²

Work-energy theorem :
The work done by a net force acting on a body is equal to change in its Kinetic Energv. Thus, W = \(\frac{1}{2}\)mv² – \(\frac{1}{2}\)mu²

Proof :
Let a body of mass‘m’ is moving with an initial velocity u.

Let a constant force F acts on the body, then it attains a final velocity v after time ‘t’

We know, Acceleration a = \(\frac{v-u}{t}\) …….. (1)
Displacement S = Average velocity × time ⇒ S = (\(\frac{v+u}{2}\))t …………. (2)
Now, Work done W = Force × displacement ⇒ W = FS = (ma)S …………. (3)
Substituting (1) and (2) in (3), we get

Question 2.
What arc collisions? Explain the possible types of collisions? Develop the theory of one dimensional elastic collision. [TS 22; AP 19, 20, 22; Imp.Q]
Answer:
Collision:
A strong interaction between two bodies that occur for a very short interval of time during which redistribution of momenta occurs is called collision.
If two bodies colllide in a short time interval then
(i) Velocities of colliding bodies change
(ii) Total Momentum will be conserved
(iii) Total Kinetic energy will be changed or conserved

Types of collisions :
(i) Elastic collision
(ii) Inelastic collision (Perfect, Partial)

(i) Elastic collision :
The collision in which both Momentum and Kinetic energy are conserved, is known as elastic collision.
Ex : Collision between two smooth Billiard balls, collisions between gas molecules

(ii) Inelastic collision :
The collision in which only Momentum is conserved but not kinetic energy, is called as inelastic collision.
Ex: Flitting a ball with a bat, collision between a bullet and its target

One dimensional elastic collisions :
If the velocities of colliding bodies, are confined along a straight line, before and after the collision, then such collisions are called “one dimensional collisions” (or) “head-on collisions”.

Consider two spheres of masses m₁ and m₂ moving with initial velocities u₁ and u₂ undergo an elastic collision. Let v₁, v₂ be the velocities after collision.

As the collision is elastic, law of conservation of linear momentum and kinetic energy are obeyed. From the law of conservation of momentum.

Total momentum before collision = Total momentum after collision.
⇒ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ⇒ m₁(u₁ – v₁) = m₂(v₂ – u₂) —— (1)
From the law of conservation of K.E,

Question 3.
State and prove law of conservation of energy in case of freely failing body. [Mar 13, May 13; AP, TS 15, 16, 17, 18, 19, 20]
Answer:
Law of conservation of energy :
Energy can neither be created nor be destroyed. It can be converted from one form to another form, (or)
Total energy remains constant in a closed system.

Proof :
Consider a freely falling body of mass ‘m’ released from a point ‘A’.
It is at a height ‘h’ above the ground, as shown in the figure.

At Point A:
Potential Energy EE = mgh
At A, velocity is v_A = u = 0
Here, Kinetic Energy K.E = \(\frac{1}{2}\) mv²_A = \(\frac{1}{2}\)m(0)² = 0
∴ Total Energy T.E = P.E + K.E = mgh + 0 = mgh ……… (1)

At Point B:
Let the body travels a distance x and reaches the point B.
Now, the height of the body from the ground is (h – x)
∴ P.E = mg(h – x) = mgh – mgx
At the point B, we have, s = x, u = 0, v = v_B, a = +g

At Point C:
Let the body hits the ground at C Here, h = 0
∴ P.E = mgh = mg (0) = 0,
At the point C, we have, s=h, u=0, v=vc, a= +g

∴ T.E = P.E + K.E = 0 + mgh = mgh ……… (3)
From (1) ,(2), (3) it is clear that the total energy T.E is always constant.
Thus, the law of conservation of energy is verified.

Solved Problems

Question 1.
Find the angle between force \(\overline{\mathrm{F}}=(3\overline{\mathrm{i}}+4\overline{\mathrm{j}}-5\overline{\mathrm{k}})\) unit and displacement \(\overline{\mathrm{d}}=(5\overline{\mathrm{i}}+4\overline{\mathrm{j}}+3\overline{\mathrm{k}})\) unit. Also find the projection of \(\overline{\mathrm{F}}on\overline{\mathrm{d}}\)
Solution:

Question 2.
A cyclist conies to a skidding stop in 10m. During this process, the force on the cycle due to the road is 200N and is directly opposed to the motion, (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Solution:
(a) The stopping force and the displacement make an angle of 180° with each other.
Thus, work done by the road,
W_r = Fd cosθ = 200 × 10 × cos 180° = -2000 J

(b) From Newton’s Third law, an equal and opposite force act on the road due to the cycle. Its magnitude is 200N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.

Question 3.
In a ballistics demonstration, a police officer fires a bullet of mass 50.0g with speed 200 ms^-1 on soft plywood of thickness 2.00cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet? [TS 15]
Solution:
The initial kinetic energy of the bullet = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × (50 × 10^-3) × (200)² = 1000J
Final kinetic energy = 0.1 × 1000= 100J.
If v_f is the emergent speed of the bullet, then

Question 4.
An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms^-1. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watt as well as in horse power. [TS 17]
Solution:
The downward force on the elevator is F = mg + F_f = (1800 × 10) + 4000 = 22000N
The motor must supply enough power to balance this force.
Hence, P = F.v = (22000)2 = 44000 W = \(\frac{44000}{746}\) ≈ 59HP ( 1 hp = 746W)

Exercise Problems

Question 1.
A test tube of mass 10 grams closed with a cork of mass 1 gram contains some ether. When the test tube is heated the cork Hies out under the pressure of the ether gas. The test tube is suspended horizontally by a weight less rigid bar of length 5cm. What is the minimum velocity with which the cork should fly out of the tube, so that test tube describing a full vertical circle about the point O. Neglect the mass of ether.
Solution:
Mass of test tube, mT = 10 g = 10 × 10^-3 kg
Mass of cork, mc = 1 g = 1 × 10^-3 kg
length of rod = radius of circular orbit = r = 5cm = 5 × 10^-2 m
The test tube will complete a full vertical circle if its velocity (V_T) is minimum and equal to √5gr

Before heating, cork and test tube are at rest. So their total linear momentum is zero. According to law of conservation of linear momentum, their total linear momentum after heating must also be equal to zero.
momentum of cork = momentum of test tube

Question 2.
A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms^-1. If the mass of each bullet is 5gm, find the power of the machine gun? [IPE’ 13, 13, 14][AP 15, 16, 18, 19; TS 18]
Solution:
Mass of each bullet (m) = 5gm = 5 × 10^-3 kg
Velocity of each bullet (v) = 600 ms^-1
No. of bullets (n) = 360; Time (t) = 1 minute = 60 s; Power (p) =?

Question 3.
Find the useful power used in pumping 3425m³ of water per hour from a well 8m deep to the surface, supposing 40% of the horse power during pumping is wasted. What is the horse power of the engine?
Solution:
Volume of water lifted, V = 3425 m³.
Density of water ρ = 1000 kg/m³ = 10³ kg/m³.
mass of water lifted, m = V × ρ = 3425 × 10³ kg
Height through which water was raised (h) = 8m
Acceleration due to gravity (g) = 9.8 ms^-2
Work done by the pump in one hour = mgh = 3425 × 10³ × 9.8 × 8 = 268520 × 10³J

Question 4.
A pump is required to lift 600kg of water per minute from a well 25m deep and to eject it with a speed of 50ms^-1. Calculate the power required to perform the above task? [AP 15, 18; TS 16, 19, 20, 22]
Solution:
Mass of water lifted (m) = 600kg
depth of well (h) = 25m
Work done to lift water (W₁) = mgh = 600 × 9.8 × 25 = 147000 J
Speed of water (v) = 50ms^-1
Mass of water (m) = 600kg

Question 5.
A block of mass 5kg initially at rest at the origin is acted on by a force along the X- positive direction represented bv F = (20 + 5x)N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.
Solution:
Mass of the body (m) = 5kg
Force on the body (F) = (20 + 5x)N
Amount of work done to move the body by a small distance dx is dw = Fdx
Total work done to move the body from x = 0 to x = 4 is

Question 6.
A block of mass 5kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum?
Solution:

Given that m = 5 kg, K = 600 N/m, g = 10m/s², compression x = ?
From the diagram, sinθ = 3/5
From Newton’s third law, force due to block on spring F_B
= – Restoring force of spring F_R [∵ Action = – Reaction]
Hence, F_B = F_R ⇒ mgsinθ = Kx.
⇒ 5 × 10 × \(\frac{3}{5}\) = 600 × x
⇒ 5 × 10 × \(\frac{3}{5}\) = 600 × x ⇒ x = \(\frac{30}{600}\) = 0.05m = 5cm

Question 7.
A force F = –\(\frac{k}{x^2}\)(x ≠ 0) acts on a particle along the X-axis. Find the work done by the force in displacing the particle from x = +a to x = +2a. Take K as a positive constant.
Solution:
Given force F = \(\frac{-k}{x^2}\)(x ≠ 0)

Question 8.
A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = -a to x = +2a?

Solution:
Total work done = Area of the graph
= Area of triangle OAB + Area of triangle OCD

Question 9.
From a height of 20m above a horizontal floor, a ball is thrown down with initial velocity 20 m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor? (g = 10m/s²)
Solution:
Initial velocity of the ball while coming downwards, u=20ms^-1.
distance travelled, s = h = 20m
Velocity of the ball at the time of hitting the ground v = ?
v² – u² = 2as ⇒ v² – 400 = 2 × 10 × 20

Question 10.
A ball falls from a height of 10m onto a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 1/√2 . What is the total distance travelled by the ball before it ceases to rebound?
Solustion:

Question 11.
Find the total energy of a body of 5kg mass, which is at a height of tOni from the earth and falling downwards straightly with a velocity of 20m/s.
(acceleration due to gravity as 10m/s²) [TS 16]
Solution:
Mass of the body (m) = 5kg;
Height of the body (h) = 10m
Velocity of the body (v)=20m/s,
Acceleration due to gravity (g) = 10 m/s
Potential Energy mgh= 5 × 10 × 10 = 500 J

Kinetic Energy = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 5 × (20)² = \(\frac{1}{2}\) × 5 × 400 = 1000J
∴ Total energy = P.E + K.E = 500 + 1000 = 1500J

Multiple Choice Questions

Question 1.
The power of a pump which can pump 100kg of water in 10s to a height of 100m is
1) 9800 W
2) 980 W
3) 98 W
4) 8800 W
Answer:
1) 9800 W

Question 2.
The work done in holding 15 Kg suitcase while waiting for a bus for 15 minutes is
1) 0
2) 1 J
3) 2 .1
4) 4 J
Answer:
1) 0

Question 3.
There are ……….. ergs in a Kwh
1) 3.6 × 10¹⁰
2) 3.6 × 10¹¹
3) 3.6 × 10¹²
4) 3.6 × 10¹³
Answer:
4) 3.6 × 10¹³

Question 4.
Two bodies of masses m₁ and m₂ have equal momenta. Their kinetic energies E₁ and E₂ are in the ratio

Answer:
3

Question 5.
A stone is projected vertically up to reach a maximum height h. The ratio of its kinetic energy to potential energy at a height 4h/5 will be
1) 5 : 4
2) 4 : 5
3) 1 : 4
4) 4 : 1
Answer:
3) 1 : 4

Question 6.
Two bodies of masses 2m and m have their kinetic energies in the ratio 8 : 1, then the ratio of their momenta is
1) 1 : 1
2) 2 : 1
3) 4 : 1
4) 8 : 1
Answer:
3) 4 : 1

Question 7.
If the speed of a vehicle increases by 2ms^-1, its kinetic energy is doubled, then original speed of the vehicle is

Answer:
3

Question 8.
One-fourth chain is hanging down from a table. Working done to bring the hanging part of the chain on the table is (mass of chain = M and length = L)

Answer:
1

Question 9.
A shell initially at rest explodes into two pieces of equal mass. The two pieces will
1) be at rest
2) move with different velocities in different directions.
3) move with same velocity in the opposite directions
4) move with same velocity in the same direction
Answer:
3) move with same velocity in the opposite directions

Question 10.
A 2Kg ball moving at 10ms^-1 collides with another 2Kg ball at rest. After collision they move together with
1) 5ms^-1 in the opposite direction
2) 5ms^-1 in the same direction
3) 10ms^-1 in the oppossite direction
4) 10ms^-1 in the same direction
Answer:
2) 5ms^-1 in the same direction

Question 11.
A particle falls from a height ‘h’ up on a fixed horizontal plane and rebounds. If ‘e’ is the coefficient of restitution, the total distance travelled before rebounding has

Answer:
1

Question 12.
A body falling from height of 10m rebounds from hard floor. If it loses 20% energy in impact, the coefficient of restitution is
1) 0.89
2) 0.56
3) 0.23
4)0.18
Answer:
1) 0.89

Question 13.
If a vector \(2\hat{i}+3\hat{j}+8\hat{k}\) is perpendicular to the vector \(4\hat{i}+4\hat{j}+\alpha\hat{k}\) then the value of α is
1) 1/2
2) -1/2
3) 1
4) -1
Answer:
2) -1/2

Question 14.
The angle between the two vectors \(\overrightarrow{A}=3\hat{i}+4\hat{j}+5\hat{k}\) and \(\overrightarrow{B}=3\hat{i}+4\hat{j}-5\hat{k}\) will be
1) 90°
2) 180°
3) zero
4)45°.
Answer:
1) 90°

Question 15.
A particle moves from a point \((-2\hat{i}+5\hat{j})\) to \((4\hat{j}+3\hat{k})\) when a force of \((4\hat{i}+3\hat{j})\)N is applied. How much work has been done by the force?
1) 8 J
2) 11 J
3) 5 J
4) 2 J
Answer:
3) 5 J

Question 16.
A uniform force of \((3\hat{i}+\hat{j})\) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position \((2\hat{i}+\hat{k})\) metre to position \((4\hat{i}+3\hat{j}-3\hat{k})\) metre. The work done by the force on the particle is
1) 13 J
2) 15 J
3) 9 J
4) 6 J
Answer:
3) 9 J

Question 17.
A body moves a distance of 10 m along a straight line under the action of a 5 N force. If the w ork done is 25.1, then angle between the force and direction of motion of the body is
1) 60°
2) 75°
3) 30°
4) 456
Answer:
1) 60°

Question 18.
If \(\overrightarrow{F}=(60\hat{i}+15\hat{j}-3\hat{k})\) N and \(\overrightarrow{v}=(2\hat{i}-4\hat{j}+5\hat{k})\) ms^-1, then instantaneous power is
1) 195 watt
2) 45 watt
3) 75 watt
4) 100 watt
Answer:
2) 45 watt

Question 19.
Flow much water a pump of 2 kW can raise in one minute to a height of 10 m? (take g = 10 ni/s²)
1) 1000 litres
2) 1200 litres
3) 100 litres
4) 2000 litres
Answer:
2) 1200 litres

Question 20.
300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (Take g = 10 m/s²)
1) 1000J
2) 200 J
3) 100 J
4) zero.
Answer:
3) 100 J

Question 21.
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s²)
1) 7.0 kW
2)10.2 kW
3) 8. 1 kW
4) 12.3 kW
Answer:
3) 8. 1 kW

Question 22.
Two bodies with kinetic energies in the ratio of 4:1 are moving with equal linear momentum. The ratio of their masses is
1) 4 : 1
2) 1 : 1
3) 1 : 2
4) 1 : 4
Answer:
4) 1 : 4

Question 23.
Two bodies of masses m and 4m are moving with equal kinetic energies. The ratio of their linear momenta is
1) 1 : 2
2) 1 : 4
3) 4 : 1
4) 1 : 1.
Answer:
1) 1 : 2

Question 24.
Two masses of 1 g and 9 g are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is
1) 1 : 9
2) 9 : 1
3) 1 : 3
4) 3 : 1
Answer:
3) 1 : 3

Question 25.
A stationary particle explodes into two particles of masses m₁ and m₂ which move in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies E₁/E₂ is
1) m₂/m₁
2) m₁/m₂
3) 1
4) m₁v₂/m₂v₁
Answer:
1) m₂/m₁

Question 26.
A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of
1) √2 : 1
2) 1 : 4
3) 1 : 2
4) 1 : √2
Answer:
3) 1 : 2

Question 27.
A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively If m₁ > m₂ then

Answer:
1

Question 28.
A particle is projected making an angle of 45° with horizontal having kinetic energy K, The kinetic energy at highest point will be
1) K/√2
2) K/2
3) 2K
4) K
Answer:
2) K/2

Question 29.
A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is
1) 40 ms^-1
2) 120 ms^-1
3) 100 ms^-1
4) 180 ms^-1
Answer:
3) 100 ms^-1

Question 30.
A bullet of mass 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the earth at the same level with a velocity of 500 m/s. The work done in joule for overcoming the resistance of air will be
1) 375
2) 3750
3) 5000
4) 500
Answer:
2) 3750

Question 31.
A position-dependent force, F = (7 – 2x + 3x²) N acts on a small body of mass 2 kg and displaces it from x = 0 to y = 5 m. The work done in joule is
1) 135
2) 270
3) 35
4) 70
Answer:
1) 135

Question 32.
A force acts on a 3 g particle in such a way that the position of the particle as a function of time is given by x = 3t – 4t² + t³ where x is in metres and t is in seconds. The work done during the first 4 second is
1) 490 mJ
2) 450 mJ
3) 240 mJ
4) 530 mJ
Answer:
3) 240 mJ

Question 33.
Force F on a particle moving in a straight line varies with distance d as shown in figure. The work done on the particle during its displacement of 12m is

1) 18 J
2) 21 J
3) 26 J
4) 13 J
Answer:
4) 13 J

Question 34.
When a body moves with a constant speed along a circle
1) no work is done on it
2) no acceleration is produced in it
3) its velocity remains constant
4) no force acts on it.
Answer:
1) no work is done on it

Question 35.
The potential energy of a system increases if work is done
1) upon the system by a nonconservative force
2) by the system against a conservative force
3) by the system against a nonconservative force
4) upon the system by a conservative force.
Answer:
2) by the system against a conservative force

Question 36.
The coefficient of restitution e for a perfectly elastic collision is
1) 1
2) 0
3) ∞
4) -1
Answer:
1) 1

Question 37.
Two equal masses m₁ and m₂ moving along the same straight line with velocities + 3 m/s and -5 m/s respectively collide elastically. ‘Their velocities after the collision will be respectively
1) – 4 m/s and +4 m/s
2) +4 m/s for both
3) -3 m/s and +5 m/s
4) -5 m/s and +3 m/s
Answer:
4) -5 m/s and +3 m/s

Students get through AP Inter 1st Year Physics Important Questions 4th Lesson Motion in a Plane which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 4th Lesson Motion in a Plane

Very Short Answer Questions

Question 1.
The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with X-axis? [TS 18; AP 19; lmp.Q]
Answer:
Vns Let ‘θ’ be angle made by the vector \(\overrightarrow{R}\) with the X-axis.
Vertical component of \(\overrightarrow{R}\) = R sinθ
Horizontal component of \(\overrightarrow{R}\) = R cosθ
∴ Rcosθ = Rsinθ ⇒ tanθ = 1 ⇒ θ = 45°

Question 2.
A vector \(\overline{\mathrm{v}}\) makes an angle e with the horizontal.
The vector is rotated through an angle e. Does this rotation change the vector ‘v\
Answer:
By rotating the vector \(\overline{\mathrm{v}}\) through angle V it’s magnitude does not change, but its horizontal and vertical components change. Also direction of the vector changes. So the rotation changes the vector \(\overline{\mathrm{v}}\).

Question 3.
Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant? |AP 15,16,17||TS 22|
Answer:
Here, P = 3 and Q = 5 and θ = 60° ⇒ cosθ = cos60° = 1/2

Question 4.
If \(\overrightarrow{A}\) = \(\overrightarrow{i}+\overrightarrow{j}\) What is the angle between vector \(\overrightarrow{A}\) with x -axis? [IPE’ 13, 13, 14; AP 20, 22; TS 17, 20]
Answer:
Comparing the vector \(\overrightarrow{i}+\overrightarrow{j}\) with \(x\overrightarrow{i}+y\overrightarrow{j}\). we get x = 1 and y = 1
If \(\overline{\mathrm{A}}\) = \(x\overline{\mathrm{i}}+y\overline{\mathrm{j}}\) makes an angle θ with the x-axis then tan θ = \(\frac{y}{x}=\frac{1}{1}\) = 1 (∵ θ = 45°)

Question 5.
When two right-angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant? [IPE’ 14; AP 16,18]
Answer:
Herc, P = 7 and Q = 24 and θ = 90° ⇒ cosθ = cos90° = 0

Question 6.
If \(\overrightarrow{P}=2\overrightarrow{i}+4\overrightarrow{j}+14\overrightarrow{k}\) and \(\overrightarrow{Q}=4\overrightarrow{i}+4\overrightarrow{j}+10\overrightarrow{k}\) find the magnitude of P + Q. [TS 15, 16, 19, 22]
Answer:

Question 7.
a vector of magnitude zero have a non zero components?
Answer:
No. A vector of magnitude zero cannot have non-zero components.

Question 8.
What is the acceleration of a projectile at the top of its projectory? [AP, TS 19]
Answer:
At the top of its projectory, the direction of acceleration is vertically downwards and its value is 9.8ms^-2.

Question 9.
Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors?
Answer:
No, the addition of two un equal vectors cannot give a zero vector

But three un equal vectors lying in a plane can give a zero vector, if triangle law is satisfied.

Short Answer Questions

Question 1.
State Parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector. [AP 22; TS 16, 17, 20, 22]
Answer:
Parallelogram law :
‘If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point then their resultant is represented in magnitude and direction by the diagonal passing through the same point’.

Let two vectors \(\overrightarrow{P}\) = \(\overrightarrow{OA}\) and \(\overrightarrow{Q}\) = \(\overrightarrow{OB}\) be acting simultaneously at a point ‘O’.
Let ‘θ’ be the angle between the two vectors \(\overrightarrow{P} and \overrightarrow{Q}\) and Q.
Now, the parallelogram OACB is completed.
Then, the diagonal \(\overrightarrow{OC}\) represents the resultant vector \(\overrightarrow{R}\).
Now OA is produced and a perpendicular CD is drawn.
Then ∠CAD = θ
Also, |\(\overrightarrow{OA}\)| = OA = P and |\(\overrightarrow{OB}\)| = OB = AC = Q

Magnitude of the Resultant vector \(\overrightarrow{R}\) :
From the right angled triangle COD,
OC² = OD² + CD²
⇒ OC² = (OA+AD)² + CD² (Since, OD = OA + AD)
⇒ OC² = OA² + AD² + 2OA . AD + CD² ⇒ OC² = OA² + (AD² + CD²) + 2OA . AD ——— (1)
Now, from the right angled triangle C AD . AC² = AD² + CD² ——- (2)

Direction of the resultant :
Let the resultant \(\overrightarrow{R}\) makes an angle ‘α’ with \(\overrightarrow{P}\).

Question 2.
What is relative motion. Explain it?
Answer:
Relative motion :
The motion of a body with respect to another body is called relative motion between them.

Consider two bodies A,B moving with Velocities V_A and V_B.
(i) If A and B are moving in the same direction then
the relative velocity of A with respect to B is \(\overrightarrow{V}_{AB}=\overrightarrow{V}_{A}-\overrightarrow{V}_{B}\)
The relative velocity of B with respect to A is \(\overrightarrow{V}_{BA}=\overrightarrow{V}_{B}-\overrightarrow{V}_{A}\)

(ii) If A and B are moving in opposite directions then
the relative velocity of A with respect to B is \(\overrightarrow{V}_{AB}=\overrightarrow{V}_{A}-(-\overrightarrow{V}_{B})=\overrightarrow{V}_{A}+\overrightarrow{V}_{B}\).

Question 3.
Show that a boat must move at an angle of 900 with respect to river water in order to cross the river in a minimum time?
Answer:
Consider a boat starts at a point A on one bank of the river and intends to reach the other bank, as shown in the Fig.

If the velocity of boat in still water is V_b, and velocity of water in the river is V_w then Resultant velocity is

In the above expression, the denominator V_R is constant (since V_b, V_w are fixed /given)
Hence ‘t’ becomes minimum when AC becomes minimum. The minimum value of AC is nothing but AB, which is equal to the width ‘d’ of the river. Here AB is perpendicular to V_w.
Thus, the boat must move at an angle of 90° with respect to the river water.

Question 4.
Define unit vector, Null vector and position vector [AP 15]
Answer:
Unit Vector :
A vector of magnitude one unit is called a unit vector.

Null Vector :
A vector of zero magnitude and arbitrary direction is called a zero vector or Null vector.

Position Vector :
The position vector of a point P is the vector from the origin ‘O’ of the coordinate system to the position of the point. It is denoted by \(\overrightarrow{OP}\).

If (x,y,z) are the co-ordinates of a point ‘P’ and ‘O’ is the origin of coordinate system then \(\overrightarrow{OP}=\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\) is the position vector of P.
Magnitude of \(\overrightarrow{r}\) is |\(\overrightarrow{r}\)| = \(\sqrt{x^2+y^2+z^2}\)

Question 5.
If |\(\overrightarrow{a}+\overrightarrow{b}\)| = |\(\overrightarrow{a}-\overrightarrow{b}\)| then what is the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\)? [AP 19; TS 18, 22]
Answer:

Question 6.
Show that the trajectory of an object thrown at a certain angle with the horizontal is a parabola. [AP 20; IPE’ 13,14; AP, TS 15, 16, 17, 18, 19, 22]
Answer:
1) Suppose an object is projected from the origin ‘O’ with an initial velocity u at an angle 0 with the horizontal.
Horizontal component of u is u_x = ucosθ
Vertical component of u is u_y = usinθ

2) Motion along horizontal: The horizontal component u_x= ucosθ and a_x = 0.

Thus, (iv) represents the equation of a parabola.
So, the trajectory of a projectile is a parabola.

Question 7.
Explain the terms average velocity and instantaneous velocity. When are they equal?
Answer:
Average velocity :
Suppose a body is at a point x₁ at time t₁ and at some other point x₂ at time t₂, then the displacement of the body is x₂ – x₁. Hence its average velocity is given by

Instantaneous velocity :
The velocity of an object at a particular point of its path or at a particular instant of time is called as instantaneous velocity.

During uniform motion, the average and instantaneous velocity are always same because the velocity during uniform motion is same at each point of its path or at each instant.

Question 8.
Show that maximum height and range of projectile are \(\frac{u^2 \sin ^2 \theta}{2 g}, \frac{u^2 \sin 2 \theta}{g}\) respectively when the terms have their regular meaning. [Imp.Q|[TS 16]
Answer:
Maximum height :
It is the maximum vertical distance travelled by the projectile where its vertical velocity component becomes zero.
Let a body be projected with a velocity ‘u’ at an angle ‘θ’ with the horizontal.

In the vertical direction:
Initial vertical velocity u = u sin θ
At maximum height, final vertical velocity v = 0
Acceleration a = -g.
From the equation v² – u²= 2as, we get 0 – (usinθ)² = -2gh_max

Horizontal Range (R) of a projectile :
It is the horizontal distance travelled by a projectile during its time of flight.
Range = Horizontal velocity x time of flight

If θ is 45° then the horizontal range of a projectile is maximum.

Question 9.
If the trajectory of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else it can be?
Answer:
It will not be parabolic from the second reference frame. It appears to move in a straight line.
Exampler :
Suppose a bomb is dropped from an aeroplane flying at a height with a constant speed. For a stationary observer on the earth, the path of the bomb will be parabola. But for the pilot in the plane, the path of the bomb will look like a vertical straight line.

Question 10.
A force \(2\hat{i}+\hat{j}-\hat{k}\) newton acts on a body which is initially at rest. At the end of 20 seconds, the velocity of the body is \(4\hat{i}+2\hat{j}-2\hat{k}\)ms^-1. What is the mass of the body? [AP 16]
Answer:

Solved Problems

Question 1.
Rain is falling vertically with a speed of 35ms^-1. Wind starts blowing after sometime with a speed of 12 ms^-1 in east to west direction. In which direction should a boy waiting at a bus stop, hold his umbrella? [Imp.Q]
Answer:

Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19° with the vertical towards the east.

Question 2.
Rain is falling vertically with a speed of 35ms^-1. A woman rides a bicycle with a speed of 12ms^-1 in east to west direction. What is the direction in which she should hold her umbrella? [TS 15]
Answer:

The velocity of rain is vp the velocity of bicycle is v_b. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the velocity of the bicycle. That is
v_rb = v_r – v_b
This relative velocity vector makes an angle θ with the vertical.
It is given by tan θ = \(\frac{v_b}{v_r}=\frac{12}{35}\) = 0.343 ⇒ θ = Tan^-1 (0.343) = 19°

Therefore, the woman should hold her umbrella at an angle of about 19° with the vertical towards the west.

Question 3.
The position of a particle is given by r = 3.0t\(\overline{\mathrm{i}}\) + 2.0t²\(\overline{\mathrm{j}}\) +5.0\(\overline{\mathrm{k}}\) where t is in seconds and the coefficients have the proper units for r to be in meters, (a) Find v(t) and a(t) of the particle, (b) Find the magnitude and direction of v(t) at t = 1.0s.
Solution:

Question 4.
Prove the statement “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. [IMP.Q]
Solution:
For a projectile launched with velocity v0 at an angle 90, the range is R = \(\frac{\mathrm{v}_0^2 \sin \left(2 \theta_0\right)}{\mathrm{g}}\)
Now, for angles (45°+ α) and (45°- α) 2θ₀ is (90°+ 2α) and (90°- 2α) respectively.
Now, sin(90°+2α) = cos2α and sin(90°-2α) = cos2α
Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.

Question 5.
A cricket ball is thrown at a speed of 28 ms^-1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the. same level. [IMP.Q]
Solution:

Exercise Problems

Question 1.
Ship A is 10km due west of ship B. Ship A is heading directly north at a speed of 30 km/h, while jihip B is heading in a direction 60“ west of north at a speed of 20km/h.
(i) Determine the magnitude of the velocity of ship B relative to ship A.
(ii) What will be their distance of closest approach?
Solution:
Assume that ships A and B lie along the X-axis as shown in the diagram.

From the concept of relative velocity, ship A may be assumed to be at rest and ship B is moving with a velocity

The closest distance is the perpendicular distance of the relative velocity vector from the fixed position of ship A.

Question 2.
If θ is angle of projection, R the range, h the maximum height, T the time of flight then . 4h u gT2
show that (a) tan θ = \(\frac{4h}{R}\) (b) h = \(\frac{8T^2}{8}\).
Solution:
If θ is angle of projection and u is the velocity of projection, then we know that

Question 3.
A projectile is fired at angle of 60° to the horizontal with an initial velocity of 800ms-1.
(i) find the time of flight before if hits the ground
(ii) find the distance if travels before it hits the ground (range)
(iii) find the time of flight for the projectile to reach maximum height.
Solution:
Given angle of projection, θ = 60°
Velocity of projection, u = 800 ms^-1.

Question 4.
For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to he √2 limes the maximum height reached by it. Show that the angle of projection is Tan^-1(2).
Solution:
Let P be the maximum height point. Then its coordinates will be (R/2, h). So its position vector

Hence, the angle of projection (θ) = Tan^-1(2)

Question 5.
An object is launched from a cliff 20 m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground? (g=10 m/s²)
Solution:
Total horizontal distance travelled, R= R₁ +R₂.

At the point B, in vertical direction, initial velocity (u) = usinθ
= 30 × sin30° = 30 × \(\frac{1}{2}\) = 15ms^-1
Acceleration (a) = g = 10 ms^-2; distance travelled (s) = 20m
Time taken (t) = ?
s = ut + \(\frac{1}{2}\) at² ⇒ 20 = 15 × t + \(\frac{1}{2}\) × 10 × t² ⇒ 20 = 15t + 5t² ⇒ 5t² + 15t – 20 = 0
⇒ t² + 3t – 4 = 0 ⇒ t² + 4t – t – 4 = 0 ⇒ t(t+4) – 1(t+4) = 0 ⇒ (t+4)(t-1) = 0 ⇒ t =1 (4 cannot be taken)

Duringthis 1 s, horizontal distance travelled, R₂ = ucosθ × t = 30 cos30° × 1 = 30 × \(\frac{\sqrt{3}}{2}\) = 15√3m
∴ Total horizontal distance travelled, R = R₁ + R₂ =45√3 + 15√3 =60√3 m

Question 6.
O is a point on the ground chosen as origin. A body first suffers a displacement of 10√2m North-East, next 10m North and finally 10√2m in North-West. How far it is from the origin? [TS 19]
Solution:

∴ The body is at a distance of 30m from the origin in north direction.

Question 7.
From a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Solustion:
For maximum range, the angle of projection is 45°.
The body is projected with a velocity u making an angle 45° with the horizontal. So u can be

Question 8.
A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7:5m above the ground while it travels a horizontal distance of 10m from the point of projection. Find the initial speed of projection. (g = 10m/s²).
Solution:
Let u be the velocity of projection.
Let t be the time taken by the body to travel 10m in the horizontal direction.
But we know that, horizontal distance travelled = uniform velocity × time

Question 9.
Wind is blowing from the south at 5ms^-1. To a cyclist is appears to be blowing from the east at 5ms^-1. Show that the velocity of the cyclist is 5√2 ms^-1 towards north-east.
Solution:

Velocity of wind w.r.to ground V_WG = 5ms^-1 from South
Velocity of wind w.r.to cyclist, V_WG = 5ms^-1 from East
Velocity of cyclist w.r.to ground, V_CG = ?

Question 10.
A person walking at 4 m/s finds rain drops falling slantwise into his face with a speed of 4m/s at an angle of 30° with the vertical. Show that the actual speed of the rain drops is 4 m/s.
Solution:

Velocity of rain w.r.to man = V_RM = 4 ms^-1,
Velocity of man w.r.to ground = V_MG = 4ms^-1
Velocity of rain w.r.to ground = V_RG = ?
(actual velocity of rain)

Multiple Choice Questions

1. If \(\overline{\mathrm{A}}=5\overline{\mathrm{i}}+7\overline{\mathrm{j}}-3\overline{\mathrm{k}}\) and \(\overline{\mathrm{B}}=2\overline{\mathrm{i}}+2\overline{\mathrm{j}}-c\overline{\mathrm{k}}\)B-2i + 2j-ck are mutually perpendicular, then the value of ‘c’ is
1) 2
2) √14
3) √10
4) √5
Answer:
2) √14

2. \(\overline{\mathrm{A}}=2\hat{i}-2\hat{j}+\hat{k}\); B = \(3\hat{i}+6\hat{j}+n\hat{k}\). What is the value of ‘n’ if \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are perpendicular?
1) 6
2) -6
3) 18
4) -18
Answer:
1) 6

3. The angle made by the vector \(\overline{\mathrm{A}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}\) with X-axis is
1) 30°
2) 60°
3) 90°
4) 120°
Answer:
2) 60°

4. Force is \(6\overline{\mathrm{i}}+C\overline{\mathrm{j}}-2\overline{\mathrm{k}}\) and displacement \(\overline{\mathrm{i}}+2\overline{\mathrm{j}}+6\overline{\mathrm{k}}\). If the work done is 6\(\overline{\mathrm{j}}\), the value of ‘C’ is
1) 6
2) 4
3) 8
4) 9
Answer:
1) 6

5. The angle between the vectors (\(\overline{\mathrm{i}}+\overline{\mathrm{j}}\)) and (\(\overline{\mathrm{j}}+\overline{\mathrm{k}}\) + k) is
1) 30°
2) 45°
3) 60°
4) 90°
Answer:
3) 60°

6. A cricket ball is hit at 45″ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
1) K
2) K/2
3) Ksin45°
4) 0
Answer:
2) K/2

7. The equations of motion of a projectile are given by x = 36t meter and 2y = 96t – 9.8t² meter. The angle of projection is

Answer:
1

8. The speed of a projectile at its maximum height is √3/2 times its initial speed. If the range of the projectile is p times the maximum height attained by it, then p =
1) 4/3
2) 2√3
3) 4√3
4) 3/4
Answer:
3) 4√3

9. A person throws a bottle into a dustbin at the same height as he is 2m away at an angle of 45°. The velocity of thrown bottle is [EAM Q]
1) g
2) √g
3) 2g
4) √2g
Answer:
4) √2g

10. Identify the vector quantity among the following.
1) Distance
2) Angular momentum
3) Heat
4) Energy
Answer:
2) Angular momentum

11. If a unit vector is represented by \(0.5\hat{i}-0.8\hat{j}+C\hat{k}\) then the value of c is

Answer:
2) √0.11

12. The vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are such that |\(\overrightarrow{A}+\overrightarrow{B}\)| = |\(\overrightarrow{A}-\overrightarrow{B}\)|. The angle between the two vectors is
1) 45°
2) 90°
3) 60°
4) 75°
Answer:
2) 90°

13. If |\(\overrightarrow{A}+\overrightarrow{B}\)| = \(|\overrightarrow{A}|+|\overrightarrow{B}|\)| then angle between A and B will be
1) 90°
2) 120°
3) 0°
4) 60°
Answer:
3) 0°

14. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
1) 45°
2) 180°
3) 0°
4) 90°
Answer:
4) 90°

15. The magnitude of vectors \(\overrightarrow{A},\overrightarrow{B}and\overrightarrow{C}\) are 3, 4 and 5 units respectively. If \(\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C}\), the angle between \(\overrightarrow{A}and\overrightarrow{B}\) is
1) π/2
2) cos^-1(0.6)
3) tan^-1(7/5)
4) π/4
Answer:
1) π/2

16. A particle has initial velocity (\(2\hat{i}+3\hat{j}\)) and acceleration (\(0.3\hat{i}+0.2\hat{j}\)). The magnitude of velocity after 10 seconds will be
1) 9√2 units
2) 5√2 units
3) 5 units
4) 9 units
Answer:
2) 5√2 units

17. A particle has initial velocity (\(3\hat{i}+4\hat{j}\)) and has acceleration (\(0.4\hat{i}+0.3\hat{j}\)). Its speed after 10 s is
1) 7 units
2) 7√2 units
3) 8.5 units
4) 10 units
Answer:
2) 7√2 units

18. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
1) θ = tan^-1 (1/4)
2) θ = tan^-1(4)
3) θ = tan^-1(2)
4) θ = 45°
Answer:
2) θ = tan^-1(4)

19. For angles of projection of a projectile at angle (45°-0) and (45°+ 0), the horizontal range described by the projectile are in the ratio of
1) 2 : 1
2) 1 : 1
3) 2 : 3
4) 1 : 2.
Answer:
2) 1 : 1

20. A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s², the range of the missile is
1) 40 m
2) 50 m
3) 60 m
4) 20 m
Answer:
1) 40 m

21. Two projectiles of same mass and with same velocity are thrown at an angle 60° and 30° with the horizontal, then which will remain same
1) time of flight
2) range of projectile
3) maximum height acquired
4) all of them.
Answer:
2) range of projectile

22. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
1) 60°
2) 15°
3) 30°
4) 45°
Answer:
1) 60°

23. The maximum range of a gun of horizontal terrain is 16 km. If g = 10 m s^-2, then muzzle velocity of a shell must be
1) 160 ms^-1
2) 200√2ms^-1
3) 400 ms^-1
4) 800 ms^-1
Answer:
3) 400 ms^-1

24. The velocity of a projectile at the initial point A is (\(2\hat{i}+3\hat{j}\))m/s. Its velocity (in m/s) at point B is
1) \(2\hat{i}-3\hat{j}\)
2) \(2\hat{i}+3\hat{j}\)
3) \(-2\hat{i}-3\hat{j}\)
4) \(-2\hat{i}+3\hat{j}\)
Answer:
1) \(2\hat{i}-3\hat{j}\)

25. A boat is sent across a river with a velocity of 8 kmh^-1. If the resultant velocity of boat is 10 kmh^-1, then velocity of river is
1) 12.8 kmh^-1
2) 6 km h^-1
3) 8 kmh^-1
4) 10 km h^-1
Answer:
2) 6 km h^-1

26. The width of river is 1 km. The velocity of boat is 5 km/hr. The boat covered the width of river in shortest time 15 min. Then the velocity of river stream is
1) 3 km/hr
2) 4 km/hr
3) √29 km/hr
4) √41 km/hr
Answer:
1) 3 km/hr

27. A person aiming to reach exactly opposite point on the hank of a stream is swimming with a speed of 0.5 m/s at an angle of 120° with the direction of flow of water. The speed of water in the stream, is
1) 0.25 m/s
2) 0.5 m/s
3) 1.0 m/s
4) 0.433 m/s
Answer:
1) 0.25 m/s

28. The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is, given by
1) 45° west
2) 30° west
3) 0°
4) 60° west
Answer:
2) 30° west

29. The angular speed of a flywheel making 120 revolutions/minute is
1) 4π rad/s
2) 4π² rad/s
3) π rad/s
4) 4π rad/s
Answer:
1) 4π rad/s

30. Two particles A and B are moving in uniform circular motion in concentric circles of radii r_A and r_B with speed v_A and v_B respectively. Their lime period of rotation is the same. ’The ratio of angular speed of A to that of B will be
1) 1 : 1
2) r_A : r_B
3) v_A : v_B
4) r_B : r_A
Answer:
1) 1 : 1

31. A particle moves in x-y plane according to rule x = asinωt and y = acosωt. The particle follows
1) an elliptical path
2) a circular path
3) a parabolic path
4) a straight line path inclined equally to x and y-axes
Answer:
2) a circular path

32. A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is
1) 1 m/s²
2 ) 7 m/s²
3) √7 m/s²
4) 5 m/s²
Answer:
4) 5 m/s²

33. The x and y coordinates of the particle at any time are x = 5t – 2t² and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is
1) 5 m s^-2
2) -4m s^-2
3) -8 m s^-2
4) 0
Answer:
2) -4m s^-2

34. A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s²)
1) 20√2 m/s, 10 m/s²
2) 20 m/s, 5 m/s²
3) 20 m/s, 0
4) 20√2 m / s, 0
Answer:
1) 20√2 m/s, 10 m/s²

Students get through AP Inter 1st Year Physics Important Questions 3rd Lesson Motion in a Straight Line which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 3rd Lesson Motion in a Straight Line

Very Short Answer Questions

Question 1.
The states of rest and motion are relative. Explain. [Imp.Q]
Answer:
For a person travelling in a train, the co-passengers appear to be at rest, but objects outside the train like electric poles, trees etc, appear to be moving backwards. Where as, for a person on the platform the electric poles, trees etc. appear to be at rest but the train appears to be moving. So the states of rest and motion are relative.

Question 2.
How is average velocity different from instantaneous velocity? [Mar 13]
Answer:
The average velocity during a time interval does not tell us how fast or in what direction the body is moving at any given time during the time interval. Where as instantaneous velocity gives the idea of fastness and the direction of motion of the body. In uniform motion, both are equal.

Question 3.
Give an example of a case, where the velocity of an object is zero, but its acceleration is not zero. [|Imp.Q| Mar 13]
Answer:
In the case of a vertically projected body at the Maximum height, its velocity is zero but it has acceleration due to gravity in the downward direction.

Question 4.
A vehicle travels half the distance L with speed v₁ and the other half with speed v₂ What is the average speed?
Answer:

Question 5.
A lift coming down is just about to reach tjie ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
(a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0 (c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a > 0
Answer:

If x denotes the displacement of the lift, then it will be negative.
i.e, x < 0 (displacement = final reading – initial reading)
As the lift is moving downwards, its velocity is directed downwards.
So v < 0 When the lift is about to touch the ground floor, its velocity gets decreased. So its acceleration is in the upward direction, i.e., a > 0. (a) is correct

Question 6.
A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time, taking the acceleration in the backward direction as positive.
Answer:
The graph for the given motion is as shown in the tig.

Question 7.
Give example of one-dimensional motion, where the particle moving along positive x-direction comes to rest periodically and moves forward.
Answer:
Consider the position of the particle at any time ‘t’ is given by x = t – sint
⇒ velocity (v) = \(\frac{dx}{dt}\) = 1 – cos t
Since cost< 1, v is positive i.e, the particle moves forward.
When t=0, v=0, i.e, the particle comes to rest

Question 8.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity?
Answer:
When velocity becomes constant, acceleration, a = 0 ⇒ 0 = g -bv ⇒ bv = g (or) v = g/b
∴ The constant velocity is v = \(\frac{g}{b}\)

9. If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame? if not, what can it be?
Answer:
It will not be parabolic from the second reference frame. It appears to move in a straight line.

Reason:
The horizontal component of projectile and the velocity of second reference frame are constants. So the person in the second reference frame will observe only the vertical component of velocity of the projectile. As it decreases till the projectile reaches maximum height and then increases, the body will appear as though it is thrown vertically upwards with velocity usinθ where u is the velocity of projection and θ is the angle of projection.

Question 10.
A spring with one end attached to a mass and the other to a rigid support is stretched and released. When is the magnitude of acceleration a maximum?
Answer:
When the spring is stretched and released, the mass executes S.H.M. In S.H.M. Acceleration is directly proportional to displacement. Hence the mass will have maximum acceleration at extreme positions.

Short Answer Questions

Question 1.
Can the equations of kinematics he used when t lie acceleration varies with time? If not, what form would these equations take?
Answer:
No. The equations of kinematics [v = v₀ + at, s = v₀t + \(\frac{1}{2}\)at², v² -v₀² = 2as, s_n = v₀ + a(n – \(\frac{1}{2}\))]cannot be used when the acceleration varies. Because in all these equations, acceleration is constant.

If the acceleration of a body is not constant, the motion of such a body is called non-uniformly accelerated motion. For such a case, let acceleration, a ∝ = t_n
Thus a = kt_n or \(\frac{dv}{dt}\) = kt_n ⇒ dv = kt_ndt ………. (1)
When t = 0, u = v₀ (initial velocity) and where k is a constant, when t = t, v = v (final velocity)

Question 2.
Derive the equation S = ut + \(\frac{1}{2}\)at² from v – t graph.
Answer:

A graph is drawn between v & t for uniformly accelerated motion. The area under v-t curve gives the displacement.
So Area between the instants 0 to t = Area of triangle ABC + Area of rectangle OACD
S = ut + \(\frac{1}{2}\)(v -u)t
S = ut + \(\frac{1}{2}\)(at)t (∵ v – u = at)
∴ S = ut + \(\frac{1}{2}\)at²

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t = t is v₂. The average velocity of the particle in this time interval is \(\frac{v_1+v_2}{2}\). Is this correct? substantiate your answer.
Answer:

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object? If so, give an example. [TS 22]
Answer:
Yes, the velocity of an object can be in a different direction other than that of acceleration.
Ex :
1) When a body is projected vertically upwards, the direction of velocity is upwards where as the direction of acceleration is downwards.
2) Incaseof oblique projectile motion, velocity is always tangential to the path followed by the projectile, whereas the gravitational acceleration is always downward.

Question 4.
A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1km above ground. Describe his motion. [TS 17]
Answer:

A parachutist flying in an aeroplane jumps when it is at a height ot 3km above the ground. Upto 2 km he travels under gravity. At a height lkm above the ground the parachute opens. Once the parachute ( pens air resistive force acts on parachutist in upward direction and grav itational force acts on him in downwards. After travelling certaind istance both forces are equal and net force on him is zero. So his acceleration is zero. At which he attains constant velocity called “Terminal Velocity”. Then the parachutist reaches the ground safely.

Question 5.
A bird holds a fruit in its beak and flies parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by (a) the bird and (b) a person on the ground. [TS 22]
Answer:
(a) To the bird, the fruit appears to fall in a straight line path.
Reason :
In horizontal direction, the bird and the fruit will have same velocity. So they will travel equal distances in equal intervals of time. So the fruit always appears just below the bird.

(b) To a person on the ground, the fruit will appear to follow a parabolic path.
Reason:
When the bird, flying horizontally, leaves the fruit, the fruit will have a velocity in horizontal direction. This velocity remains constant because no forces are acting on it in horizontal direction. But in vertical direction it has acceleration due to gravity. So the fruit appears to follow a parabola path.

Question 6.
A man runs across the roof of a tali building and jumps horizontally on to the lower roof of an adjacent building. If his speed is 9ms^-1 and the horizontal distance between the buildings is 10m and the height difference between the roofs is 9m, will he be able to land on the nest building? (g = 10ms^-2) [TS 18]
Answer:

If his horizontal range is greater than 10m, then he can land safely on the other building.
In vertical direction,
initial velocity (u) = 0
uniform acceleration (a) = +g = + 10ms^-2
distance travelled (s) = height difference = 9m
time taken (t) = ?
From S = ut + \(\frac{1}{2}\)at²

Now horizontal range = horizontal velocity × time of fall = 9 × 1.34- 12.08m ≈ 12.1 m
Since he can jump upto nearly 12.1m, so he can safely land on the roof of adjacent building.

Question 7.
A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball Sands first? Explain your answer. [TS 15]
Answer:
For a dropped body:

If a body is dropped from a tower of height ‘h’
Initial velocity u = 0, accelerations a = + g, displacement s = h
Let time of jouney be t₁ ; from s = ut + \(\frac{1}{2}\)at²h = 0 + \(\frac{1}{2}\)gt₁² t₁ = \(\sqrt{\frac{2 h}{g}}\)

For horizantally projected body:

If a body is projected horizontally from a tower of height ‘h’
Initial vertical velocity u_y = 0, Initial vertical displacement s = h
acceleration a_y = g, Let time of journey bt t₂
from s = ut + \(\frac{1}{2}\)at² ⇒ h = 0 + \(\frac{1}{2}\)gt₂² ⇒ t₂ = \(\sqrt{\frac{2 h}{g}}\)
As t₁ = t₂ both the bodies reach to the ground simultaneously

Question 8.
A ball is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change in relative velocities of the balls as a function of time. [lmp.Q]
Answer:
Suppose the ball ‘A’ is falling freely. So its initial velocity, u_A = 0
Suppose the ball ‘B’ is projected upwards with a velocity ‘u’,

so its initial velocity, u_B = u
Relative velocity of A w.r.t. B is u_AB = u_A – u_B = 0 – u = -u
After some time t, the velocity of A will be v_A = u + at
v_A = 0 + gt
v_A = gt

the velocity of B will be v_B = u – gt
Now, Relative velocity of A w.r.t. B is v_AB = V_A – V_B
= gt- u + gt = 2gt – u

Change in relative velocity of A w.r.t. B in time t is given by
(2gt – u) – (-u) = 2gt – u + u = 2gt
∴ Their change in relative velocities will be 2gt

Question 9.
A typical raindrop is about 4mm in diameter. If a raindrop falls from a cloud which is at I km above the ground, estimate its momentum when it hits the ground.
Answer:
If the rain drop is supposed not attained terminal velocity, when it is falling from the cloud, then the velocity with which it hits the ground can be calculated as follow:
Initial velocity (u) = 0; acceleration (a) = g = 9.8ms^-2
distance travelled (s) = 1km = 1000m; Final velocity (v) = ?
v² – u² = 2as ⇒ v = \(\sqrt{2gh}\) = \(\sqrt{2\times9.8\times1000}\) = 140ms^-1
Now radius of rain drop, r = \(\frac{4mm}{2}\) = 2mm = 2 × 10^-3m
Volume of the rain drop (Spherical in shape), V = \(\frac{4}{3}\)πr³
V = \(\frac{4}{3}\times\frac{22}{7}\) × (2 × 10^-3)³ = 33.52 × 10^-9m³

Let us take, density of water (d) = 1000 kg /m³
∴ Mass of rain drop, m = volume of drop × density = 33.52 × 10^-9 × 1000 = 33.52 × 10^-6 kg
= 33.52 × 10^-6 × 140 = 4692 × 10^-6
= 0.00469 ku ms^-1

Question 10.
Show that the maximum height reached by a projectile launched at an angle 45° is one quarter of the range. [AP 16,19]
Answer:
The horizontal distance travelled by a projectile during its time of flight is called Range.

Thus , maximum Fleight is one quarter of the Range.
NOTE: Please note that, (10) problem is related to the next chapter (Motion in a Plane)

Exercise Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh^-1. What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 min. [AP 18,19][TS 20]
Solution:

Question 2.
A car travels the first third of a distance with a speed of lOkmph, the second third at 20kmph and the last third at 60kmph. What is its mean speed over the entire distance? [AP 18, 22][IPE’ 14][TS 16, 18]
Solution:
Let the total distance = 3S km

Question 3.
A bullet moving with a speed of 150 ms-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree? |May 13]
Solution:
Initial speed of the bullet, (u) = 150 ms^-1.
Distance travelled in the tree before stopping (s) = 3.5 cm = 3.5 × 10^-2 m
Final velocity (v) =0; acceleration (a) =?; time taken to stop (t) = ?
Now, v² – u² = 2as ⇒ (0)² – (150)² = 2 × a × 3.5 × 10^-2.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stops for 15min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity?
Solution:
Distance travelled during 30 minutes, S₁ = speed × time = \(\frac{85km}{hr}\) × 30min = \(\frac{85km}{hr}\) × \(\frac{1}{2}\)hr = 42.5km

Distance travelled during 2 hours, S₂ = 130 km
Since motorist travelled in the same direction i.e., north,
total displacement(s) = total distance travelled
⇒ s = 42.5 + 130 = 172.5 km

Question 5.
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur?
Solution:

Question 6.
Drops of wafer fall at regular intervals from the roof of a building of height 16m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:

Distance between 1^st and 2^nd drops = S₁ – S₂ = 16 – 9 = 7 m
Distance between 2^nd and 3^rd drops = S₂ – S₃ = 9 – 4 = 5 m
Distance between 3^rd and 4^th drops = S₃ – S₄ = 4 – 1 = 3 m
Distance between 4^th and 5^th drops = S₄ – 0 = 1 – 0 = lm

Question 7.
A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the bullet. Explain why the monkey made a wrong move.
Solution:

The bullet is like a horizontal projectile and its path traces the parabolic motion, instead of a straight line.
If the monkey jumps down from the branch then the bullet may hit the monkey at some instant as shown in the diagram.

Question 8.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Solution:
Let the origin of the co-ordinate system is on the ground, directly under the position of the plane when it drops the bomb.

(i) The time t taken by the food packet to reach the ground will be determined from the vertical motion of the bomb. In the vertical direction, initial velocity (u)=0,
distance travelled (s) = h = 500m acceleration (a) = g = 10ms^-2 time taken (t) = ?
Now, s = ut + \(\frac{1}{2}\)at² ⇒ 500 = 0 + \(\frac{1}{2}\) × 10 × t² ⇒ 500 = 5t² ⇒ t² = 100 ⇒ t = 10 s

(ii) The initial horizontal velocity of food packet remains constant.
Horizontal distance = Horizontal velocity × time of travel = 100 × 10 = 1000m

Question 9.
A ball is tossed from the window of a building with an initial velocity of 8 ms^-1 at an angle of 20″ below the horizontal. It strikes the ground 3s later. From what height was the ball thrown? flow far from the base of the building does the ball strike the ground?
Solution:
In the vertical direction, initial velocity (u) = 8 sin20° = 8 × 0.3420 = 2.736 ms^-1.
Time (t) = 3 s acceleration (a) = g = 9.8 ms^-2. Required height = distance s = ?
Now s = ut + \(\frac{1}{2}\)at² = 2.736 × 3 + \(\frac{1}{2}\) × 9.8 × 9 = 8.208 + 44.1 = 52.3m
∴ height of the window = 52. 3m

The distance of the point where the hall strikes on the ground, from the base of building
= 8cos20° × 3 = 8 × 0.9397 × 3 = 7.5176 × 3 = 22.6 m

Question 10.
Two balls are projected from the same point in directions 30° and 60″ with respect to the horizontal. What is the ratio of their initial velocities if they (a) attain the same height? (b) have the same range? [AP 17]
Solution:
(a) Let ut and u₂ be their initial velocities.
Given, maximum height of 1^st body = maximum height of 2^nd body

Question 11.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown in 25.0m from the ground.(a) How high will the ball rise? (b)How long will it be before the ball hits the ground. Take g = 10 ms^-2 [Actual value of ‘g’ is 9.8 ms^-2] [AP 15; TS 15, 19]
Solution:

Question 12.
A car moving along a straight high way with speed of I26km/h is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop? [AP 20]
Solution:

Multiple Choice Questions

1. A car covers the first half of the distance between two places at 40 km/h and another half at 60 km/h. The average speed of the car is
1)40 km/h
2) 48 km/h
3) 50 km/h
4) 60 km/h
Answer:
2) 48 km/h

2. A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. The value of v is
1) 56 km/h
2) 60 km/h
3) 50 km/h
4) 48 km/h
Answer:
2) 60 km/h

3. A particle covers half of its total distance with speed v₁ and the next half distance with speed v₂. Its average speed during the complete journey is

Answer:
3

4. A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and at last one-third at 60 km/h. The average speed of the bus is
1) 9 km/h
2) 16 km/h
3) 18 km/h
4) 48 km/h.
Answer:
3) 18 km/h

5. A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is
1) 10 m/s, 0
2) 0, 0
3) 0, 10 m/s
4) 10 m/s, 10 m/s
Answer:
3) 0, 10 m/s

6. The distance travelled by a freely falling body in first, second and third second of its fall are in the ratio
1) 1:2:3
2) 1:3:5
3) 1:4:9
4) 1:1:1
Answer:
2) 1:3:5

7. In the presence of considerable resistance of air, a stone is thrown up. If the time of ascent is t, and that of descent is t2, then
1) t₁ > t₂
2) t₁ = 2t₂
3) t₁ < t₂
4) t₁ = t₂
Answer:
1) t₁ > t₂

8. From a place where g = 9.8ms^-2, a stone is thrown up with a velocity of 4.9ms^-1. The time taken for the stone to return to the earth is
1) 1 s
2) 1.5 s
3) 3 s
4) 2.5s
Answer:
1) 1 s

9. A body falling for 2s covers a distance S equals to that covered in next second. If g = 10ms^-2, S =
1) 30m
2) 10 m
3) 60 m
4) 20 m
Answer:
1) 30m

10. If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?
1) 1 cm
2) 2 cm
3) 3 cm
4) 4 cm
Answer:
1) 1 cm

11. A car moves along a straight line whose motion is given by S = 12t + 3t² – 2t³ where S is in metres and’t’ is in seconds. The velocity of the car at the start will be
1) 7 m/s
2) 9 m/s
3) 12 m/s
4) 16 m/s
Answer:
3) 12 m/s

12. A particle moves along a straight line such that its displacement at any time t is given by s = (t³ – 6t² + 3t + 4) metres. The velocity when the acceleration is zero is
1) 3 m/s
2) 42 m/s
3) -9 m/s
4) -15 m/s
Answer:
3) -9 m/s

13. The position x of a particle varies with time, (t) as x = at² – bt³. The acceleration will be zero at time t is equal to
1) a/3b
2) zero
3) 2a/3b
4) a/b
Answer:
1) a/3b

14. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of
1) 1440 cm
2) 2980 cm
3) 20 m
4) 400 m
Answer:
4) 400 m

15. A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s, then the maximum height attained by it (g = 10 m/s²)
1) 8m
2) 20 m
3) 10 m
4) 16 m
Answer:
3) 10 m

16. The velocity of train increases uniformly from 20 km/h to 60 km/ h in 4 hours. The distance travelled by the train during this period is [NEET Q]
1) 160 km
2) 180 km
3) 100 km
4) 120 km
Answer:
1) 160 km

17. The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms^-2, in the third second is f
1) 10/3m
2) 19/3 m
3) 6 m
4) 4 m
Answer:
1) 10/3m

18. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S₁ and that covered in the first 20 seconds is S₂, then
1) S₂ = 3S₁
2) S₂ = 4S₁
3) S₂ = S₁
4) S₂ = 2S₁
Answer:
2) S₂ = 4S₁

19. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms^-1 to 20 ms^-1 while passing through a distance 135 m in t second. The value of t is
1) 12
2) 9
3) 10
4) 1.8
Answer:
2) 9

20. A train of 150 metre length is going towards north direction at a speed of 10 m/s. A parrot flies at the speed of 5 m/s towards south direction parallel to the railways track. The time taken by the parrot to cross the train is
1) 12 s
2) 8 s
3) 15 s
4) 10 s
Answer:
4) 10 s

21. A body starts from rest, what is the ratio of the distance travelled by the body during the 4lh and 3rd second?
1) 7/5
2) 5/7
3) 7/3
4) 3/7
Answer:
1) 7/5

22. What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5,h seconds of journey?
1) 4:5
2) 7:9
3) 16:25
4) 1:1.
Answer:
2) 7:9

23. A boy standing at the top of a tower of 20m height drops a stone. Assuming g = 10 m s^-2, the velocity wish which it hits the ground is
1) 10.0 m/s
2) 20.0 m/s
3) 40.0 m/s
4) 5.0 m/s
Answer:
2) 20.0 m/s

24. A ball is thrown vertically upward. It has a speed of 10 m/sec when it has reached one half of its maximum height. How high does the ball rise? (Take g = 10 m/s²)
1) 10 m
2) 5m
3) 15 m
4) 20 m
Answer:
1) 10 m

25. Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is
1) 4/5
2) 5/4
3) 12/5
4) 5/12
Answer:
1) 4/5

Students get through AP Inter 1st Year Physics Important Questions 5th Lesson Laws of Motion which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 5th Lesson Laws of Motion

Very Short Answer Questions

Question 1.
What is inertia? What gives the measure of inertia? [AP 19][IPE’ 14][TS 17,19]
Answer:
Inertia :
Inertia is an inherent property of a body, by which it resists any change in its state. Mass is the measure of inertia.

Question 2.
According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place? [TS 19][AP 15,18]
Answer:
Action and reaction always take place on two different bodies. Hence, they never cancel each other. So movement of bodies is possible.

Question 3.
When a Bullet is fired from a Gun, the Gun gives a kick in fhe backward direction. Explain. [Imp.Q][AP 15]
Answer:
This is due to the ‘law of conservation of momentum’. When a Bullet is fired from a gun, the momentum of the Gun becomes equal and opposite to that of Bullet. So, when the Bullet moves forward, the Gun kicks back. This motion of Gun is called Recoil of the Gun.

Question 4.
Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridge? [Imp.Q]
Answer:
Recoil of the rifle V = \(\frac{mu}{M}\). Here, mass M of rifle
For a heavy rifle, mass M is more which is in the denominator. Hence, its recoil velocity V is less.

Question 5.
If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain. [TS 15,16,18,22]
Answer:
This is due to the ‘law of conservation of momentum’. Bomb at rest has zero linear momentum. So, when the bomb explodes into two pieces, the sum of their momenta must be equal to zero. So, the two pieces must have equal and opposite momenta. Therefore the pieces must travel in opposite directions.

Question 6.
Define force. What are the basic forces in nature? [Imp.Q]
Answer:
Force is that physical quantity which changes or tries to change the state of a body.
The basic forces in nature are

1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force.

Question 7.
Can the coefficient of friction be greater than one? [TS 18]
Answer:
Yes. In general coefficient of friction is less than one. If the surfaces are polished heavily then adhesive forces between the molecules increase and then coefficient of friction will be greater than one.

Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres? [AP 20][May 13]
Answer:
A flattened tyre gets more deformation. So rolling friction becomes more to that tyre. Hence a car with flattened tyre comes to rest sooner than; car with inflated tyres.

Question 9.
A horse has to exert a greater force during the start of the motion than later. Explain. [AP 16,18,22]
Answer:
Before starting the motior, Horse has static friction and later it will be converted into kinetic friction. Bui, Static friction is more than kinetic friction Hence, Horse has to exert a greater force during the start of the motion than later.

Question 10.
What happens to the coefficient of friction if weight of the body is doubled. [TS’ 19,22] [AP 16,19]
Answer:
Coefficient of friction is independent of weight of the body. So it remains constant.

Short Answer Questions

Question 1.
A stone of mass 0.1kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone (a) during its upward motion (b) during its downward motion (c) at the highest point, where it momentarily comes to rest.
Answer:
The acceleration due to gravity (g = 9.8ms^-2) is always downward whether the body is rising or falling or it is momentarily at rest at maximum height. Therefore, the net force on the stone in all the three cases is vertically downwards.
m = 0.1kg, g = 9.8 ms^-2.
∴ F = mg = 0.1 × 9.8 = 0.98 N vertically downward

Question 2.
Define the terms momentum and impulse. State and explain the law of conservation of momentum. Give example. [TS 15,18J[AP 20|
Answer:
Momentum :
The product of mass and velocity of a body is called its momentum.
If m is the mass of a body moving with a velocity v then its momentum, p = mv.

Impulse :
The product of the force and time of action of the force that produces a finite change in momentum of the body is called “Impulse”.
I = ∫ Fdt
i.e., Impulse = F_avg × ∆t

Law of conservation of Momentum :
When there is ‘no resultant external force’, the total momentum of all the interacting bodies in a system, remains constant.
In other words, Total momentum before collision = Total momentum after collision

Thus, Total momentum before collision = Total momentum after collision

Ex 1. Explosion of bomb :
When a stationary bomb explodes into two pieces, then according to law of conservation of linear momentum,
Total momentum before explosion = Total momentum after explosion
⇒ 0 = m₁v₁ + m₂v₂ ⇒ m₁v₁ = – m₂v₂
Thus, the momenta of the two pieces after explosion are equal and opposite.

Ex 2. Motion of a Rocket and a Jet plane is according to the law of conservation of momentum.

Question 3.
Why are shock absorbers used in motorcycles and cars?
Answer:
When vehicles move on uneven roads, shock absorbers increase the time of impulse. As a result, impulsive force decreases. Hence, passengers get better comfort.

Question 4.
Explain the terms limiting friction, dynamic friction, and rolling friction. [Imp.Q]
Answer:
1) Limiting friction (f_L) :
It is the maximum value of static friction between the contact surfaces when the body is just ready to slide over a surface.

2) Kinetic/Sliding/dynamic friction (f_k) :
The resistive force encountered by a sliding body on the surface is known as kinetic friction (or) sliding friction.

3) Rolling friction (f_r) :
The resistive force encountered by a rolling body on the surface is known as rolling friction.

Question 5.
Explain the advantages and disadvantages of friction [AP 15,19][TS 15, 17,22]
Answer:
Advantages :

1. We cannot walk without friction between feet and ground.
2. ‘Brakes’ are able to stop the vehicles only due to friction between brake shoes and the inner surface of the brake drum.
3. We are able to pick up a Book only because of friction between Hand and the Book.
4. In a Machine, power is transmitted from the Motor to different parts of the Machine only by using friction between the Belt and the Wheels.

Disadvantages :

1. A part of the energy delivered by the Motor is wasted to overcome friction.
2. Friction between different parts of the Machine produces uneven wear and tear’.

Question 6.
Mention the methods used to decrease friction. [AP 18,22] [TS 16,19]
Answer:
1) Polishing :
By polishing the surfaces, frictional force can be reduced.

2) Using Lubricants :
A lubricant forms a thin layer between two surfaces in contact and it reduces the friction.

3) Using ball bearings :
The wheels of motor vehicles, cycles and shafts of motors, dynamos are provided with ball bearings to reduce friction.

4) Stream lining :
Aeroplanes and Auto mobiles have special construction to reduce friction due to air.

Question 7.
State the laws of rolling friction. [TS 20,22][Imp.Q]
Answer:
1) Rolling friction (fr) is directly proportional to the normal reaction (N)
i.e., f_r ∝ N

Rolling friction (fr is inversely proportional to the radius(r) of the rolling wheel.
i.e., f_r ∝ \(\frac{1}{r}\)
∴ Rolling friction, f_r = \(\frac{\mu_r N}{r}\)
where µ_R = coefficient of rolling friction.

Question 8.
Why is pulling the lawn roller preferred to pushing it? [lmp,Q]
Answer:
Suppose a lawn roller is pulled by a force F making an angle ‘θ’ with the horizontal.

This force can be resolved into two components Fcosθ and Fsinθ.
Total upward force = Total downward force
N + Fsinθ = W ⇒ N = W – Fsinθ
∴ Frictional force f₁ = µN = µi(W-Fsinθ) ………. (1)

Suppose the roller is pushed by the same force F, making angle θ with the horizontal.

Here, Force can be resolved into two components Fcosθ and Fsinθ.
Total upward force = Total downward force
N = W + Fsinθ
∴ Frictional force f₂ = µN = µ(W + Fsinθ) ………. (2)
From (1) & (2), it is clear that f₁ < f₂.
Hence pulling is preferred than pushing.

Long Answer Questions

Question 1.
(a) State Newton’s Ilnd law of motion. Hence derive the equation of motion F = ma.
(b) A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body? [TS 18][Mar 13, May 13][AP 16,17,19]
Answer:
(a) Newton’s Second Law :
“The ‘rate of change of momentum of a body is directly proportional to the external force acting on the body and it is in the direction, in which the force acts”.
Thus, \(\frac{dp}{dt}\) ∝ F

Derivation of F = ma
Let m = mass of the body, v = velocity of the body,
∴ Linear momentum, p = mv
Let F = external force applied on the body in the direction of motion of the body.

Here, k can be made equal to 1, by properly selecting the units of F, m, a.
Then, F = ma.

(b) Yes, there should be a force acting on it.
Reason :
When a body is moving along a circular path, the direction of velocity at a point is along the tangent at that point. So the direction of velocity continuously changes from point to point. So the body has acceleration which is directed towards the centre of the circle. This acceleration is called centripetal acceleration. As there is acceleration, there must be a force acting on the body towards the centre of the circle. This force is called centripetal force.

Question 2.
Define angle of friction and angle of repose. Show that angle of friction is equal to angle of repose for a rough inclined plane.
A block of mass 4kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30N is applied on it. If g = 10ms^-2, find the total contact force exerted by the plane on the
Answer:

Angle of friction :
The angle which the resultant of force of limiting friction f_L and the normal reaction N makes with the normal reaction N, is called angle of friction.

From the figure, tan θ =\(\frac{AD}{OA}=\frac{f_L}{N}\) = µ_s …………. (1)

Angle of repose :
The angle of repose is the angle which an inclined plane makes with the horizontal when a body placed over it just starts sliding down.
1) Let us consider a body of weight w placed on an inclined plane OA, whose inclination Φ with the horizontal can be changed. The weight of the body w acts vertically downwards through the centre of gravity of the body.

2) Increase the angle of inclination very slowly till the body just begins to slide down the plane. This particular angle of inclination is called the” angle of sliding(or)angle of repose”, represented by Φ. The value of Φ depends upon the material and the nature of the two surfaces in contact.

3) The weight w of the body can be resolved into two components (a) wcosΦ acting perpendicular to the plane and (b) wsinΦ acting parallel to the plane. The component wcosΦ balances the normal reaction N, while the component wsinΦ is equal to the limiting friction f_L.

Hence the tangent of angle of repose is equal to coefficient of static friction.
But µ_s = tanθ, where θ is the angle of friction
∴ tanθ = tanΦ ⇒ θ = Φ
Thus angle of friction is equal to angle of repose.

Problem:

Given m = 4kg and g = 10ms^-2.
∴ weight of the body, w = mg = 4 × 10
= 40 N = normal reaction (N)

Maximum static frictional force, f_L = F = 3 ON
The total contact force (F) on the block by the plane is equal to resultant of N and f_L.

Solved Problems

Question 1.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100ms^-2. What is the acceleration of the astronaut at the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him).
Solution:
Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.

Question 2.
A bullet of mass 0.04kg moving with a speed of 90 ms-1 enters a heavy wooden block and is stopped after a distance of 60cm. What is the average resistive force exerted by the block on the bullet?
Solution:
The retardation ‘a’ of the bullet (assumed constant) is given by

The retarding force, by the second law of motion, is 0.04kg × 6750 ms^-2 = 270N
The actual resistive force and therefore retardation of the bullet may not be uniform.
The answer, therefore, only indicates the average resistive force.

Question 3.
The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\)gt². Find the force acting on the particle.
Solution:
We know y = ut + \(\frac{1}{2}\)gt² Now, v = \(\frac{dy}{dt}\) = u + gt, acceleration, a = \(\frac{dv}{dt}\) = g
Then the force is given by F = ma = mg

Question 4.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 mr1. If the mass of the ball is 0.15kg, determine the impulse im¬parted to the ball. (Assume linear motion of the ball) [ AP 17][TS 19,20]
Solution:
Change in momentum = (0.15 × 12) – (-0.15 × 12) = 3.6 Ns
Impulse = 3.6 Ns in the direction from the batsman to the bowler.

Question 5.
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.
Solution:
The acceleration of the box is due to the static friction.
∴ a_max = µ_sg = 0.15 × 10ms^-2 = 1.5 ms^-2

Question 6.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
Solution:
The condition for the cyclist not to slip is given by v² ≤ µ_sRg ………… (1)
Now, v = 18 km/h = 5 ms^-1 ⇒ v² = 25 m²s^-2.
Also, R = 3m, g = 9.8 ms^-2, µ_s = 0.1 ⇒ µ_sRg = 2.94 m²s^-2.
The condition (1) is not obeyed. The cyclist will slip while taking the circular turn.

Exercise Problems

Question 1.
The linear momentum of a particle as a function of time t is given by p = a + bt, where ‘a’ and ‘b’ are positive constants. What is the force acting on the particle?
Solution:
Given linear momentum, p = a + bt
Force, F = ?
Force = rate of change of momentum
F = \(\frac{dp}{dt}\) = \(\frac{d}{dt}\)(a + bt) = b
∴ Force on the body = b

Question 2.
Calculate the time needed for a net force of 5N to change the velocity of a 10kg mass by 2 m/s. [TS 16]
Solution:
Given net force on the body, F = 5N
mass of the body, m = 10kg
change in velocity, v – u = 2 ms^-1
time t = ?

Question 3.
A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity, what is the impulse received by the ball due to gravity force during its flight? (neglect air resistance)
Solution:
Ball reaches the maximum height h. So its initial velocity of projection will be u = \(\sqrt{2gh}\)
Initial momentum, Pj = mu = m \(\sqrt{2gh}\) (in the upward direction)
When it reaches the ground, it will have the same velocity, u, but in opposite direction. Hence final momentum, P_f = – mu = – m\(\sqrt{2gh}\)
∴ Magnitude of change in momentum = P_i – P_f
= m\(\sqrt{2gh}\) – (-m\(\sqrt{2gh}\)) = 2m\(\sqrt{2gh}\) = \(\sqrt{8m^2gh}\)
∴ Impulse received by the ball due to gravity force
= change in momentum = \(\sqrt{8m^2gh}\)

Question 4.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
Given mass of the body, m = 3kg
Initial velocity, u = 2 ms^-1
Final velocity, v = 3.5 ms^-1
Time t = 25 sec
Force F = ?

Since velocity of the body is increasing, the direction of force is in the direction of motion of the body.

Question 5.
A man in a lift feels an apparent weight VV when the lift is moving up with a uniform acceleration of 1/3_rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight?
Solution:
Let mg be the true weight of the man. When the lift is moving upwards with an acceleration ’a’ then the apparent weight will be mg+ma
In the first case, a = \(\frac{g}{3}\) and apparent weight is w ⇒ w = mg \(\frac{mg}{3}=\frac{4mg}{3}\) ……… (1)

When the lift is moving downwards with an acceleration a then the apparent weight is mg – ma
Given a = g/2.

Question 6.
A container of mass 200kg rests on the back of an open truck. If the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?
Solution:
The box is accelerating along with the truck.
Therefore, force on the box is F = ma = 200 × 1.5 = 300 N
Since the truck is accelerating in the forward direction the reaction of the container is equal to F = 300 N, but acts in the backward direction.
The backward motion of the container is opposed by the limiting friction between the container and the floor of the truck.

Question 7.
A bomb initially at rest at a height of 40m above the ground suddenly explodes in to two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10ms², what is the separation between the fragments 2s after the explosion?
Solution:
The bomb has exploded into two pieces and one piece is moving vertically downwards with a velocity of 10ms^-1. So, according to law of conservation of momentum, the other piece of same mass should travel in the upward direction with a velocity of 10ms^-1.
For that piece, time of ascent (ta) = \(\frac{u}{g}=\frac{10}{10}\) = 1 sec
So in 2sec, it will reach its original position, i.e., it will be at a height of 40m from the ground. The piece which is moving downwards has initial velocity (u) = 10ms^-1
acceleration (a) = g = 10ms^-2
time (t) = 2sec
distance travelled (s) = ?

Now, s = ut + \(\frac{1}{2}\)at² ⇒ s = 10 × 2 + \(\frac{1}{2}\) × 10 × 4 = 20 + 20 = 40m
∴ Distance of separation between the two pieces = 40m

Question 8.
A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3kg on the other side. Another 3kg is hung from the other 3kg as shown with another light string. If the system is released from rest, find the common acceleration? (g = 10ms²)

Solution:
When the system is released, A will move upwards and B and C move downwards with common acceleration. Let it be ‘a’.

From eq (2); T₁ + 30 – 40 – 4a = 3a ⇒ T₁ – 4a = 3a + 10
⇒ T₁ – 7a = 10 ……… (4)
From eq (3) T₁ = 30 – 3a
Substituting the value of T| in equaiton (4); T₁ – 7a = 10
⇒ 30 – 3a – 7a = 10
⇒ 30 – 10a = 10 ⇒ 10a = 20 ⇒ a = 2ms^-2.

Question 9.
A block of mass of 2kg slides on an inclined plane that makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √3/2.
(a) What force should be applied to the block so that it moves down without any acceleration?
(b) What force should be applied to the block so that it moves up without any acceleration?
Solution:

(a) mg sin θ = 2 × 9.8 × sin 30° = 2 × 9.8 × \(\frac{1}{2}\) = 9.8N
Maximum static friction, f_L = µ_sN = µ_smg cosθ
= \(\frac{\sqrt{3}}{2}\) × 2 × 9.8 × cos 30° = \(\frac{\sqrt{3}}{2}\) × 2 × 9.8 × \(\frac{\sqrt{3}}{2}\) = 3 × 4.9 = 14.7N

If F₁ is the force to be applied on the body so that it moves downwards without any acceleration, then
F₁ + mg sin θ = f_L
⇒ F₁ + 9.8 = 14.7
⇒ F₁ = 4.9 N

(b) In this case, the frictional force f_L and mg sinθ act downwards. So the minimum force F required to move the body upwards without acceleration is
F = mg sinθ + f_L = 9.8 + 14.7 = 24.5 N

Question 10.
A block is placed on a ramp of parabolic shape given by the equation y = x²/20. If µ_s = 0.5, what is the maximum height above the ground at which the block can be placed without slipping?
Solution:
Given that µ_s = 0.5

Question 11.
A block of metal of mass 2kg on a horizontal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fail. The coefficient of sliding friction between the block and table is 0.2.

Calculate (a) the initial acceleration (b) the tension in the string (c) the distance the block would continue to move if, after 2s of motion, the string should break.
Solution:
Let M₁ = 2kg, M₂ = 0.5kg
Given that µ = 0.2
(a) For M₁, T – µM₁g = M₁a ……(1)
For M₂, M₂g – T = M₂a ……….. (2)

⇒ T = f + m₂a = |µmg + m₂a = 0.2 × 2 × 9.8 + 2(0.2) = 4.32N

(c) We have u=0, a = 0.2 m/s², t = 2s
The velocity of the block before the string breaks is given by v = u + at
⇒ v = 0 + 0.2 × 2 = 0.4 m/s²
After break if the block travels a distance s then workdone = change in K.E

Question 12.
One a smooth horizontal surface a block A of mass !0kg is kept. On this block a second block B of mass 5kg is kept. The coefficient of friction between the tw o blocks is 0.4. A horizontal force of 30N is applied on the lower block as shown. What is the force of friction between the blocks?

Solution:
The maximum force that can be applied on the lower block so that the two blocks move together is
F_max = µ_sg(m_A + m_B) = 0.4 × 10 (10 + 5) = 4 × 15 = 60N
Since the applied force F = 30N < F_max the blocks move together with a common acceleration ‘a’.

Frictional force between the blocks f = m_Ba = 5 × 2 = 10N

Multiple Choice Questions

1. Gravel is dropped into a conveyor belt at rate of 0.5 kg/sec. The extra force required in newtons, to keep the belt in moving at 2m/s is
1) 1
2) 2
3) 4
4) 5
Answer:
1) 1

2. A body of 50 kg in lift moving down with an acceleration of 9.8ms^-2. The apparent weight of the body is (g = 9.8ms^-2)
1) 509.8N
2) 0
3) 50 N
4) \(\frac{50}{9.8}\) N
Answer:
2) 0

3. A machine gun fires 20 bullets per second into a target. Each bullet weighs 150g and has a speed of 800 m/s. The force necessary to hold the gun is
1) 800 N
2) 1000 N
3) 1200 N
4) 2400N
Answer:
4) 2400N

4. A body of mass 2kg is acted upon by two forces each of magnitude 1 newton making an angle of 60″ with each other. The net acceleration of the body is
1) 0.5
2) 1.0
3) √3/2
4) √2/3
Answer:
3) √3/2

5. A man of mass 20kg standing in an elevator whose cable broke suddenly. If the elevator falls freely, the force exerted by the floor of the elevator on the man is
1) 20 N
2) 10 N
3) 0
4) 6 N
Answer:
3) 0

6. A gun fires a bullet of mass 50gm with a velocity of 30 m/s. This gun is pushed back with a velocity of Im/s, then mass of the gun is
1) 3.5 kg
2) 30 kg
3) 1.5 kg
4) 20 kg
Answer:
3) 1.5 kg

7. A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity the gun rebounds backward?
1) 1 m/s
2) 0.01 m/s
3) 0.1 m/s
4) 10 m/s
Answer:
1) 1 m/s

8. Physical independence of force is a consequence of
1) third law’ of motion
2) second law of motion
3) first law of motion
4) all of these laws
Answer:
3) first law of motion

9. A shell, in flight, explodes into four unequal parts. Which of the following is conserved?
1) Potential energy
2) Momentum
3) Kinetic energy
4) Both (1) and (3).
Answer:
2) Momentum

10. A 10 N force is applied on a body produce in it an acceleration of 1 m/s². The mass of the body is
1) 15 kg
2) 20 kg
3) 10 kg
4) 5 kg
Answer:
3) Kinetic energy

11. A body, under the action of a force F = \(6\hat{i}-8\hat{j}+10\hat{k}\) acquires an acceleration of 1 m/s². The mass of this body must be
1) 10 kg
2) 20 kg
3) 10√2 kg
4) 2√10 kg
Answer:
3) 10√2 kg

12. A force vector applied on a mass is represented as \(\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}\) and accelerates with 1 m/s². What will be the mass of the body?
1) 10 kg
2) 20 kg
3) 10√2 kE
4) 2√10 kg
Answer:
3) 10√2 kE

13. A cricketer catches a ball of mass 150 gm in 0.1 sec moving with speed 20 m/s, then he experiences force of
1) 300 N
2) 30 N
3) 3 N
4) 0.3 N
Answer:
2) 30 N

14. A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is
1) MV
2) 1.5MV
3) 2MV
4) zero
Answer:
3) 2MV

15. A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the adjacent figure. The value of impulse imparted by the wall on the ball will be

1) mV
2) 2 mV
3) mV/2
4) mV/3
Answer:
2) 2 mV

16. A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly
1) 1.4 kg m/s
2) 0 kg m/s
3) 4.2 kg m/s
4) 2.1 kg m/s
Answer:
3) 4.2 kg m/s

17. The force F acting on a particle of mass m is indicated by the force-time graph as shown.

The change in momentum of the particle over the time interval from zero to 8 s is
1) 24 Ns
2) 20 Ns
3) 12 Ns
4) 6 N s
Answer:
3) 12 Ns

18. A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by
1) 68%
2) 41%
3) 200%
4) 100%
Answer:
2) 41%

19. A mass of 1 kg is thrown up with a velocity 100m/s. After 5 seconds, it explodes into two parts. One part of mass 400 g comes down with a velocity 25 m/s. The velocity of other part is (Take g = 10 ms^-2)
1) 40 m/s
2) 80 m/s
3) 100 m/s
4) 60 m/s
Answer:
3) 100 m/s

20. A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km/hr. The centripetal force is
1) 1000 N
2) 750 N
3) 250 N
4) 1200N
Answer:
1) 1000 N

21. A block slides down a rough inclined plane of inclination 45″. If coefficient of kinetic friction is 0.5, then acceleration of the sliding block is

Answer:
1

22. A body of weight 50 N is placed on a surface. If the force required to move the body on the surface is 30N, the coefficient of friction is
1) 0.6
2) 1.2
3) 0.3
4) 1.67
Answer:
1) 0.6

23. A particle is projected up along a rough inclined plane of inclination 45° with the horizontal. If the coefficient of friction is 0.5, the retardation is

Answer:
3

24. A car of mass rn is moving on a level circular track of radius R. If u represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

Answer:
4

25. A conveyor belt is moving at a constant speed of 2ms^-1. A box is gently dropped on it. The coefficient of friction between them is p = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 ms^-2, is
1) 0.4 m
2) 1.2 m
3) 0.6 m
4) zero
Answer:
1) 0.4 m

26. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a friction less road. If the banking angle is 45°, the speed of the car is
1) 20ms^-1
2) 30 ms^-1
3) 5 ms^-1
4) 10 ms^-1
Answer:
2) 30 ms^-1

27. A car is moving in a circular horizontal track radius 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1,0 m. The angle made by the wire with the vertical is
1) π/3
2) π/6
3) π/4
4) 0°
Answer:
3) π/4

28. A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between
1) 16 m/s and 17 m/s
2) 13 m/s and 14 m/s
3) 14 m/s and 15 m/s
4) 15 m/s and 16 m/s
Answer:
3) 14 m/s and 15 m/s

29. When milk is churned, cream gets separated due to
1) centripetal force
2) centrifugal force
3) frictional force
4) gravitational force
Answer:
2) centrifugal force

30. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let S_n be the distance travelled by the block in the interval t = n – 1 to t = n. Then, the ratio \(\frac{S_n}{S_{n+1}}\)

Answer:
3

31. A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s². What would be the reading on the scale? (g = 10 m/s²)
1) zero
2) 400 N
3) 800 N
4) 1200 N
Answer:
4) 1200 N

Students get through AP Inter 1st Year Physics Important Questions 2nd Lesson Units and Measurements which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 2nd Lesson Units and Measurements

Very Short Answer Questions

Question 1.
Distinguish between Accuracy and Precision. [Mar 13, May 13, 11; AP, TS 15, 16,18]
Answer:

Accuracy	Precision
1) Accuracy denotes how close by a quantity can be measured near to the True value.	1) Precision denotes how precisely (sharply) a quantity can be measured.
2) It varies with Errors in the measurements	2) It varies with least count of the measuring instruments

Question 2.
What are the different types of errors that can occur in a measurement?
Answer:
Types of Errors:

1. Systematic errors
2. Random errors

Systematic errors are further classified as:
(a) Instrumental errors
(b) Personal errors
(c) Imperfection in experimental technique.

Question 3.
How can systematic errors be minimised or eliminated? [IPE’ 14; AP,TS 17]
Answer:
The Systematic errors can be minimized by

1. selecting ‘better instruments’ with ‘higher resolution’
2. avoiding personal bias in taking reading.
3. improving the ’experimental techniques’.

Question 4.
Illustrate how the result of a measurement is to be reported indicating the error involved.
Answer:
1) Suppose the values obtained in several measurements of a quantity are a₁, a₂, a₃, …..a_n.
Calculate their arithmetic mean amean as follows:

2) Now Calculate absolute errors in the measurements as shown below|
|∆a₁| = |a_mean – a₁|, |∆a₂| = |a_mean – a₂|,……|∆a_n| = |a_mean – a_n|
3) Next find mean absolute error as follows:

The final result of measuring the physical quantity is given in the form ∆a_mean =(a_mean ± ∆a_mean)

Question 5.
What are significant figures and what do they represent when reporting the result of a measurement? [TS 18]
Answer:
Significant figures:
The digits of a number that are ‘known reliably’, plus ‘one uncertain digit’, are called “significant figures”.
The significant figures indicate the precision of measurement.

Question 6.
Distinguish between fundamental units and derived units. [IPE’ 14; TS 16,22]
Answer:

Fundamental Units	Derived Units
1) The units of the fundamental quantities are called fundamental units. Ex: metre, kilogram, second.	1) The units of the derived quantities are called derived units. Ex: ms-1, newton.
2) There are 7 fundamental quantities and hence 7 types of fundamental units exist.	2) There are several derived quantities and hence several derived units exist.

Question 7.
Why do we have different units for the same physical quantity? [TS 15, 16, 22]
Answer:
A physical quantity possesses a wide range of magnitudes.

For example, Mass ranges from 10^-30 kg (for an electron) to 10⁵³ kg (for the known universe). Hence, we need different units for different ranges of the same physical quantity.
Ex: mg, gm, kg etc., for mass ; mm, cm, m, km for length

Question 8.
What is dimensional analysis?
Answer:
Dimensional analysis is the process of analysing a physical problem with the help of dimensional equations.

Question 9.
How many orders of magnitude greater is the radius of the atom as compared to that of the nucleus?
Answer:
Radius of atom is of the order of 10^-10 m
Radius of nucleus is of the order of 10^-4 m
So radius of atom is 10⁴ times greater than that of nucleus.
Hence Radius of atom is 4 orders greater than that of nucleus.

Question 10.
Express unified atomic mass unit in Kg. [TS 19, 22]
Answer:
1 unified atomic unit (u) = 1.67 × 10^-27 kg

Short Answer Questions

Question 1.
The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance?
Answer:
Minimum inaccuracy = least count of vernier callipers

Question 2.
In a system of units, the unit of force is 100N, unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system?
Answer:

Question 3.
The distance of a galaxy from Earth is of the order of 10²⁵m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:

∴ The time taken by light to reach us from the galaxy is in the order of 1016 s with order of magnitude 16 (As 3.33 lies between 0 and 10)

Question 4.
The Earth-Moon distance is about 6(1 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:
Imagine 2R_E as the arc of a circle of radius 60R_E.
Then length of arc = radius x angle subtended by the arc in radians.
⇒ 2R_E = 60 R_E × θ

Question 5.
Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6s, t₂ = 39.9s and t₃ = 39.5s. What is the precision in the measurements? What is the accuracy of the measurements?
Answer:
The precision denotes upto what resolution (number of decimal places) the quantity can be measured using the given instrument.

From the given measurements , it is clear that the precision is the least count of watch i.e., 0.1 s
Average time for the 3 measurements = \(\frac{39.6+39.9+39.5}{3}\) = 39.66 s = 39.7 s
Accuracy is the closeness of measured value with true value.
Hence 39.6 s is the accurate of the given 3 measurements.

Question 6.
1 calorie = 4.2 J where 1J = 1 m²s^-2. Suppose we employ a system of units of mass is a kg, the unit of length is \(\hat{a}\) m and the unit of time is \(\hat{a}\) s. Show that a calorie has a magnitude 4.2 \(\dot{a}^{-1} \hat{\mathrm{a}}^{-2} \tilde{\mathrm{a}}^2\) in the new system.
Answer:
Calorie is unit of heat energy. The dimensional formula for energy is ML²T^-2

In S.I system:	New’ system:
Unit of mass, [M1] = 1kg	Unit of mass, [M2] = \(\dot{a}\) kg
Unit of length, [L1] = lm	Unit of length, [L2] = \(\hat{\mathrm{a}}\) m
Unit of time, [T1] = 1 s	Unit of time, [T2] = \(\tilde{\mathrm{a}}\)s
Numerical value, n1 = 4.2	Numerical value, n2 = ?

Question 7.
A new unit of length is chosen so that the speed of light in vacuum is 1 new unit of length /see. If light takes 8 min and 20 s to cover the distance between Sun and Earth, what is the distance between the Sun and Earth in terms of the new unit?
Answer:

Question 8.
A student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5mm. What is the estimate of the thickness of hair?
Answer:

Question 9.
A physical quantity X is related to four measurable quantities a,b,c and d as follows: x = a² b³c^5/2d^-2. The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X?
Answer:
Given x = a² b³c^5/2d^-2

Question 10.
The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in Si what are the units of A, B and C?
Answer:
From the principle of homogeneity of dimensions, the units of V, At², Bt, C are equal.
∴ SI unit of At² = SI unit of V ⇒ A s² = m/s ⇒ unit of A = m/s³
SI unit of Bt = SI unit of V ⇒ Bs = m/s ⇒ unit of B = m/s²
SI unit of C = SI unit of V ⇒ unit of C = m/s

Solved Problems

Question 1.
The moon is observed from two diametrically opposite points A and B on Earth. The angle G subtended at the moon by the two directions of observation is 1°54′. Given, the diameter of the Earth to be about 1.276 × 10⁷ m. Compute the distance of the moon from the Earth.

Solution:
We have θ = 1°54′ = 60′ + 54′ = 114′ [∵ 1° = 60′]
= (114 × 60)” × (4.85 × 10^-6) rad [∵ 1″ = 4.85 × 10^-6 rad]
Given that AB = 1.276 × 10⁷m
We have AB = (d)(θ) [∵ Arc length= Radius × Angle]

Question 2.
The Sun’s angular diameter is measured to be 1920″. The distance d of the Sun from the Earth is 1.496 × 10¹¹ m. What is the diameter of the Sun?

Solution:
Sun’s angular diameter α = 1920″
= 1920 × 4.85 × 10^-6 rad = 9.31 × 10^-3 rad
Sun’s diameter D = (d)(α) [∵ Arc length= Radius × Angle]
= (9.31 × 10^-3) × (1.496 × 10¹¹)m = 1.39 × 10⁹ m

Question 3.
If the size of a nucleus (in the range of 10^-15 to 10^-14 m) is scaled up to the tip of a sharp pin. What roughly is the size of an atom? Assume tip of the pin to be in the range 10^-5m to 10^-4 m.
Solution:
The size of a nucleus is in the range of 10^-15 m and 10^-14m. The tip of a sharp pin is taken to be in the range of 10^-5 m and 10^-4 m. Thus we are scaling up by a factor of 10¹⁰. Ar. atom roughly of size 10^-10 m will be scaled up to a size of lm. Thus a nucleus in an atom is as small in size, the tip of a sharp pin placed at the centre of a sphere of radius about a meter long.

Question 4.
The resistance R = V/I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R.
Solution:

Relative error in R = 0.07
Percentage error in R = 0.7 × 100 = 7%

Question 5.
The period of oscillation of a simple pendulum is T = 2π\(\sqrt{\frac{L}{g}}\). Measured value of L is 20.0cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of Is resolution. What is the accuracy in the determination of g?
Solution:

Question 6.
5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Solution:
There are 3 significant figures in the measured mass 5.74 g, whereas there are only 2 significant figures in the measured volume 1.2cm³. Hence the density should be expressed to only 2 significant figures.
Density = \(\frac{5.74}{1.2}\)gcm^-3 = 4.8 gem^-3

Question 7.
Let us consider an equation \(\frac{1}{2}\)mv² = nigh, where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct. .
Solution:
The dimensions of LHS are [M][LT^-1]² = [M][L²T^-2] = [ML²T^-2]
The dimensions of RHS are [M][LT^-2][L] = [M][L²T^-2] = [ML²T^-2]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

Question 8.
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (I), mass of the bob (m) acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Solution:
We have T = kl^xg^ym^z. By considering dimensions on both sides, we have
[L°M°T¹] = [L¹]^x[L¹T^-2]^y[M¹]^z = L^x+yT^-2yM^z.
On equating the dimensions on both sides, we have x + y = 0; -2y = 1; z = 0
Hence, x = 1/2, y = -1/2, z = 0.
Then T = kl^1/2 g^-1/2 ⇒ T = k\(\sqrt{\frac{l}{g}}\). The value of k is practically found to be 2π.

Exercise Problems

Question 1.
In the expression P = El²m^-5G^-2 the quantities E, l, m and G denote energy, angular momentum, mass and gravitational constant respectively. Show that P is a dimensionless quantity.
Solution:
The D.F of energy, E is ML²T^-2.
The D.F of angular momentum, l is ML²T^-1.
The D.F of mass, m is M
The D.F of universal gravitational constant, G is M^-1L³T^-2.
ρ = El² m^-5G^-2
= [ML²T^-2] [ML²T^-1]² [M]^-5 [ M^-1L³T^-2 ]^-2
= ML²T^-2 × M²L⁴T^-2 × M^-5 × M²L^-6T⁴ = M^1+2-5+2L^2+4-6T^-2-2+4 = M°L°T°
Hence, P = El²m^-5G^-2 is a dimensionless quantity.

Question 2.
If the velocity of light c, Planck’s constant h and the gravitational constant G are taken as fundamental quantities; express mass, length and time in terms of dimensions of these quantities.
Solution:
Let mass [M] = c^xh^yG^z ⇒ M¹L°T° = [LT^-1]^x [ML²T^-1]^y[M^-1L³T^-2]^z
⇒ M¹L°T° = L^xT^-x × M^yL^2yT^-y × M^-zL^3zT^-2z
⇒ M¹L°T° = M^y-zL^x+2y+3zT^-x-y-2z
y – z = 1 ………. (1); x + 2y + 3z = 0 ……… (2); -x – y – 2z = 0 ………. (3)

Question 3.
An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite T = \({\frac{k}{R}}\sqrt{\frac{r^3}{g}}\) where k is a dimensionless constant and g is acceleration due to gravity.
Solution:
The dependence of time period (T) of satellite on the quantities r, G, and M as a product mass be written as T = kr^aG^bM^c Where k is dimensionless constant By considering dimensions on both sides
[M°L°T¹] = [L]^a [M^-1L3T^-2]^b [M]^c ⇒ [M°L°T¹] = [L]^a [M^-bL^3bT^-2b] [M]^c
⇒ [M°L°T¹]= [M^-b+c L^a+3b T^-2b] on equating dimensions on both sides, we have -b + c = 0,

Question 4.
Stale the number of significant figures in the following:
(a) 6729 (b) 0.024 (c) 0.08240 (d) 6.032 (e) 4.57 × 10⁸.
Solution:
(a) 6729: The number of significant figures is 4
Reason: All the non-zero digits are significant

(b) 0.024: The number of significant figures is 2. They are 2 and 4
Reason: If the number is less than 1, the zero(s) on the right of decimal point but to the left of the first non-zero digit are not significant.

(c) 0.08240: The number of significant figures is 4. They are 8,2,4, and 0
Reason: The trailing zero(s) in a number with a decimal point are significant.

(d) 6.032: The number of significant figures is 4. They are 6,0,3 and 2.
Reason: All the zeroes between two non-zero digits are significant, no matter where the decimal point is, if at all.

(e) 4.57 × 10⁸: The number of significant figures is 3. They are 4,5 and 7.
Reason: The power of 10 (here it is 8) is irrelevant to the determination of significant figures.

Question 5.
A stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end to end what is the total length? If the two sticks are placed side by side, what is the difference in their length?
Solution:
Let l₁ = 12.132 cm and l₂ = 12.4 cm.
Total length when placed end to end = l₁ + l₂ = 12.132 + 12.4 = 24.532 cm = 24.5 cm
When placed side by side, difference in lengths = l₂ – l₁ = 12.4 – 12.132 = 0.268 = 0.3 cm

Reason :
In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places.

Question 6.
Each side of a cube is measured to be 7.203m. What is (i) the total surface area and (ii) the volume of the cube, to appropriate significant figures?
Solution:
Side of a cube, s=7.203 m
Total surface area of cube, A = 6s² = 6(7.203)² = 6×51.883209 = 311.299254 m² = 311.3 m²
The answer should contain only 4 significant digits as side of the cube has 4 significant figures Volume of the cube, V = s³ = (7.203)3 = 373.7147544m³ = 373.7 m³.

Question 7.
The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01g and 0.1cm³. Find the maximum error in density.
Solution:
Measured mass, m = 2.42 g
Measured volume, V = 4.7 cm³
Error in mass, ∆m = 0.01 g
Error in volume, ∆V = 0.1 cm³

Percentage error in density = Maximum error × 100 = 0.0254 × 100 = 2.54
In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures.

Question 8.
The error in measurement of radius of a .sphere is 1%. What is the error in the measurement of volume? [AP 19, 19]
Solution:

∴ Error in the measurement of volume is 3%.

Question 9.
The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities? [TS 20] [AP 18,20]
Solution:

Question 10.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 (molar volume). If the size of the hydrogen molecule is about 1A. What is the ratio of molar volume to the atomic volume of the mole of hydrogen?
Solution:
Diameter of hydrogen molecule = 1Å = 1 × 10^-10 m

Question 11.
Find the relative error in Z if Z = \(\frac{A^4 B^{1 / 3}}{C D^{3 / 2}}\) [TS May 19]
Solution:

Multiple Choice Questions

1. The S.l unit of Moment of Inertia is
1 ) Kg/m²
2) Kgm²
3) N/m²
4) Nm²
Answer:
2) Kgm²

2. The unit of angular momentum in C.GS system
1) J-s
2) erg-s
3) ergs^-1
4) cals^-1
Answer:
2) erg-s

3. The unit of thermal conductivity is
1) W m^-1 K^-1
2) J m K^-1
3) J m^-1 K^-11
4) W m K^-1
Answer:
W m^-1 K^-1

4. The dimensional formula ML²T^-2 represents
1) Moment of a force
2) Force
3) Acceleration
4) Momentum
Answer:
1) Moment of a force

5. Which of the following has the dimensions of pressure?
1) [MLT^-2]
2) [ML^-1T^-2]
3) [ML^-2T^-1]
4) [M^-1L^-1]
Answer:
2) [ML^-1T^-2]

6. The dimensional formula of torque is
1) [ML²T^-2]
2) [MLT^-2]
3) [ML^-1T^-2]
4) [ML^-2T^-2]
Answer:
1) [ML²T^-2]

7. The dimensional formula for angular velocity is
1) [M^-1L¹T°]
2) [M°L^-1T^-1]
3) [M^-1L^-1T°]
4) [M°L°T^-1]
Answer:
4) [M°L°T^-1]

8. The dimensional formula of angular momentum is
1) [ML²T^-2]
2) [ML^-2T^-1]
3) [MLT^-1]
4) [ML²T^-1]
Answer:
4) [ML²T^-1]

9. Dimensions of stress are
1) [MLT^-2]
2) [ML²T^-2]
3) [ML°T^-2]
4) [ML^-1T^-2]
Answer:
4) [ML^-1T^-2]

10. The dimensions of universal gravitational constant are
1) [M^-1L³T^-2]
2) [ML²T^-1]
3) [M^-2L³T^-2]
4) [M^-2L²T^-1]
Answer:
1) [M^-1L³T^-2]

11. The dimensional formula for latent heat is
1) MLT^-2
2) ML²T²
3) M°L²T^-2
4) MLT^-1
Answer:
3) M°L²T^-2

12. Which one of the following represents the correct dimensions of the coefficient of viscosity?
1) ML^-1T^-2
2) MLT^-1
3) ML^-1T^-1
4) ML^-2T^-2
Answer:
3) ML^-1T^-1

13. The dimensions of (µ₀ε₀)^-1/2 are
1) [L^1/2T^-1/2]
2) [L^-1T]
3) [LT^-1]
4) [L^1/2T^1/2]
Answer:
3) [LT^-1]

14. The dimensions of Planck’s constant equals to that of
1) energy
2) momentum
3) angular momentum
4) power
Answer:
3) angular momentum

15. The ratio of the dimensions of Planck’s constant and that of moment of inertia is the dimensions of
1) time
2) frequency
3) angular momentum
4) velocity
Answer:
2) frequency

16. The dimensions of impulse are equal to that of
1) pressure
2) linear momentum
3) force
4) angular momentum
Answer:
2) linear momentum

17. The pair of quantities having same dimensions is
1) Impulse and Surface Tension
2) Angular momentum and Work
3) Work and Torque
4) Young’s modulus and Energy
Answer:
3) Work and Torque

18. Which pair do not have equal dimensions?
1) Energy and torque
2) Force and impulse
3) Angular momentum and Planck’s constant
4) Elastic modulus and pressure.
Answer:
2) Force and impulse

19. If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formula of the force is
1) PV²T²
2) P^-1V²W^-2
3) PVT^-2
4) P^-1VT^-2
Answer:
1) PV²T²

20. If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are
1) [FVT^-1]
2) [FVT^-2]
3) [FV^-1T^-1]
4) [FV^-1T]
Answer:
4) [FV^-1T]

21. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
1) [F][A^-1][T]
2) [F][A][T]
3) [F] [A] [T²]
4) [F] [A] [T^-1]
Answer:
3) [F] [A] [T²]

22. If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of
1) [M²][L^-2][T^-1]
2) [M²][L^-1][T°]
3) [M][L^-1][T^-1]
4) [M][L°][T°]
Answer:
2) [M²][L^-1][T°]

23. If the error in the measurement of radius of is 2%, then the error in the determination of of the sphere will be
1) 8%
2) 2%
3) 4%
4) 6%
Answer:
3) 4%

24. Percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in the estimate of kinetic energy obtained by measuring ma. and speed will be
1) 8%
2) 2%
3) 12%
4) 10%.
Answer:
1) 8%

25. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and lengths are 3% and 2% respectively, the maximum error in the measurement of density would be
1) 12%
2) 14%
3) 7%
4) 9%.
Answer:
4) 9%.

26. In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, %, 3% and 4% respectively. Then (lie maximum percentage of error in the measurement X, where X = \(\frac{\mathrm{A}^2 \mathrm{~B}^{1 / 2}}{\mathrm{C}^{1 / 3} \mathrm{D}^3}\), will be
1) 10%
2) (3/13)%
3) 16%
4) -10%
Answer:
3) 16%

27. If the dimensions of a physical quantity are given by M^aL^bT^c, then the physical quantity will be
1) velocity if a = 1, b = 0, c = -1
2) acceleration if a = 1, b = 1, c = -2
3) force if a = 0, b = -1, c = -2
4) pressure if a = 1, b = -1, c = -2
Answer:
4) pressure if a = 1, b = -1, c = -2

28. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are
1) kg m s^-1
2) kg m s^-2
3) kg s^-1
4) kg s
Answer:
3) kg s^-1

29. Taking into account of the significant figures, what is the value of 9.99 m – 0.0099 m?
1) 9.9801 in
2) 9.98 m
3) 9.980 m
4) 9.9 m
Answer:
2) 9.98 m

30. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is
1) 0.01 mm
2) 0.25 mm
3) 0.5 mm
4) 1.0 mm
Answer:
3) 0.5 mm

Students get through AP Inter 1st Year Physics Important Questions 1st Lesson Physical World which are most likely to be asked in the exam.

AP Inter 1st Year Physics Important Questions 1st Lesson Physical World

Very Short Answer Questions

Question 1.
What is physics? [AP, TS 15,16]
Answer:
Physics is a branch of science which deals with the study of the basic laws of nature and their demonstration in different phenomena.

Question 2.
What is the discovery of C.V. Raman? [AP 16,18,19,20,2211; TS 18,20]
Answer:
One of the major discovery of C.V. Raman is Raman effect. It deals with scattering of light by molecules of a medium when they are excited to vibrational energy levels.

Question 3.
What are the fundamental forces in nature? [AP 19; TS 18,19,22][Imp.Q]
(or)
Arrange the fundamental forces in descending order according to relative strength.
Answer:
Four fundamental forces in nature:

1. Strong nuclear forces(F_s)
2. Electro Magnetic forces(F_e)
3. Weak Nuclear forces(F_w)
4. Gravitational force (F_g)

Descending order of forces according to relative strength: F_s : F_e : F_w : F_g = 10₃₉ : 10₃₇ : 10₂₆ : 1

Question 4.
Which of the following has symmetry?
(a) acceleration due to gravity
(b) law of gravitation
Answer:
Law of gravitation, because it holds good everywhere in the universe.

Question 5.
What is the contribution of S. Chandra Sekhar to physics? [AP, TS 15, 16, 17, 19]
Answer:
Chandra Sekhar limit, structure and evolution of stars.

Additional Very Short Answer Questions

Question 6.
What are the discoveries of Albert Einstein?
Answer:
The main discoveries of Albert Einstein are
1) Explanation to Photo electric effect
2) Theory of Brownian motion
3) Theory of relativity
4) Mass energy equivalence relation i.e, E = mc².

Question 7.
Which year is. declared as International year of physics?
Answer:
2005 was declared as International year of physics, in recognition of Einstein’s monumental contribution to physics.

Question 8.
What are the fundamental laws of physics?
Answer:
Conservation of energy, momentum, angular momentum, charge etc., are considered to be fundamental laws in physics.

Question 9.
What are the contributions of S.Bose to physics?
Answer:
Bose gave a new derivation of Planck’s law, treating radiation as a gas of photons. He employed new statistical methods of counting photon states.

Question 10.
What is beta (β) decay? Which force is a function of it? [TS 16]
Answer:
The nucleus emits an electron and an uncharged particle called neutrino. This process is called beta (P) decay. Weak Nuclear force is a function of it.