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Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 5th Lesson Work, Energy and Power Class 11 Textbook Exercise Questions and Answers.

Work, Energy and Power Class 11 Questions and Answers AP Inter 1st Year Physics 5th Lesson

I. Work, Energy and Power Multiple Choice Questions (1 Mark)

Question 1.
A man lifted a box of 50 N weight steadily from a ground to a height of 8m. Then the work done by him on the box is……
(1) 25 J
(2) 100 J
(3) 200 J
(4) 400 J
Answer:
(4) 400 J

Question 2.
Two masses 2 kg and 8 kg are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
(1) 1: 4
(2) 4: 1
(3) 1: 2
(4) 2: 1
Answer:
(3) 1: 2

Question 3.
Internal forces can change…………
(1) the linear momentum, not kinetic energy.
(2) the kinetic energy, not the linear momentum.
(3) linear momentum as well as kinetic energy.
(4) neither the linear momentum nor the kinetic energy.
Answer:
(2) the kinetic energy, not the linear momentum.

Question 4.
The kinetic energy of a body is 4 times its momentum. Then the speed of the body is………….
(1) 8m/s
(2) 4 m/s
(3) 2 m/s
(4) 10 m/s
Answer:
(1) 8m/s

Question 5.
A man increases the speed of his car from 20m/s to 40m/s on a smooth road. The ratio of final kinetic energy to the initial kinetic energy is…………
(1) 1: 4
(2) 4: 1
(3) 1: 2
(4) 2: 1
Answer:
(2) 4: 1

Question 6.
The work energy theorem states that the change in kinetic energy of a body is equal to the network done by……..
(1) all conservative and non conservative forces only
(2) all external and internal forces only
(3) all pseudo forces only
(4) all types of forces
Answer:
(4) all types of forces

Question 7.
A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
(1) Kinetic energy
(2) Potential energy
(3) Total mechanical energy
(4) Total linear momentum
Answer:
(3) Total mechanical energy

Question 8.
The work done by weight lifter holding a 150 kg mass steadily on his shoulder in a time of 30 second is
(1) 450 J
(2) 180 J
(3) 0 J
(4) 50 J
Answer:
(3) 0 J

Question 9.
A person fires a bullet of mass 50 gram with a speed of 200 m/s on a soft plywood of thickness 2 cm . The bullet emerges with 10 % of its initial kinetic energy. Then its final kinetic energy is
(1) 500 J
(2) 1000 J
(3) 2000 J
(4) 100 J
Answer:
(4) 100 J

Question 10.
A rain drop of mass 1 gm is falling from a height of 1 km . It hits the ground with a speed of 50m/s. The work done by the gravitational force is (Take acceleration due to gravity g =10 m/s²)
(1) 5 J
(2) 10 J
(3) 15 J
(4) 20 J
Answer:
(2) 10 J

Question 11.
The elastic potential energy of a spring of force constant K for an extension x is………….
(1) Kx
(2) 1/2 Kx²
(3) \(\frac{1}{2}\) Kx
(4) 1/2 k² x²
Answer:
(2) 1/2 Kx²

Question 12.
A bob of mass M is suspended by light string of length L. The speed of the bob at the lowest point of vertical circular path is ………………
(1) \(\sqrt{\mathrm{gL}}\)
(2) \(\sqrt{2 \mathrm{gL}}\)
(3) \(\sqrt{3 \mathrm{gL}}\)
(4) \(\sqrt{5 \mathrm{gL}}\)
Answer:
(4) \(\sqrt{5 \mathrm{gL}}\)

II. Work, Energy and Power Fill in the Blanks (1 Mark)

Question 1.
The work done by a conservative force in moving a body along a closed path is …………
Answer:
zero

Question 2.
Kinetic energy is lost in case of ………… collisions.
Answer:
inelastic

Question 3.
The rate at which work is done is called …………
Answer:
power

Question 4.
5 kWh= ………… joules.
Answer:
18× 10⁶

Question 5.
The power utilized when work of 1000 J is done in 2 seconds is …………
Answer:
500 watt

Question 6.
The compressed or elongated spring possesses ………… energy.
Answer:
potential

Question 7.
A force f= \(5\hat{i}+3 \hat{j}+2 \hat{k} \mathrm{~N}\) is applied over a particle which displaces it from its origin to the point r= \((3 \hat{i}-2 \hat{j})\)m. The work done on the particle is ………….
Answer:
9J

Question 8.
Sailing ship employs………… energy of the wind.
Answer:
kinetic

Question 9.
The area under the force-displacement curve is …………
Answer:
work done

Question 10.
When two equal masses undergo oblique elastic collision with one of them at rest, after the collision they will move at ………… to each other.
Answer:
right angles

Question 11.
1 horse power = ………… watt.
Answer:
746

III. Work, Energy and Power One Word Answer Questions (1 Mark)

Question 1.
When a body of mass M is lifted up through a height h, what is the work done by gravitational force?
Answer:
Negative

Question 2.
Which physical quantity is conserved in all kinds of collisions?
Answer:
Momentum

Question 3.
When two bodies undergo perfectly inelastic collision, what is their relative velocity after collision?
Answer:
Zero

Question 4.
A man carrying a bag of sand walks on a horizontal road with uniform velocity. What is the work done by him if the road is smooth?
Answer:
Zero

Question 5.
If two vectors \(2\hat{i}+2 \hat{j}+\hat{k}\) and \(3 \hat{i}-6 \hat{j}+n \hat{k}\) are perpendicular to each other then what is the value of n ?
Answer:
Given vectors are:

Give two vectors are perpendicular if their dot product is zero, so \(\overline{\mathrm{A}} \cdot \overline{\mathrm{~B}}\)= 0
⇒ -12+n =0
∴ n =12

Question 6.
A motor boat is going in a river with a velocity v=7 \(\hat{i}+2 \hat{j}-5 \hat{k} \mathrm{~m} / \mathrm{s}\). If the resisting force due to stream is \(\mathrm{F}=9 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \mathrm{N}\). What is the power of the boat?
Answer:
Given V= \(7 \bar{i}+2 \bar{j}-5 \bar{k} \mathrm{~m} / \mathrm{s}\)
F = 9 \(\overline{\mathrm{i}}+3 \overline{\mathrm{j}}-3 \overline{\mathrm{k}} \mathrm{~N}\)
Power P = \(\overline{\mathrm{F}} . \overline{\mathrm{V}}\)
⇒ P=9(7)+2(3)+(-5)(-3)
⇒ P=63+6+15
∴ P=84 W
∴ The power of the boat is 84 W

Question 7.
Find the angle between F= \((\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\) N and displacement d=\((\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\) m.
Answer:

Question 8.
A body is dragged along a horizontal floor by a rope. The rope makes an angle of 60° with the horizontal. If the tension in the rope is 100 N and the body is moved through a distance of 10 m, then what is the work done?
Answer:
Given Angle (θ)=60°
The tension in the rope is F=100 N
Distance (d) =10m
The workdone is, W=Fd cos θ
⇒ W=100 × 10 × cos 60°
⇒ W= \(100 \times 510 \times \frac{1}{z_1^{\prime}}\)
∴ W=500 J
∴ The workdone is 500 J

Question 9.
What is the dimensional formula for force constant or spring constant ?
Answer:
MT^-2

IV. Work, Energy and Power Very Short Answer Questions (2 Marks)

Question 1.
State the conditions under which a force does not work.
Answer:
A force does no work under the following conditions.
Workdone,
W=\(\bar{F} \cdot \bar{d}\)
W=Fd cos θ

• If the displacement (s) of the body is zero, then W=0.
• If the angle (θ) between the force and displacement is 90°, then W=0.

Question 2.
Define work. power and energy. State their SI units.
Answer:
Work: Work is said to be done by a force when a body undergoes displacement parallel to the line of action of the force.
Work, W=Fd cosθ
SI unit : Joule (J)
Power: Power is the rate of doing work.
Power, P = \(\frac{\mathrm{W}}{\mathrm{t}}\)
SI Unit : Joule / sec (or) watt (W)
Energy : Energy is defined as the ability or capacity of doing work.
SI Unit : Joule (J)

Question 3.
State the relation between the kinetic energy and momentum of a bods.
Answer:
K E= \(\frac{P^2}{2m}\)
Where, P-Momentum
m – Mass of the body
KE – Kinetic Energy

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) Positive.
(b) Negative.

Question 5.
State the sign of work done by a force in the following
(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) Negative
(b) Positive

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 the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
(a) Positive.
(b) Negative.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer
(a) Total energy of a system is always conserved, no matter what internal and external 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 is always conserved. No matter what internal and external forces on the body are present.

(b) True. The work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero. The gravitational force is always perpendicular to its direction of motion.
W = Fd cos 90°
∴ W =0

V. Work, Energy and Power Short Answer Questions (4 Marks)

Question 1.
What is potential energy ? Derive an expression for the gravitational potential energy.
Answer:
Potential Energy : Potential energy is defined as the energy possessed by a body by virtue of its position or configuration.
Examples: Energy possessed by

• Bird sitting on a branch of tree.
• Water stored in a dam.
• Compressed spring.

Expression for gravitational potential energy : Consider, a body of mass ‘ m ‘ on the ground. Let the body be lifted vertically upwards through height ‘h’.

The minimum force required to lift the body is equal to its weight, F=mg
Work done to lift the body
W=Fs cosθ
Since lifting force F & s are in the same direction θ = 0° and cos 0°=1
W = Fs
W = mgh
This workdone is stored as potential energy in the body.
∴ Potential Energy =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?
Answer:
Case-i:
From Newton’s second law
F= \(\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \Rightarrow \Delta \mathrm{t}=\frac{\Delta \mathrm{P}}{\mathrm{~F}}\)
Given that ΔP and F are same. Hence Δt is also same.
∴ Both car and lorry comes to rest in same interval time.

Case – ii :
From work energy theorem, workdone W= change in kinetic energy.
⇒ F.s =Δ KE
⇒ Fs= \(\frac{\mathrm{p}^2}{2 \mathrm{~m}}\)
Given P and F are same.
Hence S α \(\frac{1}{\mathrm{~m}}\)
i.e., distance travelled α \(\frac{1}{\text { mass }}\)
∴ Lorry (more mass) travel less distance before coming to rest.

Question 3.
Distinguish between conservative and non-conservative forces with one example each.
Answer:
Conservative Force :

• A force is conservative, if the work done by the force on a body along any closed path is zero.
• The workdone by the conservative force on a body depends only on the initial and final positions and does not depend on the path followed.

Example: Gravitational force, electrostatic force, magnetic-force and spring force.
Non – Conservative Force :

• A force is non-conservative, if the workdone by the force on a body along a closed path is not zero.
• The workdone by the non-conservative force on a body depends on the path followed.
  Example : Frictional force, air resistance viscous force.

Question 4.
Show that two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest.
Answer:
Consider two spheres of equal masses. Let the initial velocity of first sphere is u₁ and that of second one u₂=0 (at rest). Let these spheres collide elastically and be moving in different directions which are making angles α and β respectively with respect to initial line of motion.
Where, m₁= m₂=m
u₁=u and u₂=0

According to the law of conservation of linear momentum.
Total linear momentum of the system before collision = Total linear momentum of system after collision.

Since, α+β is the angle between \(\bar{v}_1, \bar{v}_2\).
From the law of conservation of kinetic energy.
K.E of the system before collision = KE of the system after collision


∴ They move at right angles each other after the elastic collision.

VI. Work, Energy and Power Long Type Questions (8 Marks)

Question 1.
Define work and kinetic energy. State and prove work energy theorem for constant force.
Answer:
1. Work: Work is said to be done by a force when a body under goes displacement parallel to the line of action of force.
\(\mathrm{W}=\overline{\mathrm{F}} \cdot \overline{\mathrm{~s}}=\mathrm{Fs} \cos \theta\)

Kinetic Energy : Kinetic energy is defined as the energy possessed by a body virtue of its motion.
KE=\(\frac{1}{2}\) mv²

Example:

• A vehicle in motion.
• Water flowing in a river, etc.

Work – Energy Theorem :
Statement : The work done on a particle by a resultant force is equal to the change in its kinetic energy.
Proof: Consider a particle of mass ‘m ‘ moving with an initial velocity ‘ u ‘. When a constant resultant force ‘ F’ acts on it, it moves with uniform acceleration ‘a’ and attains velocity ‘ v ‘ after t sec.
Let ‘s’ be the displacement of the particle. From equations of motion
v²-u²=2 as
Multiplying the above equation with \(\frac{\mathrm{m}}{2}\) on both sides.

The above equation gives relation between Work and kinetic energy.
k_f-k_i=w
Where k_i and k_f are the initial and final kinetic energies of the object respectively.

Question 2.
Define elastic and inelastic collisions. Derive expressions for the final velocities of bodies in one dimensional elastic collision, when the body collides with other body which is initially at rest.
Answer:
A strong interaction between bodies which involves exchange of momenta is called collision. They are two types.

• Elastic collision,
• Inelastic collision.

Elastic collision : The collisions in which both momentum and kinetic energy are conserved are known as elastic collisions.

Inelastic collisions : The collisions in which kinetic energy is not conserved but momentum is conserved are known as inelastic collisions. Here the loss of kinetic energy appears in the form of heat or other forms of energy.
One – Dimensional

Elastic Collision : If the velocities of the objects involved in collision are along the same straight line before and after collisions then such collisions are known as one dimensional collisions.

Consider two smooth spheres moving along a straight line joining their centres. Let m₁ and m₂ are the masses of the two bodies. Suppose they undergo one dimensional elastic collision. Before collision, Let u₁ and u₂ are their velocities. After collision, let v₁ and v₂ are their final velocities. Assume that u₁ > u₂
From the law of conservation of linear momentum.
Momentum of the system before collision = momentum of the system after collision.
m₁ u₁+m₂u₂=m₁v₁+m₂v₂
⇒ m₁(u₁-v₁)=m₂(v₂-u₂)
In case of elastic collision, kinetic energy is also conserved.
Hence,
kinetic energy of the system before collision = kinetic energy of the system after collision.

From equations (5) & (6), It is concluded that the final velocities of both the bodies depend on their initial velocities and masses.

Question 3.
State and prove law of conservation of mechanical energy in case of a freely falling body.
Answer:
Statement: Energy can neither be created nor be destroyed. It can be converted from one form of energy to another but total energy remains constant.
Proof:
Let us consider a body of mass ‘m’ is falling freely through a height ‘h’ above the ground.

At B :
Travelled distance =s=x, u=0 and velocity (v)=v_B, Height =h-x
PE & =mgh
= mg(h-x)
PE & =mgh – mgx

At C : Let the body hits the ground at ‘ c ‘.
u=0 and final velocity (v)=v_c
Height (h)=0, distance (s)=h

From the equations (1) (2) & (3)
The total energy at A, B and C are equal.
E_A = E_B = E_c = mgh
Hence proved the law of conservation of energy incase of freely falling body.

Question 4.
An object is tied to a string and rotated in a vertical circle. Obtain expressions for speeds of the object at the highest and lowest points, when the string becomes slack only at the highest point.
Answer:
When a string rotates an object in a vertical circle, the object’s speed is minimal at the highest point and maximal at the lowest point. If the string becomes slack at the highest point, it implies the tension is zero.
The veloctiy at the lowest point (v-b) is √5gr, where g is the acceleration due to gravity and ‘r’ is the radius of the circle.

Explanation:
1. At the highest point (Tension = 0):

• When the string becomes slack, the tension (T) in the string is zero.
• The only force acting on the object at the highest point is gravity, which acts as downwards.
• For circular motion, the centripetal force (mv²/r) must be provided by the component of gravity acting towards the center.
• ∴ mg = mv – t²/r
  Where ‘m’ is the mass of the object and v-t is the velocity at the highest point.

2. Conservation of Energy: The total mechanical energy of the object is conserved throughout the circular motion.
∴ Energy at the highest point = Energy at the lowest point.

VII. Problems 

Question 1.
A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600m/s. If the mass of each bullet is 5 g, find the power of the machine gun.
Answer:
Let velocity of each bullet be v = 600 m
Power = ?
Mass of each bullet,
m=5 gm
m=5 x 10^-3kg
n=360
No. of bullets fired
Time t=1 minute =60
Power =P= \(\frac{1}{2} \times \frac{m n v^2}{t}\)
P= \(\frac{1}{2} \times \frac{0.005 \times 360 \times 600 \times 600}{60}\)
Power =5400 W
∴ Power (P)=5.4 kW

Question 2.
Find the useful power used in pumping 3425 m³ of water per hour from a well 8 m deep to the surface, supposing 40% of the horse power during pumping is wasted. What is the horse power of the engine?
Answer:
Volume of the water to be pumped is V = 3425 m³
Density of water,
d=1000 kg/m³
Mass of the water that has to be pumped is
M=(volume) density
M=(3425) 10³kg
g=10 m/s², Height h =8 m
time = 1hr =3600 S
1 hp=746 W

Question 3.
A pump is required to lift 600 kg of water per minute from a well 25 m deep and to eject it with a speed of 50 m/s. Calculate the power required for the above task. Take g=10m/s².
Answer:
Mass of water that has to be lifted is M = 600kg
Height to be lifted, h = 25 m, g = 10 m/s²
Velocity of ejection, v = 50 m/s
Time (t) = 60s

P=1500 W
∴ P=15 kW

Question 4.
A block of mass 5 kg initially at rest at the origin is acted on by a force along the X-positive direction represented by F=(20+5x) N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.
Ans.
As the force is linearly increasing with displacement, we can use the average force to calculate the work F_i=20 N, F_f=40N.
Hence the average force is 30N and displacement is 4m and the average force and displacement are in the same direction.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 4th Lesson Laws of Motion Class 11 Textbook Exercise Questions and Answers.

Laws of Motion Class 11 Questions and Answers AP Inter 1st Year Physics 4th Lesson

I. Laws of Motion Multiple Choice Questions (1 Mark)

Question 1.
When a bus starts suddenly, the passengers are pushed back. This is an example for………..
(1) Newton’s First Law
(2) Newton’s Second Law
(3) Newton’s Third Law
(4) Newton’s law of gravitation
Answer:
(1) Newton’s First Law

Question 2.
Which of the following is an example of centripetal force?
(1) Pushing door
(2) Mud flying off tire
(3) Planets orbiting the sun
(4) Churning of butter in butter milk
Answer:
(3) Planets orbiting the sun

Question 3.
A non zero net force exerted on a body does not change its ………………
(1) direction of motion
(2) momentum
(3) acceleration
(4) inertia
Answer:
(4) inertia

Question 4.
The rate of change of momentum is proportional to…………
(1) acceleration
(2) velocity
(3) force
(4) displacement
Answer:
(3) force

Question 5.
The net force (F) experienced by a body of mass (m) whose displacement is y= ut+\(\frac{1}{2}\) gt² is
(1) F=m a
(2) F=m g
(3) F=0
(4) F≠0
Answer:
(2) F=m g

Question 6.
The linear momentum of a particle as a function of time t is given by P=a+b t. The force acting on the particle is………..
(1) a
(2) b
(3) a/b
(4) b/a
Answer:
(2) b

Question 7.
The inertia of a moving object depends on ………….
(1) momentum of the object
(2) speed of the object
(3) shape of the object
(4) mass of the object
Answer:
(4) mass of the object

Question 8.
A force accelerates a lighter vehicle more easily than a heavier vehicle. Guess the law behind this.
(1) Newton’s First Law of motion
(2) Newton’s Second Law of motion
(3) Newton’s Third Law of motion
(4) Newton’s Law of gravitation
Answer:
(2) Newton’s Second Law of motion

Question 9.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s
(1) 2 sec
(2) 6 sec
(3) 4 sec
(4) 3 sec
Answer:
(3) 4 sec

II. Laws of Motion Fill in the Blanks (1 Mark)

Question 1.
Friction opposes the ………… between the surfaces in contact with each other.
Answer:
Relative motion

Question 2.
………… is required to keep a body in accelerated motion.
Answer:
Non zero net force

Question 3.
Kinetic friction produces ………… energy.
Answer:
heat

Question 4.
Powder is sprinkled on a carom board to ………… friction between the striker and the board surface.
Answer:
reduce

Question 5.
To every action there is always an equal and ………… reaction.
Answer:
opposite

Question 6.
Impulse is the product of force and time which equals ………..
Answer:
change in momentum

Question 7.
………… gives the measure of inertia.
Answer:
mass

Question 8.
The area under F – t graph represents …………
Answer:
impulse

Question 9.
The Newton’s Second Law is consistent with the first law only if ………….
Answer:
F=0

Question 10.
Sliding friction is ………… than the limiting value of static friction.
Answer:
less

III. Laws of Motion One Word Answer Questions (1 Mark)

Question 1.
State Newton’s First Law.
Answer:
“Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled to change that state by a net external force.”

Question 2.
State Newton’s Third Law.
Answer:
For every action, there is an equal and opposite reaction.

Question 3.
Write the mathematical equation of Newton’s Second Law.
Answer:
F=ma

Question 4.
State the SI unit of force.
Answer:
Newton (N)

Question 5.
Define centripetal force.
Answer:
Centripetal force is the force that keeps a body moving in a circular path, directed towards the center of the circle.

Question 6.
According to Newton’s Third Law every force is accompanied by an equal and opposite force. Then how can a movement ever take place?
Answer:
Because action and reaction act on different bodies, not on the same body. Thus, the net force on a single object is not canceled, allowing it to move.

Question 7.
Why does a heavy rifle not recoil as strongly as light rifle using the same cartridges?
Answer:
For the same force, a heavier object undergoes a smaller acceleration than the lighter one because a=F/ma. That is why a heavy rifle does not recoil as strongly as a light rifle using the same cartridges.

IV. Laws of Motion Very Short Answer Questions (2 Marks)

Question 1.
What is inertia ? What gives the measure of inertia ?
Answer:
Inertia is the property of an object to resist changes in its state of motion or rest. Mass of the object gives the measure of its inertia.

Question 2.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain.
Answer:
When a bullet is fired, it gains forward momentum. To conserve momentum, the gun moves backward with equal momentum (Newton’s Third Law).

Question 3.
If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
Answer:
A bomb at rest has zero total momentum. After explosion, the two pieces must move in opposite directions so that total momentum remains zero (conservation of momentum).

Question 4.
Define force. What are the basic forces in nature?
Answer:
Force is a push or pull that changes or tries to change the state of motion of a body.
Basic forces : Gravitational, Electromagnetic, Strong Nuclear, Weak Nuclear.

Question 5.
State the SI units and dimensions of lineæ momentum.
Answer:
SI Unit: kg m/s
Dimensions : M¹L¹T^-1

Question 6.
State the SI units and dimensions of impulse.
Answer:
SI Unit : N.s (Newton – second) or Kgm/s
Dimensions : M¹L¹T^-1 (same as momentum)

Question 7.
Can the coefficient of friction be greater than one ? Give the reason.
Answer:
Yes, the coefficient of friction can be greater than one.
“It is found experimentally than μ_s > μ_k, and both can be more than 1 depending on surface roughness.” This means the frictional force can be greater than the normal force in some cases.

Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres?
Answer:
A flat tyre increases the contact area and friction, which opposes motion more strongly, causing the car to stop sooner.

Question 9.
A horse has to pull harder during the start of the motion than later. Explain.
Answer:
Initially, the horse must overcome static friction, which is greater than kinetic friction encountered during motion.

Question 10.
What happens to the coefficient of friction if the weight of the body is doubled?
Answer:
The coefficient of friction remains unchanged because it is independent of normal force or weight; it depends only on the nature of surfaces.

Question 11.
Write the expressions for maximum possible speed of a car moving on (a) level road (b) banked road.
Answer:
(a) On a level road:
\(\mathrm{v}_{\max }=\sqrt{\mu \mathrm{Rg}}\)
where μ = coefficient of friction, R= radius, g= gravity

(b) On a banked road (no friction):
\(\mathrm{v}_{\max }=\sqrt{\mathrm{Rg} \tan \theta}\)
where θ = banking angle

V. Laws of Motion Short Answer Questions (4 Marks)    

Question 1.
A stone of mass 0.1 kg 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:
Let the acceleration due to gravity g =9.8m/s²
Mass of stone = 0.1 kg
Weight = m g 0.1 × 9.8=0.98 N

a) Upward motion :
Magnitude =0.98 N, Direction = Downward (only gravity acts, opposing motion).

b) Downward motion :
Magnitude =0.98 N, Direction = Downward (gravity aids motion).

c) At highest point :
Magnitude = 0.98 N, Direction = Downward (velocity is zero but force of gravity acts).

Question 2.
Define the terms momentum and impulse. State and explain the Law of Conservation of linear momentum. Give examples.
Answer:
Momentum : Product of mass and velocity; p = mv, a vector quantity.
Impulse : Change in momentum; Impulse = Force × Time
Impulse =Δp = F Δ t

“The total momentum of an isolated system of interacting particles is conserved.” If two bodies collide :
m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂

Examples :

• Recoil of a gun when a bullet is fired.
• Explosion of a bomb at rest into fragments that move in opposite directions.
• A person jumping off a boat causes the boat to move in the opposite direction.

Question 3.
Why are shock absorbers used in motorcycles and cars ?
Answer:
“Shock absorbers reduce the impact of impulsive forces by increasing the time over which the force acts.”
By increasing the time of impact during bumps or collisions, the force experienced is reduced, protecting both the vehicle and passengers.

Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:

• Limiting friction : Maximum static friction just before motion begins.
• Dynamic (Kinetic) friction : Friction acting when two surfaces are in relative motion.
• Rolling Friction : It is the resistance a body faces when rolling over a surface. It is usually much smaller than sliding friction. For example, a ball bearing rolling causes less friction than a block sliding.

Question 5.
Explain advantages and disadvantages of friction.
Answer:
Advantages of Friction :

• Essential for motion : Walking, running, or driving is possibble due to friction between feet/tires and the ground.
• Enables grip: Helps to hold objects, nails to stay in wood, etc.
• Braking systems : Vehicles stop due to friction between brakes and wheels.
• Transfers motion: Belts in machines rely on friction to transmit power.

Disadvantages of Friction :

• Causes wear and tear: Machinery and moving parts wear out over time.
• Wastes energy : Converts mechanical energy into heat, reducing efficiency.
• Requires more force : More effort is needed to move objects due to resistance.

Question 6.
Mention the methods used to decrease friction.
Answer:

• Lubrication : Applying oil or grease reduces contact between surfaces.
• Polishing surfaces : Makes them smoother, reducing resistance.
• Using ball bearings or rollers : Converts sliding friction into rolling friction (which is much lower).
• Streamlining: In vehicles and aircraft to reduce air resistance.
• Using proper materials : Like Teflon, which has a low coefficient of friction.

Question 7.
State the Laws of rolling friction.
Answer:

• Rolling friction is smaller than sliding or static friction.
• It depends on the nature of the surfaces in contact, especially the deformation of the rolling object or the surface.
• It increases with load but remains significantly less than sliding friction for the same surfaces.

Question 8.
Derive an expression for the maximum possible speed of a car moving on a banked road?
Answer:
Expression for the maximum possible speed of a car moving on a banked road : Consider a body of mass ‘m’ on an banked rough surface making an angle ‘θ’ as shown in the figure.
To find max speed from the figure.


V. Laws of Motion Long Answer Questions (8 Marks)

Question 1.
(a) State Newton’s Second Law of motion. Hence derive the equation of motion F=ma from it.
(b) A body is moving along a circular path such that its speed always remains constant. Should there be force acting on the body?
Answer:
(a) Newton’s Second Law of Motion and Derivation of F=ma:
Newton’s Second Law states that the rate of change of momentum of a body is directly proportional to the applied external force, and this change occurs in the direction of the force.
Let the mass of the body be m, and its velocity change from u to v in time t.
Change in momentum =m v-m u=m(v-u)
Rate of change of momentum = \(\frac{m(v-u)}{t}=m \frac{(v-u)}{t}=m \cdot a\)
So, force F α ma, and with proper units,
F=ma

(b) Yes, a force is ‘required’ even if the speed is constant because the direction of motion continuously changes, meaning the velocity changes (since velocity is a vector). This change in direction requires centripetal acceleration, which is provided by a centripetal force directed towards the center of the circle. Without this force, the body would move in a straight line due to inertia.

From Newton’s Second Law :
\(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}\)
Since there is acceleration (towards the center), there must be a net force directed towards the center. This is known as the centripetal force, and is given by :
F = mv²/R
Where:

• m= mass of the object
• v = constant speed
• R = radius of circular path

VII. Problem (Marks 8)

Question 1.
A force 2i+j-k Newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i+2j-2 kms^-1. What is the mass of the body?
Answer:
Given :
Force, F=2i+j-kN
Initial velocity, u=0 (body is at rest)
Final velocity, v=4i+2j-2 km/s
Time, t=20s
Using Newton’s Second Law :

Question 2.
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) ?
Answer:

Question 3.
A container of mass 200 kg 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 ?
Answer:
Given m=200kg, a=1-5ms^-2, g=9.8 ms^-2
Co-efficient of friction μs=\(\frac{\mathrm{a}}{\mathrm{g}}=\frac{1.5}{9.8}\)=0.153

Question 4.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically down-wards with an initial speed of 10m/s. If acceleration due to gravity is 10m/s², what is the separation between the fragments 2 sec after the explosion.
Answer:
Given :
One fragment moves down with u=10 m/s
Other fragment (by conservation of momentum) must move up with u=10m/s Time t=2s
Acceleration due to gravity is g=10m/s²
Using the equation of motion : s=u t+\(\frac{1}{2}\) gt²
Downward fragment : s₁=10 × 2+\(\frac{1}{2}\) ×10.4=20+20=40m (downward)
Upward fragment : s₂=10 × 2-\(\frac{1}{2}\) × 10.4=20-20=0m (still at explosion point)
So, separation =40m

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 3rd Lesson Motion in a Plane Class 11 Textbook Exercise Questions and Answers.

Motion in a Plane Class 11 Questions and Answers AP Inter 1st Year Physics 3rd Lesson

1. Motion in a Plane Multiple Choice Questions (1 Mark)

Question 1.
A particle moving with uniform speed in a circular path maintains-
(1) constant velocity
(2) constant acceleration
(3) constant velocity but varying acceleration
(4) varying velocity and varying acceleration
Answer:
(4) varying velocity and varying acceleration

Question 2.
The projection or component of the vector \(3\hat{i}+4 \hat{k}\) on y-axis is
(1) 7
(2) 5
(3) 4
(4) 0
Answer:
(4) 0

Question 3.
The angle made by the vector \(\overrightarrow{\mathrm{A}}=[\hat{\mathrm{i}}]\) with x -axis is
(1) 22.5°
(2) 30°
(3) 45°
(4) 90°
Answer:
(3) 45°

Question 4.
A body is projected into air from the ground at an angle 90° to the horizontal. On reaching the highest point above the ground
(1) its velocity is zero, acceleration is not zero
(2) its acceleration is zero, velocity is not zero
(3) both the velocity and acceleration are not equal to zero
(4) both the velocity and acceleration are equal to zero
Answer:
(1) its velocity is zero, acceleration is not zero

Question 5.
Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ration of their initial velocities if they attain the same height ?
(1) √3: 1
(2) 1: √3
(3) 1 : 1
(4) 3: 1
Answer:
(1) √3: 1

Question 6.
An aeroplane is dropping a bomb while moving horizontally at a constant speed. When air resistance is taken into consideration, then the bomb
(1) falls alongside the plane
(2) falls on earth behind the plane
(3) falls on earth ahead of the plane
(4) falls on earth exactly below the plane
Answer:
(2) falls on earth behind the plane

Question 7.
Which of the following is the vector quantity ?
(1) Distance
(2) Speed
(3) Acceleration
(4) Temperature
Answer:
(3) Acceleration

Question 8.
Which of the following is a scalar quantity ?
(1) Momentum
(2) Mass
(3) Velocity
(4) Force
Answer:
(2) Mass

Question 9.
Which of the following is not a vector quantity ?
(1) Weight
(2) Torque
(3) Momentum
(4) Potential energy
Answer:
(4) Potential energy

Question 10.
Which of the following physical quantity can be negative ?
(1) Speed
(2) Mass
(3) Velocity
(4) Distance
Answer:
(3) Velocity

Question 11.
Vector addition obeys
(1) commutative law only
(2) associative law only
(3) both commutative and associative laws
(4) neither commutative nor associative laws
Answer:
(3) both commutative and associative laws

Question 12.
The relation between the time of flight of projection T_f and the time to reach the maximum height t_m is ………..
(1) T_f = t_m
(2) T_f =2 t_m
(3) T_f =t_m / 2
(4) T_f =√2t_m
Answer:
(2) T_f =2 t_m

Question 13.
Which of the following is not a projectile motion
(1) A stone is thrown horizontally from a building
(2) A built fired from a gun at angle 45° w.r.t. ground
(3) The jump of a frog
(4) A car moving in a straight line
Answer:
(4) A car moving in a straight line

Question 14.
A missile is fired for a maximum range with an initial velocity of 30m/s. If g= 10m/s², then the range of the missile will be
(1) 100 m
(2) 90 m
(3) 80 m
(4) 70 m
Answer:
(2) 90 m

Question 15.
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
(1) θ=45°
(2) θ=tan^-1 (1/4)
(3) θ=tan^-1 (4)
(4) θ=tan^-1 (2)
Answer:
(3) θ=tan^-1 (4)

II. Motion in a Plane Fill in the Blanks (1 Mark)

Question 1.
The dimensional formula for centripetal acceleration in a circular motion is__________
Answer:
M⁰L¹T²

Question 2.
If 0 is the angle of projection, g is the acceleration due to gravity, then maxi­mum height of projectile is_________
Answer:
U²Sin²θ/2g

Question 3.
The equation for the trajectory of a projectile motion is________
Answer:
y = ax bx²

Question 4.
The path of the projectile with respect to ground is______________
Answer:
parabola

Question 5.
The resultant of multiplying the scalar with a vector is a____________
Answer:
vector

Question 6.
The time required by a projectile to reach its highest point in terms of u and g is_________
Answer:
U sin θ/g

Question 7.
The position of a particle is r = 3.0ti + 2.0t²j+5.0 k, where t is in seconds, the coefficients have the proper units of r to be in metres, then v(t) =________
Answer:
\(3 \hat{i}+4 t \hat{j}\)

Question 8.
The angle between two vectors A and B with magnitudes √3 and 4 respectively, and \(\vec{A} \cdot \vec{B}=2 \sqrt{3}\) is_____________
Answer:
(π/3)

III. Motion in a Plane One Word Answer Questions (1 Mark)

Question 1.
When are the two vectors said to be equal vectors?
Answer:
Equal direction and magnitude

Question 2.
Can a vector of zero magnitude have non-zero components ?
Answer:
No

Question 3.
When will the magnitude of sum of two vectors be maximum and minimum ?
Answer:
Maximum : 0°, Minimum : 180°

Question 4.
At what angle of projection of body with respect to ground, its range is maximum?
Answer:
45°

Question 5.
What is the acceleration of a projectile at the top of its trajectory ?
Answer:
9.8 m/s² downward

Question 6.
What is the direction of velocity at any point on the path of a circularly moving object?
Answer:
Tangent

Question 7.
What is the angle between the direction of velocity and acceleration at the highest point of a projectile path ?
Answer:
90°

IV. Motion in a Plane Very Short Answer Questions (2 Marks)

Question 1.
What are scalars and vectors?
Answer:
Scalars have only magnitude (e.g., mass, time), while vectors have both magnitude and direction (e.g.: velocity, force).

Question 2.
State Triangle Law of vector addition.
Answer:
Triangle Law : If two vectors are represented as two sides of a triangle taken in order, their sum is given by the third side taken in reverse order.

Question 3.
When two right angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant ?
Answer:
Using Pythagoras :
Resultant = \(\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625} \)= 25 units

Question 4.
If \(\vec{p}=2 \hat{i}+4 \hat{j}+14 \hat{k}\) and \(\vec{Q}=4 \hat{i}+4 \hat{j}+10 \hat{k}\). find the magnitude of \(\vec{P}+\vec{Q}\).
Answer:

Question 5.
Can two vectors of unequal magnitude add up to give the zero vector ? Can three unequal vectors add up to give the zero vector ?
Answer:
Two unequal vectors cannot give zero vector, but three unequal vectors can, if they form a closed triangle when added head to tail.

Question 6.
If vertical component of a vector is equal to its horizontal component, what is the angle made by the vector with X -aixs ?
Answer:
If both components are equal, angle with X -axis is 45°

Question 7.
Two forces of magnitude 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant?
Answer:
Using vector formula:
R= \(\sqrt{3^2+5^2+2.3 .5 \cdot \cos \left(60^{\circ}\right)}=\sqrt{9+25+15}=\sqrt{49}=7 \) units

Question 8.
Why does the vertical component of velocity for a projectile change with time, where the horizontal component of the velocity remains constant in parabolic path?
Answer:
The vertical component changes due to acceleration from gravity, while the horizontal component remains constant because thére’s no acceleration in the horizontal direction.

V. Motion in a Plane Short Answer Questions (4 Marks)

Question 1.
Using parallelogram law of vectors derive an expression for the magnitude and direction of the resultant vector.
Answer:
Parallelogrm law of vectors : If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram drawn from a point then their resultant is also represented in magnitude and direction by the diagonal passing through the same point.
Expression for magnitude of the resultant :

Question 2.
Define unit vector, null vector and position vector.
Answer:
1. Unit Vector : A unit vector is a vector of unit magnitude that points in a particular direction. It has :

• No dimension and only specifles direction.
• Commonly used to indicate directions along the axes in coordinate systems.

For a unit vector \(\hat{A}\) in the direction of vector \(\vec{A}\), it is given by :
\(\hat{A}=\vec{A}|\vec{A}|\)
Unit vectors along the \(\vec{x}, \vec{y}\) , and \( \vec{z} – axes\) are denoted as \(\hat{i} ; \hat{j},\hat{k} \) respectively.

2. Null Vector (or Zero Vector): A null vector is a vector that has zero magnitude and no specific direction.
It is represented as :
\(\overrightarrow{0}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\)
In geometrical representation, it is a point at the origin and does not change position or direction.

3. Position Vector: The position vector of a particle P located in a plane with reference to the origin 0 is :
\(\overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}\)
where : x and y are the coordinates of point P,
\(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) are unit vectors along the x – and y -axes respectively.
This vector shows the position of the particle in space relative to the origin.

Question 3.
If \(\vec{a}+\vec{b}=\vec{a}-\vec{b}\), prove that the angle between \(\vec{a}\) and \(\vec{b}\) is 90°?
Ans.

Squaring on both sides
a²+b²+2 a b cos θ =a²+b²-2 a bcosθ
⇒ 4 ab cosθ = 0
⇒ cos θ =0
⇒ θ =90°

Question 4.
Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola ?
Answer:
Let a body be projected with an intial velocity ‘u’ at an angle ‘θ’ with the horizontal, from the point 0.
The path of the body is called trajectory.

Initial horizontal component of velocity
u_x=u cos θ
Initial Vertical component of velocity u_r=u sinθ
The horizontal component is constant throughout the motion of the body.
The vertical component changes both in magnitude and direction due to gravity.
Let the projectile is at the point P(x\y) after a time interval ‘t’

Hence the path of the projectile is a parabola.

Question 5.
Show that the maximum height and range of a projectile are U²sin² θ/2g and U²sin 2θ/g respectively, where the terms have their regular meanings.
Answer:
Maximum Height (H): The greatest vertical displacement of a projectile during its journey is called maximum height.
In vertical motion of a projectile

Range of a projectile: The maximum horizontal distance travelled by the projec­tile is called Range of a projectile.
In horizontal motion of a projectile p_x = Ucos0, a_x = 0, s_x = R, t = T
Using s_x = \(\mathrm{U}_{\mathrm{x}} \mathrm{t}+\frac{1}{2_{\mathrm{x}}} \mathrm{at}^2\)
R = (U cos θ)T

Question 6.
Show that the maximum height reached by a projectile launched at an angle of 45° is one quarter of its range.
Answer:
Given: Angle of projection, θ= 45°
Acceleration due to gravity, g
Initial velocity of projection, u
To Prove: H = 14R/H
Where: H = Maximum height, R = Range

Step 1: Write formula for Range R
R = u²sin²θ/g
Substitute 0 = 45°, then sin 2θ = sin90° = 1:
R = u²/g

Step 2: Write formula for Maximum
Height H H = u²sin² θ/2g
Substitute 0 = 45°, then sin² 45° = (1/√2 )² = 1/2
H = u².1/2 /2g = u²/4g

Step 3: Divide Equation (2) by Equation (1)
H/R = u²/4g/u²/g = 1/4
Hence Proved: H = 1/4R
So, the maximum height reached by a projectile launched at 45° is one-fourth of the range.

Question 7.
If the horizontal ranges described by two projectiles, projected at angles (45° + α) and (45° – α) from the same point and same velocity, then the ratio of horizontal ranges will be ?
Answer:
Given
Angle of projections: First projectile: θ₁ = 45° + α
Second projectile: θ₂ = 45° – α
Initial velocity of projection: u (same for both)
Acceleration due to gravity: g
To Find:
Ratio of horizontal ranges R₁: R₂
Formula for Range :
For a projectile: R = u²sin(2θ)/g
So,
For first projectile:
R, = u²sin[2(45° + α)]/g = u²sin(90° + 2α)/g
For second projectile: It, = u²sin[2(45° – α)]/g = u²sin(90° – 2α)/g
R₁ = u²cos(2α)/g
R₂ = u²cos(2α)/g
Therefore:
R₁/R₂= u²cos(2α)/g / u²cos(2α)/g = 1
R₁: R₂ = 1:1
So, the horizontal ranges are equal, i.e., the ratio is 1:1, when projectiles are launched at angles (45° + α) and (45° – α) with the same velocity.

Question 6.
If θ is the angle of projection, R the range, h the maximum height, T the time of flight, then show that
(a) tan θ = 4h / R and (b) h= gT²/8
Answer:

Question 9.
Define centripetal acceleration ? Derive an expression for the centripetal acceleration of a particle moving with uniform speed v along a circular path of radius r.
Answer:
Centripetal acceleration : The acceleration directed towards the centre of a circular path is called centripetal acceleration consider a particle moving in a circular path of radius ‘r’ with speed ‘v’.
This instantaneous acceleration of the particle is

VI. Problems

Question 1.
A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:
i) Find the time of flight of the projectile before it hits the ground.
ii) Find the distance it travels before it hits the ground (range).
iii) Find the time of flight for the projectile to reach its maximum height.
Answer:
Given:
Initial velocity, u = 800 m/s
Angle of projection, 0 = 60°
Acceleration due to gravity, g = 9.8 m/s²
To Find:
i. Time of flight, T
ii. Horizontal range, R
iii. Time to reach maximum height,T_up

i. Time of Flight, T:

Question 2.
A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection. (g=10m/s²)
Answer:
Given :
Vertical height, y =7.5m
Horizontal distance, x=10m
Angle of projection, θ =45°
Acceleration due to gravity, g=10m/s²

To Find:
Initial velocity of projection, u
Use the equation of the projectile path:
y = x tanθ – gx²/2u²cos²θ
Step 1: Plug in known values:
tan 45° -1, cos 45° = 1/√2 ⇒ cos²45° = 1/2
7.5 = 10.1 – 10.10²/2u² 1/2

Step 2: Simplify the equation:
7.5 = 10 -1000/u²
⇒ 1000/u² =10 – 7.5 = 2.5
⇒ u² =1000/2.5 = 400
⇒ u = √400 = 20 m/s
This is the initial speed of projection required for the particle to reach 7.5 m in height and 10 m horizontally when launched at an angle of 45°.

Question 3.
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 be √2 times the maximum height reached by it. Show that the angle of projection is Tan¹ (2).
Answer:
Let the initial speed be ‘u’ and angle of a projection is θ.
At the highest point, the horizontal displacement is

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 2nd Lesson Motion in a Straight Line Class 11 Textbook Exercise Questions and Answers.

Motion in a Straight Line Class 11 Questions and Answers AP Inter 1st Year Physics 2nd Lesson

I. Motion in a Straight Line Multiple Choice Questions (1 Mark)

Question 1.
The velocity of a body depends on time according to the equation v=10+2t². The body is undergoing
(1) Uniform acceleration
(2) uniform retardation
(3) non-uniform acceleration
(4) zero acceleration
Answer:
(3) non-uniform acceleration

Question 2.
Which of the following is true?
(1) Speed is always equal to magnitude of velocity.
(2) Speed is always greater than the magnitude of velocity.
(3) Velocity is always greater than speed.
(4) Speed in general is either equal to or greater than the magnitude of velocity
Answer:
(4) Speed in general is either equal to or greater than the magnitude of velocity

Question 3.
A bullet is dropped from a certain height, at the same time another bullet is fired horizontally from the same height. Both bullets will hit the ground.
(1) one after the other
(2) simultaneously
(3) depends on the observer
(4) depends on the velocity of the fired bullet
Answer:
(2) simultaneously

Question 4.
If a car is moving with a uniform velocity of 60 km/h. Then its acceleration is ……………
(1) 0 km/ h²
(2) 10 km/ h²
(3) 20 km/ h²
(4) 30 km/h²
Answer:
(1) 0 km/ h²

Question 5.
A ball is thrown vertically upward with an initial velocity of 50 m/s. Its maximum height is (take g=10 m/s²)
(1) 125 m
(2) 250 m
(3) 500 m
(4) 1000 m
Answer:
(1) 125 m

Question 6.
A particle moves in a straight line with an initial velocity of 10m/s and a constant acceleration of 2m/s². Its displacement after 5 seconds will be :
(1) 50 m
(2) 75 m
(3) 150 m
(4) 200 m
Answer:
(1) 50 m

Question 7.
A body starts from rest and moves with a uniform acceleration of 2 m/s². Its velocity after 10 seconds will be……………….
(1) 5 m/s
(2) 10 m/s
(3) 15 m/s
(4) 20 m/s
Answer:
(4) 20 m/s

Question 8.
For the motion with uniform velocity, the slope of the velocity-time graph is equal to……………..
(1) 1m/s
(2) zero
(3) initial velocity
(4) final velocity
Answer:
(2) zero

Question 9.
If a cyclist takes two minutes to complete half revolution on a circular path of 120 m radius, what is the average speed ?
(1) 1/s
(2) 2m/s
(3) 3.14m/s
(4) 4.14m/s
Answer:
(3) 3.14m/s

Question 10.
In a particular time, the net displacement travelled by the body is equal to………..
(1) the area which v – t graph encloses with displacement axis
(2) the area which x – t graph encloses with the time axis
(3) the area which v-t graph encloses with time axis
(4) the area which a-t graph encloses with axis
Answer:
(3) the area which v-t graph encloses with time axis

Question 11.
A particle moves along a straight line. At a time t, the distance x of the particle from the origin is given by x=20+12t-t³, where x is in meters and t in sec, the time taken for the particle for coming to rest is
(1) 1 sec
(2) 2 sec
(3) 4 sec
(4) 6 sec
Answer:
(2) 2 sec

Question 12.
A car travels half of the distance with speed v and the remaining distance with speed 2v. Its average speed is…………
(1) \(\frac{4 u}{3}\)
(2) \(\frac{3 v}{4}\)
(3) \(\frac{v}{3}\)
(4) \(\frac{2 v}{3}\)
Answer:
(1) \(\frac{4 u}{3}\)

Question 13.
A car runs at constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed of each circular lap respectively is…………
(1) 10 m/s, 0 m/s
(2) 0 m/s, 0 m/s
(3) 0 m/s, 10 m/s
(4) 10 m/s, 10 m/s
Answer:
(3) 0 m/s, 10 m/s

Question 14.
The motion of the particle is given by the equation S=(2t³+5t²+16 t+8) m. The value of acceleration of the particle at t=1 sec is…………
(1) 20 m/s²
(2) 22 m/s²
(3) 24 m/s²
(4) 26 m/s²
Answer:
(2) 22 m/s²

Question 15.
The ratio of distances travelled by a freely falling body in the 1st, 2nd, 3rd, 4th second………….
(1) 1: 2: 3: 4
(2) 1: 4: 9: 16
(3) 1: 3: 5: 7
(4) 1: 1: 1: 1
Answer:
(3) 1: 3: 5: 7

Question 16.
A stone falls freely under gravity. It covers distances h₁ h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁ h₂ and h₃ is…………
(1) h₂=3 h₁ and h₃=3h₂
(2) h₁=h₂=h₃
(3) h₁=2h₂=3h₃
(4) h₁= \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
Answer:
(4) h₁= \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)

II. Motion in a Straight Line Fill in the Blanks (1 Mark)

Question 1.
………… deals with the study of motion of objects without taking into account, the factors which cause the motion.
Answer:
Kinematics

Question 2.
The shortest path travelled by a moving object in a given time from initial to final position is called …………
Answer:
displacement

Question 3.
The distance covered by a body along a semicircular path of radius r is………….
Answer:
πr

Question 4.
The slope of velocity-time graph gives ………….
Answer:
acceleration

Question 5.
The slope of the displacement-time graph gives …………
Answer:
velocity

Question 6.
The area under acceleration-time graph gives …………
Answer:
Change in velocity

Question 7.
The acceleration of a body moving with uniform velocity is …………
Answer:
zero

Question 8.
The velocity of a body at any instant of time is called …………
Answer:
instantaneous velocity

Question 9.
The value of an acceleration of a freely falling body is ………….
Answer:
(g=9.8 m/sec²)

Question 10.
The ratio of distances travelled by a freely falling body from the rest in the first three seconds is …………
Answer:
1:3:5

III. Motion in a Straight Line One Word Answer Questions (1 Mark)

Question 1.
What is uniform motion?
Answer:
Constant (Uniform motion means constant velocity.)

Question 2.
What is acceleration?
Answer:
Rate (Acceleration is defined as the rate of change of velocity with time.)

Question 3.
What is average speed?
Answer:
Ratio (Average speed = total distance / total time)

Question 4.
What is instantaneous velocity?
Answer:
Limit (Defined as the limit of average velocity as time interval approaches zero)

Question 5.
Draw position-time graph for an object at rest.
Answer:
Horizontal-line

Question 6.
What represents the area under the curve of velocity – time graph ?
Answer:
Displacement

Question 7.
Mention one similarity between speed and velocity.
Answer:
Magnitude of motion.

Question 8.
Draw velocity-time graph for a particle moving with uniform velocity.
Answer:
Straight horizontal line.

Question 9.
What is the time taken by the body, when it is freely dropped from a height “h”?
Answer:
\(\sqrt{\frac{2 h}{g}}\)

IV. Motion in a Straight Line Very Short Answer Questions (2 Marks)

Question 1.
The states of motion and rest are relative. Explain.
Answer:
Rest and motion depend on the observer’s frame of reference. An object at rest with respect to one observer may be in motion with respect to another.

Question 2.
Is it necessary that the direction of acceleration and velocity of a body should be the same ? Give an example.
Answer:
No. For example : When a ball is thrown upward, its velocity is upward, but acceleration due to gravity is downward.

Question 3.
Is it possible for a body to have acceleration but zero velocity? Give an example.
Answer:
Yes. For example, when a ball thrown vertically upward reaches the highest point, its velocity is zero, but it still has acceleration due to gravity acting downward.

Question 4.
Which is greater, a velocity of 72 kmph or 25 m/s ?
Answer:
Convert 72km/h to m/s :
72 ×1000 / 3600=20 m/s
So, 25 m/s is greater than 72 km/h.

Question 5.
Give an example of one dimensional motion where a particle moving along a straight line comes to rest and moves backward.
Answer:
A ball thrown straight up moves upward, comes to rest at the top, and then reverses direction to move downward. This is a one-dimensional motion along a vertical line.

Question 6.
A vehicle travels half the distance L with a speed v₁ and other half with speed v₂. What is the average speed ?
Answer:
Average speed = \(\frac{2 v_1 v_2}{v_1+v_2}\)
This is valid when the distance, not time, is equally split between the two speeds.

Question 7.
A body coveres a distance of 5m along a semicircular path from its one end to another end. What is the ratio of its distance covered to its displacement ?
Answer:
Distance =5m
⇒ πr=5 ⇒ r=5/π
Displacement =2 r=2 ×5 / π=10/π =1/2
Ratio = Distance / Displacement = 5/π × π/10=1/2
Ratio =1: 2

Question 8.
Displacement travelled by a body is given by S=2t+5t² m. What is the acceleration of that body?
Answer:
S=2t+5t²
⇒ v= \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 2+10t
a= \(\frac{\mathrm{dv}}{\mathrm{dt}}\)=10 m/s²
So, acceleration =10m/s²

Question 9.
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 constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity?
Answer:
At terminal velocity, a=0 ⇒ g=bv ⇒ v= \(\frac{g}{b}\)
So, terminal velocity = g/b

Question 10.
Under what circumstances is the relationship s=vt valid ?
Answer:
This equation is valid when the object moves with uniform (constant) velocity, i.e., acceleration =0 through out the motion.

V. Motion in a Straight Line Short Answer Questions (4 Marks)

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take?
Answer:
No, kinematic equations like
v=u+at, s=ut+1/2 at², v²=u²+2as
are valid only when acceleration is constant. If acceleration varies with time, one must use calculus-based methods, such as :
v=∫a(t) dt, s=∫v(t) dt
This is implied in the book where derivations are based on uniform acceleration (constant ‘ a ‘).

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 (V₁+V₂)/2. Is this correct ? Substantiate your answer.
Answer:
Yes, but only when the acceleration is uniform (constant).
For constant acceleration, average velocity is : v_avg=\(\frac{\mathrm{v}_1+\mathrm{v}_2}{2}\)
This is derived from the linear velocity-time graph under constant acceleration. It does not apply to varying acceleration.

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.
Answer:
Yes, they can. For example, when a car is moving forward (positive velocity) but slowing down, the acceleration is in the opposite direction (negative). This occurs in retardation or deceleration scenarios, like a ball thrown upward.

Question 4.
Obtain the equation of motion S=ut+ \(\frac{1}{2}\) at² for constant acceleration using graphical method.
Answer:
Draw a velocity-time graph with initial velocity u, final velocity v, and time t.
The area under the v-t graph (a trapezium) gives displacement s
Split it into a rectangle and triangle :
s= area of rectangle + area of triangle = ut+ \(\frac{1}{2}(v-u)t\)
Since v=u+at, substitute to get :
s=ut+\(\frac{1}{2} \mathrm{at}^2\)

Question 5.
Explain the terms the average velocity and instantaneous velocity. When are they equal?
Answer:
Average velocity is the total displacement divided by total time :
\(v_{\text {avg }}=\frac{s}{t}\)
Instantaneous velocity is the velocity at a particular instant, found using calculus:
\(v=\frac{d S}{dt}\)
They are equal only when velocity is constant, i.e., when there is no acceleration.

VI. Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5km\h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5km\h^-1. What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 min .
Answer:
Distance to market =2.5 km
Speed while going =5 km/ h
Speed while returning =7.5 km/h
Time taken :

(a) Average velocity = Displacement /Time =0 /(5/6)=0 km/h
(b) Average speed Total distance/Time =\(\frac{5}{\frac{5}{6}}\) = 6 km/h

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph .What is its means speed over the entire distance?
Answer:
Let the total distance =35 km
Time taken to travel first third distance

Question 3.
A bullet moving with a speed of 150 m s ‘ 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 ?
Answer:
u =150 m/s, v = 0, s = 0.035 m using v²= u² + 2as :
0 = (150)² + 2a (0.035)
a = 22500 + 0.07a
⇒ a = 0.07-22500 = -3.21 x 10⁵m/s² = 3.214 x 10^s ms ²
∴ Magnitude of acceleration a = 3.214 x 10⁵ ms^-2
Time: Using v = u + at ⇒ 0
= 150 – (3.21 x 10⁵) t ⇒ t = 3.21 x 105150 ≈ 4.67 x 10⁴s

Question 4.
A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Answer:
First segment:
Time = 30 min = 0.5 h Speed = 85 km/h
Displacement = 85 x 0.5 = 42.585, times = 42.5 km
Stop duration:
Time = 15 min = 0.25 h
No movement, so displacement = 0 km

Second segment:
Distance = 130 km
Time = 2hours
Displacement =42.5 km+130 km=172.5 km
Total time =0.5h+0.25 h+2h=2.75 h
Average velocity = Total displacement/Total time
=172.5 /2.75
=62.72 km/h
∴ Total Displacement =172.5 km
Average Velocity =62.72 km/h

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?
Answer:
Given that velocity of Ball A is v_A = 2v_B
Let the ball A’ is dropped from a Height
‘H’ and the two balls will collide at a height ‘h’ from the ground.
For freely falling ball

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

Hence, the time interval between 5th drop and 2nd, 5th drop and 3rd, 5th drop and 4th is 1.35,0.90 and 0.45 respectively.
For second drop h₂= \(\frac{1}{2}\) gt²=\(\frac{1}{2}\) × 9.8 ×(1.35)²=8.93 m
d₁₂=16-8.93=7.06 ≈ 7m
For third drop, h₃=\(\frac{1}{2}\)× 9.8 ×(0.90)²=3.97 m
d₂₃=8.93-3.97=4.961 ≈ x 5m
For fourth drop, h₄= \(\frac{1}{2}\) × 9.8 ×(0.45)² ≈ x 0.9922 m
d₃₄=3.97-0.9922=2.9978 ≈ 3m
Similarly d₄₅=0.9922-0=0.9922 ≈ 1m

Question 7.
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.
Answer:
Horizontal speed of airplane = 360 km/h = 360 × 1000/3600 = 100 m/s
Height = 500 m
Acceleration due to gravity g = 9.8 m/s²
Initial vertical velocity u_y = 0

i. Time of Descent
Use the kinematic equation: s = ut +12gt²
500 = 0 + 12 x 9.8 x t² ⇒ t² = 500 x 29.8 = 10009.8 ≈ 102.04
⇒ t = 102.04 ≈ 10.1 seconds

ii. Horizontal Distance Travelled
Horizontal distance = horizontal speed x time
= 100 m/s x 10.1 s = 1010 m
∴ (i) Time of Descent = 10.1s
(ii) Horizontal Distance = 1010m

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 1st Lesson Units and Measurements Class 11 Textbook Exercise Questions and Answers.

Units and Measurements Class 11 Questions and Answers AP Inter 1st Year Physics 1st Lesson

I. Units and Measurements Multiple Choice Questions (1 Mark)

Question 1.
Pascal is the unit of ………….
(1) power
(2) pressure
(3) potential energy
(4) pressure gradient
Answer:
(2) pressure

Question 2.
When the number 0.046508 is reduced to four significant figures then it becomes…………
(1) 0.04651
(2) 4650
(3) 4.651
(4) 4.650
Answer:
(1) 0.04651

Question 3.
Which of the following physical quantity has no dimensions?
(1) Angular Velocity
(2) Angular Acceleration
(3) Angular Displacement
(4) Stress
Answer:
(3) Angular Displacement

Question 4.
Mention same dimensions in the dimensional formula of torque and angular momentum
(1) mass, time
(2) time, length
(3) mass, length
(4) mass
Answer:
(3) mass, length

Question 5.
The velocity of an object varies with time as V=At²+Bt+C, then the dimensions of C are…………
(1) LT^-1
(2) LT^-2
(3) LT^-3
(4) LT^-4
Answer:
(4) LT^-4

Question 6.
Which of the following equation is dimensionally correct ?
(1) \(\frac{1}{2}\) mv²=mgh
(2) V=u t+a t
(3) s=u t+a t^3
(4) t=t+u v
Answer:
(1) \(\frac{1}{2}\) mv²=mgh

Question 7.
The velocity (V) of sound in air pressure (P) and density (4) of air are related as V α P^xd^y. The value of x and y are……….
(1) 1, 1/2
(2) -1/2, 1/2
(3) \(\frac{1}{2}\),-1/2
(4) 1 / 2,1 / 2
Answer:
(3) \(\frac{1}{2}\),-1/2

Question 8.
The velocity of a body is given by V=At²+Bt+C. If V and t are expressed in SI units, then unit of A is …………….
(1) m/s
(2) m/s²
(3) m/s³
(4) ms
Answer:
(3) m/s³

Question 9.
The dimensional formula for product of two physical quantities P and Q is ML² T^-2 and dimensional formula of P/Q is MT^-2 then P is……….
(1) force
(2) displacement
(3) acceleration
(4) velocity
Answer:
(1) force

Question 10.
E, M, L and G denote energy, mass, angular momentum and gravitational constant respectively. Then the dimension of
EL² M^-5 G^-2 is ………..
(1) angle
(2) length
(3) mass
(4) time
Answer:
(1) angle

II. Units and Measurements Fill in the Blanks (1 Mark)

Question 1.
Plane angle is the ratio of length of arc to …………
Answer:
radius

Question 2.
The unit of plane angle is ………….
Answer:
radian

Question 3.
The SI unit of luminous intensity is ………….
Answer:
Candela

Question 4.
The dimensional formula for the coefficient of viscosity is …………
Answer:
ML^-1 T^-1

Question 5.
The number of significant figures in 0.02308 is ………….
Answer:
4

Question 6.
If F,M and T are the force, mass and time, then FM^-1 T² denotes …………
Answer:
length

Question 7.
A number 2.735 is rounded off to three significant figure is ………….
Answer:
2.74

Question 8.
Plank’s constant and …………….have same dimensions.
Answer:
angular momentum

Question 9.
The dimensional formula of angular impulse is …………….
Answer:
(ML²T^-1)

Question 10.
A force is represented by F=Ax²+Bt², where x= distance and t= time then the dimensions of ‘A’ is …………
Answer:
(ML^-1T^-2)

III. Units and Measurements One Word Answer Questions (1 Mark)

Question 1.
What is fundamental physical quantity?
Answer:
A fundamental physical quantity is one which cannot be expressed in terms of any other physical quantity.
Examples include: length, mass, and time.

Question 2.
What is derived physical quantity?
Answer:
A derived physical quantity is one that is expressed as a combination of fundamental quantities.
Examples include: velocity, force, and energy.

Question 3.
What is the SI unit of solid angle ?
Answer:
The SI unit of solid angle is Steradian (sr).

Question 4.
Write any one physical constant that is having dimensions.
Answer:
Gravitational constant (G) is a physical constant having dimensions.

Question 5.
Which physical quantity has negative dimensions in mass ?
Answer:
The Gravitational constant (G) has negative dimensions in mass (M^-1)

Question 6.
Which physical quantity is represented by √gR, where R is the radius of the Earth?
Answer:
It represents acceleration due to gravity (g).

Question 7.
Which physical quantity is represented by PV, where P is pressure and V is volume?
Answer:
The quantity PV represents energy (or work).

Question 8.
Name atleast one physical quantity whose SI unit is Nm^-2
Answer:
Pressure has SI unit newton per square metre (Nm^-2)

IV. Units and Measurements Very Short Answer Questions (2 Marks)

Question 1.
What are significant figures and what do they represent when reporting the result of a measurement?
Answer:
Significant figures are the reliably known digits in a measured quantity including the uncertain digit. They represent the precision of the measurement.

Question 2.
Distinguish between fundamental units and derived units.
Answer:
Fundamental units are the basic units defined for fundamental physical quantities (E.g.: metre for length, second for time).
Derived units are combinations of fundamental units used for derived physical quantities (E.g.: m/s for speed, N for force).

Question 3.
What are the two fundamental physical quantities used to measure angle?
Answer:
For plane angle : Length and radius (ratio of two lengths)
For solid angle : Area and radius (ratio of area to square of radius)

Question 4.
Why do we have different units for the same physical quantity ?
Answer:
Because different systems of units (like CGS, FPS, SI) exist due to historical and regional practices, leading to multiple units for the same quantity.

Question 5.
Define plane angle.
Answer:
Plane angle is defined as the ratio of the length of arc to the radius of the circle.

Question 6.
What is meant by dimension of a physical quantity ?
Answer:
The dimension of a physical quantity refers to the powers to which the base quantities (like mass, length, time) are raised to represent that quantity.

Question 7.
What is meant by dimensional formula ?
Answer:
A dimensional formula expresses a physical quantity in terms of base dimensions such as [M], [L], [T] with appropriate powers.

Question 8.
Write the physical quantities that are having Jm^-1 and Jm^-2 as their units.
Answer:
Jm^-1 : Surface tension
Jm^-2 : Pressure

V. Units and Measurements Short Answer Questions (4 Marks)

Question 1.
In a system of units, the unit of force is 100 N, unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system ?
Answer:
We use the dimensional formula for force:
F = ma = m.L/T²
Let: [F’] = 100N; [L’] = 10m; [T’] = 100s
We rearrange to find the new unit of mass [M’]:
[M’] = [F’]-[T]²/[L’]
= 100-(100)²/10
= 100-10⁴ /10
= 10⁵
The unit of mass in this system is 10⁵kg.

Question 2.
1 calorie = 2 J where 1J = 1 kgm²s ². Suppose we employ a system of units in which the unit of mass is a kg, the unit of length is pm, and the unit of time is ys. Show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of new units.
Answer:
Given:
1 calorie = 4.2 J, and 1J = 1 kg . m²/s²
In new system: Unit of mass = α kg; Unit of length = βm; Unit of time = γs
We rewrite 1J in the new unit system:
[J’] = α -β²γ^-2
So the numerical value of 1 calorie in new units:
= 4.2/α-β²-γ^-2
– 4.2.α^-1.β^-2.γ²
In the new system, 1 calorie = α^-1-β^-2γ²

Question 3.
The velocity of a body is given by v = At² + Bt + C, If v and t are expressed in SI units, what are the units and dimensions of A, B and C ?
Ans.
We are given: v = At² + Bt + C
where: v is velocity → SI unit: m/s, dimension: [LT¹]
t is time → SI unit: s
We compare the dimensions on both sides for each term:

i. At²: A x [T²] = [LT^-1] ⇒ [A] = [LT³]
Unit of A: m/s³

ii. Bt: B × [T] = [LT¹] ⇒ [B] = [LT²]
Unit of B : m/s²

iii. C : Since it is added to the other terms, it must have the same dimensions as velocity:
[C] = [LT¹]
Unit of C : m/s
∴ A: Unit = m/s³, Dimension = [LT³]
B: Unit = m/s², Dimension = [LT²]
C: Unit = m/s, Dimension = [LT¹]

Question 3.
What are fundamental quantities ? State their units in SI system.
Answer:
Fundamental quantities are those physical quantities that are independent and cannot be derived from other physical quantities. There are seven fundamental quantities in the SI system:

Quantity	Unit Name	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Thermodynamic temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Question 2.
Explain the uses of Dimensional methods with examples.
Answer:
Uses of Dimensional Methods:
1. To check the correctness of equations:
Using dimensional analysis, we can verify whether a physical equation is di­mensionally consistent.
Example:
To check if s = ut +1/2 at² is dimensionally correct:
LHS: [s] = [L]
RHS : [u][t] + [a][t²] = [LT^-1][T] + [LT^-2][T²] = [L] + [L] = [L]
‘Hence, it is dimensionally consistent.

2. To derive relationships between physical quantities :
If the dependency of a physical quantity is known on others, we can find its formula using dimensional analysis.
Example:
Derive the time period T of a simple pendulum:
Assume T a l^ag^b, where 1 is length, g is acceleration due to gravity.
Using dimensional method:
[T] = [L]^a[LT^-2]^b = L^a + bT – 2b ⇒ a + b = 0, -2b = 1 ⇒ b = -1/2, a = 1/2
So, T α √1/g

Question 6.
What are the rules for arithmetic operations with significant figures ?
Answer:

• Addition / Subtraction : The result should have the same number of deci­mal places as the term with the least decimal places.
• Multiplication / Division: The result must have the same number of signif­icant figures as the value with the fewest significant figures.
• Exact numbers (like counted values) do not affect significant figures. Always round the final result, not intermediate steps.

Question 7.
What is the rounding off a number and what are the rules to be followed in it ?
Answer:
Rounding off is the process of reducing the number of digits in a number while maintaining its approximate value based on significant figure rules.

Rules for Rounding Off :
1. If the digit to be dropped is less than 5, leave the preceding digit unchanged. Example : 32 → 7.3
2. If the digit is more than 5, increase the preceding digit by 1. Example : 7.38 → 7.4
3. If the digit is exactly 5:

• If the preceding digit is even, leave it unchanged. Example : 7.35 →7.3
• If the preceding digit is odd, increase it by 1. Example : 7.25 → 7.3

VI. Problems

Question 1.
If the velocity of light c, Planck’s constant h and the gravitational constant G are taken as fundamental quantities, then express mass, length and time in . terms of dimensions of these quantities.
Answer:
Here c = [LT^-1], h = [ML²/T^-1], G = [M^-1L³T^-2]
(i) Let m = c^x h^y G² ………………… (1)
⇒ [M¹L⁰T⁰] = [LT^-1]^X [ML²T^-1]^y [ML³T²]²
⇒ [M¹L⁰T⁰] = M^y-Z L^{x + 2y + 32} T^x-y-2z
Applying principle of homogenity of dimensions, we get
y- z = 1 ………………… (2)
x + 2y + 3z = 0 ………………… (3)
-x-y-2z = 0 ………………… (4)
Adding (2), (3) – and (4) 2y = 1 ⇒ y = 1/2
From (2) z = y -1 = \(\frac{1}{2}\) – 1 = \(\frac{-1}{2}\)
From (4) x = -y-2z= \(\frac{-1}{2}\) + 1 = \(\frac{1}{2}\)
Substitute values of x, y & z in eq (1) we get

(ii) Let i = c^x h^y G³ ____ (5)
⇒ [M¹L⁰T⁰] = [LT^-1]^x [ML²T^-1]^y [M^-1L³r^-2]^z
⇒ [M¹L⁰T⁰] = M^y-2 L^x=-2y+3z T^-x-y-2z
Applying principle of homogenity of dimensions, we get
y-z = 0
x + 2y + 3z = 1
-x – y – 2z = 0
On solving the above equations, we get
\(x=\frac{-3}{2}, y=\frac{1}{2}, z=\frac{1}{2}\)
Substituting the values of x, y and z in eq (5) we get
l = C^3/2 h^1/2 G^1/2
⇒ l=\(\sqrt{\frac{\mathrm{hG}}{\mathrm{c}^3}}\)

Question 2.
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
\(\mathrm{I}=\frac{\mathrm{K}}{\mathrm{R}} \sqrt{\frac{\mathrm{r}^3}{\mathrm{~g}}}\)
Where k is a dimensionless constant and g is acceleration due to gravity.
Answer:
Given that T² α r³ Or T α r^3/2
Also, T is a function of g and R
Let T α r^3/2 g^a R^b T = K r^3/2 g^a R^b  ……………… (1)
Where ‘K’ is dimensonless constant of proportionality
From equation (1) [M⁰L⁰T¹] = L^3/2 [LT²]^a[L]^b
= M⁰L^a+b+2T^-2a
Applying the principle of homogenity of dimensions, we get
a + b +p\(\frac{3}{2}\) = 0    ……………… (2)
-2a = 1 ⇒ a = -1/2
From equation (2)  \(\frac{-1}{2}+b+\frac{3}{2}=0\) ⇒ b=-1
Substituting the values of a and b in equation (1)
we get T=Kr^3/2 g^-1/2R^-1
T=\(\frac{K}{R} \sqrt{\frac{r^3}{g}}\)

Question 3.
State the number of significant figures in the following.
a) 6729
b) 0.024
c) 0.08240
d) 6.032
e) 4.57 x10^-8
Answer:
a) In 6729 all are significant figures. So number of significant figures in 6729 are
b) In 0.024 the zero’s to the left of 1^st non-zero digit of a numbers are insignifi­cant. So number of significant figures in 0.024 are 2.
c) In 0.08240 the zero’s to the left of first non-zero digit of a numbers are insig­nificant. So number of significant figures in 0.08240 are 4.
d) In 6.032 the zero between two non-zero digits is significant. So number of significant figures in 6.032 are 4.
e) In 4.57 x 10⁸ the number of significant figures are 3.

Question 4.
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 lengths ?
Answer:
a) When two sticks placed end to end
total length, l = l₁+ l₂
l₁+ l₂ = 12.132 + 12.4 = 24.532 cm.
In addition final answer must have least number of significant digits in that addition, i.e., one after decimal point. …………..
Hence, the total length is 24.5 cm.

b) If two sticks are placed side by side, the difference in lengths,
l₁– l₂ = 12.4 – 12.132 = 0.268 cm.
In subtraction final answer must be adjusted to least number of signifi­cant figure in that operation.
Here least number is one digit after decimal on rounding off, the final answer is 0.3 cm.

Question 4.
Each side of a cube is measured to be 7.203 m. What is (i) the total surface area and (ii) the volume of the cube, to appropriate significant figures ?
Answer:
Side of cube, a = 7.203 m
Here number of significant figures are four.

i. Surface area of cube = 6a² = 6 x (7.203)² = 311.299
But final answer must be rounded to least number of significant figures i.e., four digits.
So surface area of cube = 311.3 m²

ii. Volume of cube, v = a³ = (7:203)³ = 373.147
the final anwer must be limited to four significant figures.
∴ volume of sphere v = 373.1 m³

Andhra Pradesh BIEAP IPE AP Inter 1st Year Physics Study Material PDF Free Download, AP Inter 1st Year Physics Weightage BluePrint 2025-2026, AP Board Solutions Class 11 Physics, Telugu Academy Physics Intermediate 1st Year Textbook Pdf Free Download Telugu Medium, Physics Study Material Pdf Class 11, Inter First Year Physics Textbook Pdf Solutions, Intermediate 1st Year Physics Study Material Pdf are part of AP Inter 1st Year Study Material Pdf.

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AP Inter 1st Year Physics Study Material Pdf Download

Inter 1st Year Physics Study Material Pdf English Medium

AP Inter 1st Year Physics Textbook Solutions Pdf | Physics Inter 1st Year Textbook Solutions Pdf

• Chapter 1 Units and Measurements Class 11 Questions and Answers
• Chapter 2 Motion in a Straight Line Class 11 Questions and Answers
• Chapter 3 Motion in a Plane Class 11 Questions and Answers
• Chapter 4 Laws of Motion Class 11 Questions and Answers
• Chapter 5 Work, Energy and Power Class 11 Questions and Answers
• Chapter 6 System of Particles and Rotational Motion Class 11 Questions and Answers
• Chapter 7 Gravitation Class 11 Questions and Answers
• Chapter 8 Mechanical Properties of Solids Class 11 Questions and Answers
• Chapter 9 Mechanical Properties of Fluids Class 11 Questions and Answers
• Chapter 10 Thermal Properties of Matter Class 11 Questions and Answers
• Chapter 11 Thermodynamics Class 11 Questions and Answers
• Chapter 12 Kinetic Theory Class 11 Questions and Answers
• Chapter 13 Oscillations Class 11 Questions and Answers
• Chapter 14 Waves Class 11 Questions and Answers

AP Inter 1st Year Physics Study Material Pdf Telugu Medium

AP Intermediate Physics Textbook Solutions Telugu Medium Pdf Free Download

• Chapter 1 భౌతిక ప్రపంచం
• Chapter 2 ప్రమాణాలు, కొలత
• Chapter 3 సరళరేఖాత్మక గమనం
• Chapter 4 సమతలంలో చలనం
• Chapter 5 గమన నియమాలు
• Chapter 6 పని, శక్తి, సామర్ధ్యం
• Chapter 7 కణాల వ్యవస్థలు, భ్రమణ గమనం
• Chapter 8 డోలనాలు
• Chapter 9 గురుత్వాకర్షణ
• Chapter 10 ఘనపదార్ధాల యాంత్రిక ధర్మాలు
• Chapter 11 ప్రవాహుల యాంత్రిక ధర్మాలు
• Chapter 12 పదార్ధ ఉష్ణ ధర్మాలు
• Chapter 13 ఉష్ణోగతిక శాస్త్రం
• Chapter 14 అణుచలన సిద్ధాంతం

AP Inter 1st Year Physics Weightage Blue Print 2025-2026

AP Inter 1st Year Physics Weightage 2025-2026 | AP Inter 1st Year Physics BluePrint 2026

Intermediate 1st Year Physics Syllabus

AP Inter 1st Year Physics Syllabus

Chapter 1 Units and Measurements

• 1.1 Introduction
• 1.2 The International system of units
• 1.3 Significant figures
• 1.4 Dimensions of Physical Quantities
• 1.5 Dimensional formulae and dimensional equations
• 1.6 Dimensional analysis and its applications

Chapter 2 Motion in a Straight Line

• 2.1 Introduction
• 2.2 Instantaneous velocity and speed
• 2.3 Acceleration
• 2.4 Kinematic equations for uniformly accelerated motion

Chapter 3 Motion in a Plane

• 3.1 Introduction
• 3.2 Scalars and vectors
• 3.3 Multiplication of vectors by real members
• 3.4 Addition and subtraction of vectors – graphical Method
• 3.5 Resolution of vectors
• 3.6 Vector addition – Analytical method
• 3.7 Motion in a plane
• 3.8 Motion in a plane with constant acceleration
• 3.9 Projectile motion
• 3.10 Uniform circular motion

Chapter 4 Laws of Motion

• 4.1 Introduction
• 4.2 Aristotle’s fallacy
• 4.3 The law of inertia
• 4.4 Newton’s first law of motion
• 4.5 Newton’s second law of motion
• 4.6 Newton’s third law of motion
• 4.7 Conservation of momentum
• 4.8 Equilibrium of a particle
• 4.9 Common forces in mechanics, friction
• 4.10 Circular motion
• 4.11 Solving problems in mechanics

Chapter 5 Work, Energy, and Power

• 5.1 Introduction
• 5.2 Notions of work and kinetic energy: The work-energy theorem
• 5.3 Work
• 5.4 Kinetic Energy
• 5.5 Work done by a variable force
• 5.6 The work-energy theorem for a variable force
• 5.7 The concept of Potential energy
• 5.8 The conservation of mechanical energy
• 5.9 The Potential energy of a spring
• 5.10 Power
• 5.11 Collisions

Chapter 6 System of Particles and Rotational Motion

• 6.1 Introduction
• 6.2 Centre of mass
• 6.3 Motion of Centre of Mass
• 6.4 Linear momentum of a system of particles
• 6.5 Vector product of two vectors
• 6.6 Angular velocity and its relation with linear velocity
• 6.7 Torque and angular momentum
• 6.8 Equilibrium of a rigid body and centre of gravity
• 6.9 Moment of inertia
• 6.10 Kinematics of rotational motion about a fixed axis
• 6.11 Dynamics of rotational motion about a fixed axis
• 6.12 Angular momentum in the case of rotations about a fixed axis

Chapter 7 Gravitation

• 7.1 Introduction
• 7.2 Kepler’s laws
• 7.3 Universal law of gravitation
• 7.4 The gravitational constant
• 7.5 Acceleration due to gravity of the Earth
• 7.6 Acceleration due to gravity below and above the surface of Earth
• 7.7 Gravitational potential energy
• 7.8 Escape speed
• 7.9 Earth Satellites
• 7.10 Energy of an orbiting satellite

Chapter 8 Mechanical Properties of Solids

• 8.1 Introduction
• 8.2 Stress and strain
• 8.3 Hooke’s law
• 8.4 Stress – strain curve
• 8.5 Elastic moduli
• 8.6 Applications of the elastic behavior of materials

Chapter 9 Mechanical Properties of Fluids

• 9.1 Introduction
• 9.2 Pressure
• 9.3 Streamline flow
• 9.4 Bernoulli’s principle
• 9.5 Viscosity
• 9.6 Surface tension

Chapter 10 Thermal Properties of Matter

• 10.1 Introduction
• 10.2 Temperature and Heat
• 10.3 Measurement of temperature
• 10.4 Ideal-gas equation and absolute temperature
• 10.5 Thermal expansion
• 10.6 Specific Heat Capacity
• 10.7 Calorimetry
• 10.8 Change of state
• 10.9 Heat transfer
• 10.10 Newton’s law of cooling

Chapter 11 Thermodynamics

• 11.1 Introduction
• 11.2 Thermal equilibrium
• 11.3 Zeroth law of thermodynamics
• 11.4 Heat, internal energy, and work
• 11.5 First law of thermodynamics
• 11.6 Specific heat capacity
• 11.7 Thermodynamic state variables and equation of state
• 11.8 Thermodynamic processes
• 11.9 Second law of thermodynamics
• 11.10 Reversible and irreversible processes
• 11.11 Carnot engine

Chapter 12 Kinetic Theory

• 12.1 Introduction
• 12.2 Molecular nature of matter
• 12.3 Behavior of gases
• 12.4 Kinetic theory of an ideal gas
• 12.5 Laws of equipartition of energy
• 12.6 Specific heat capacity
• 12.7 Mean free path

Chapter 13 Oscillations

• 13.1 Introduction
• 13.2 Periodic and oscillatory motions
• 13.3 Simple Harmonic motion
• 13.4 Simple Harmonic motion and uniform circular motion
• 13.5 Velocity and acceleration in simple harmonic motion
• 13.6 Force law for simple harmonic motion
• 13.7 Energy in simple harmonic motion
• 13.8 The Simple Pendulum

Chapter 14 Waves

• 14.1 Introduction
• 14.2 Transverse and longitudinal waves
• 14.3 Displacement relation in a progressive wave
• 14.4 The speed of a travelling wave
• 14.5 The principle of superposition of waves
• 14.6 Reflection of waves
• 14.7 Beats

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