Category: Class 11 Physics Objective Questions

Physics Online Test for Class 11 on “Thermodynamics – Heat, Internal Energy and Work”.

1. Internal energy of a system is defined as?
a) The sum of kinetic energies of all molecules of the system
b) The sum of kinetic and potential energies of all molecules of the system
c) The sum of potential energies of the system
d) The average kinetic energy of all molecules
Answer: b
Clarification: The internal energy of a system corresponds to the energy possessed by all molecules. Thus it is the sum of kinetic and potential energies of all molecules in the system considered. Also note that potential energy is frame dependent, so we choose a frame in which the centre of mass is at rest.

2. Select the correct statement.
a) Internal energy is a path variable
b) Heat is a path variable
c) Work done is a state variable
d) Internal energy is a microscopic variable
Answer: b
Clarification: Internal energy corresponds to the mechanical energy of molecules. It is a state variable as it doesn’t depend on the path taken. While heat is the energy in transit, so it is a path variable. Internal energy is a thermodynamic variable and is therefore macroscopic as thermodynamics deals with bulk systems.

3. A box contains 10⁵ molecules. The average kinetic energy of each molecule is 0.5*10^-6J. With respect to an observer at rest the box, having a mass of 1kg, is moving with a velocity of 2m/s. What is the value of kinetic energy that will contribute to the internal energy?
a) 0.05J
b) 2.05J
c) 0
d) 2J
Answer: a
Clarification: When calculating internal energy of molecules, the system is to be studied from a frame in which the centre of mass is at rest. So the kinetic energy due to movement of the entire box will not be considered. Only the individual kinetic energies of molecules will be considered, and their value is: 10⁵*0.5*10^-6
= 0.05J.

4. If some heat is provided to a system and work is done on it, it is possible that internal energy will decrease. True or False?
a) True
b) False
Answer: b
Clarification: Internal energy corresponds to the mechanical energy of the molecules. When these molecules are provided with heat and are compressed their mechanical energy will increase. Both factors support the increase of internal energy hence the given statement is false.

5. Which of the following types of kinetic energies contribute to internal energy?
a) Translational
b) Rotational
c) Vibrational
d) Translational, Vibrational & Rotational
Answer: d
Clarification: Internal energy consists of the total kinetic energy of all molecules, which means all types, namely translational, rotational & vibrational kinetic energies contribute to internal energy.

6. A body has 15J of heat energy in a particular state. At the same state it has an internal energy of 30J. Assume no work is done, what will happen to the internal energy if 5J of heat is added to the system?
a) Heat energy = 20J, Internal energy = 35J
b) Heat energy = 15J, Internal energy = 30J
c) Heat energy = 15J, Internal energy = 35J
d) A body cannot have heat energy in a particular state
Answer: d
Clarification: Heat energy is the energy in transit. A system’s state cannot be defined by heat energy, it is defined by internal energy. Heat and work are path functions. Heat can be supplied or removed from a system but can’t define a system in a state.

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Physics Quiz Online for Class 11 on “Oscillations – Velocity and Acceleration in Simple Harmonic Motion”.

1. A particle is executing SHM and currently going towards the amplitude. If it is at A/2, what is the relation between the direction of velocity and acceleration?
a) Both vectors point towards the amplitude
b) Velocity is towards amplitude & acceleration is towards mean position
c) Velocity is towards mean position & acceleration is towards the amplitude
d) Both vectors point towards the mean position
Answer: b
Clarification: The force on a particle in SHM is always towards the mean position. The particle is currently going towards an extreme, thus velocity will be towards the amplitude.

2. In a SHM, for what value of w, will the magnitude of maximum acceleration be greater than the magnitude of maximum velocity? Here, w is angular frequency.
a) w > 1
b) w < 1
c) w = 0
d) Not possible for any value of w
Answer: a
Clarification: Let the velocity equation of SHM be: v = Awcos(wt),
acceleration will be given by: a = -Aw²sin(wt).
Maximum magnitude of vel = Aw
Maximum magnitude of acc = Aw².
For, Aw² > Aw
w > 1.

3. A particle of mass m starts from the mean position of a SHM, at t=0, and goes towards -A. If the angular frequency of SHM is w, find the force acting on it as a function of time.
a) mAw²sin(wt)
b) mAw²sin(wt+π)
c) -mAw²cos(wt+π)
d) -mAw²cos(wt)
Answer: a
Clarification: The displacement equation will be given by: x = -Asin(wt).
On taking its derivative, we get:
v = -Awcos(wt).
Further, we get: a = Aw²sin(wt).
Thus, force is given by: F(t) = mAw²sin(wt).

4. The graph of acceleration vs time for a SHM is given. FInd the equation of position of the particle as a function of time. Assume that the particle is oscillating on the x-axis about the origin. And consider RHS of origin to be positive & LHS to be negative.

a) 8cos(πt)
b) -8cos(πt)
c) 0.81sin(πt)
d) -0.81sin(πt)
Answer: d
Clarification: The particle has zero acceleration at t=0, so it starts from the mean position. Then the acceleration becomes positive which implies that the particle is moving towards -A ( negative extreme). The equation of acceleration can be given by: a = 8sinwt.
The time period is 2s,
so w = 2π/2 = πs^-1.
∴ a = 8sin(πt).
∴ v = -8/π cos(πt)
∴ x = -8/pi²sin(πt) = -0.81sin(πt).

5. A particle is undergoing a SHM of amplitude 10cm. What should be the minimum value of acceleration at an extreme position for maximum speed at centre to be 5m/s?
a) 20m/s²
b) 5m/s²
c) 0
d) 250m/s²
Answer: d
Clarification: v_max = Aw.
5 = 0.1w.
w = 50s^-1.
a_min at extreme = Aw²
= 0.1*50*50
= 250m/s².

6. A particle in uniform circular motion is projected on its diameter. The motion of projection will be simple harmonic. Select the correct option regarding speed and acceleration of the particle in circular motion.
a) Speed is constant, acceleration is zero
b) Speed is constant, acceleration is non-zero
c) Speed changes, acceleration is non-zero
d) Speed changes, acceleration is zero
Answer: a
Clarification: For the particle in uniform circular motion, magnitude of velocity is constant, but direction is continuously changing. Thus, speed is constant but acceleration is non-zero.

7. The curves given below are that of position(x), velocity(v) & acceleration(a) vs time. What can we infer about the value of angular frequency ‘f’ from the given curves?

a) f > 1
b) f < 1
c) f = 0
d) We need the time period in order to find ‘f’
Answer: a
Clarification: Amplitude of position = A
Amplitude of velocity = Af
Amplitude of acceleration = Af².
From the graph we deduce that the amplitude of a is greater than the amplitude of v which in turn is greater than the amplitude of x.
Thus, angular frequency will be greater than 1.

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Physics Multiple Choice Questions on “Power”.

1. What are the units of power?
a) Newton
b) Joule
c) Watt
d) No units
Answer: c
Clarification: The SI unit of power is “Watt”. It is named after the famous inventor “James Watt” who is widely remembered for his improvements on the steam engine.
1 Watt = 1 Joule per second.

2. A machine gun fires 360 bullets per minute. Each bullet has a mass of 5 grams and travels at 600 m/s. What is the power of the gun? (Assume no loss of energy and 100% power transmission)
a) 300 W
b) 600 W
c) 900 W
d) 1800 W
Answer: c
Clarification: No. of bullets fired per second = 360/60 = 6
Power of the gun is completely transmitted to bullets.
Power (P) = Power of bullets in 1 second
= [(1/2 x m x v²) x 6] / 1
m = 5 x 10^-3 kg
v = 600 m/s
P = (1/2 x 5 x 10^-3 x 600² x 6) / 1
= 900 W.

3. A motor is used to deliver water through a pipe. Let the motor have a power P initially. To double the rate of flow of water through the pipe, the power is increased to P’. What is the value of P’/P?
a) 2
b) 4
c) 6
d) 8
Answer: d
Clarification: Power = Force x Velocity
= (Mass/Time) x Velocity²
= Rate of flow x Velocity²
Rate of flow = Area x Velocity
Doubling the rate of flow will double the velocity.
P = Rate of flow x Velocity²
P’ = (2 x Rate of flow) x (2 x Velocity) ²
= 8P
P/P’ = 8.

4. A motor is used to pump water from a depth of 5 m to fill a volume of 10 cubic meters in 5 minutes. If 50% of the power is wasted, what is the power of the motor? (Assume g = 10m/s²)
a) 1000/3 W
b) 3000/3 W
c) 5000/3 W
d) 10000/3 W
Answer: d
Clarification: Density of water = 1000 kg/m³
Mass of water in 10m³ = 1000 x 10
= 10,000 kg
Volume of water filled in 1s = 10/(5×60)
= 1/30 m³
Mass filled in 1s = 1/30 x 1000
= 100/3 kg
Power = 100/3 x 10 x 5; [g = 10m/s²]
= 5000/3 W
Since 50% of power is wasted, power of motor = 5000/3 x 2
= 10000/3 W.

5. A machine has a power of 20kW. How long will it take for it to lift a body of mass 10kg from the ground to a height of 100m? (Assume g = 10m/s²)
a) 1 second
b) 2 seconds
c) 3 seconds
d) 4 seconds
Answer: b
Clarification: Power (P) = Energy (E) x Time (t)
20,000 = (m x g x h) x t
= (10 x 10 x 100) x t
t = 2 seconds.

6. A 2-ton vehicle travels at 20m/s on a road where the frictional force is 10% of the weight of the vehicle. What is the power required? (Assume g = 10m/s²)
a) 220 kW
b) 240 kW
c) 420 kW
d) 440 kW
Answer: d
Clarification:Weight of car = m x g
= 2,000 x 10
= 20,000 N
Frictional force = (10/100) x (20,000)
= 2,000 N
Total force (F) = 20,000 + 2,000
= 22,000 N
Velocity (v) = 20 m/s
P = F x v
= 22,000 x 20
= 4,40,000 W
= 440 kW.

7. 1000 kW of power is supplied to a motor. 90% of this is transmitted to a machine operating at 80% efficiency. The machine lifts an object of 10-tons with a velocity _____ m/s. (Assume g = 10m/s²)
a) 2
b) 4
c) 6
d) 8
Answer: c
Clarification: Power transmitted to machine = (90/100) x (1000) kW
= 900 kW
Power used to lift object = (80/100) x 900 KW
= 720 kW
7,20,000 = F x v
= (1/2 x m x v²) x v
= 10,000/2 x v³
v³ = 216
v = 6 m/s.

8. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 10 m/s. How much power is delivered by the engine? (Assume g = 10m/s²)
a) 10 kW
b) 20 kW
c) 30 kW
d) 40 kW
Answer: b
Clarification: Weight of car = m x g
= 2,00 x 10
= 2,000 N
Total force (F) = 2,000
Velocity (v) = 10 m/s
P = F x v
= 2000 x 10
= 20,000 W
= 20 kW.

9. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 36 km/h. How much power is delivered by the engine if the frictional force is 25% of the weight of the car? (Assume g = 10m/s²)
a) 25 kW
b) 50 kW
c) 100 kW
d)125 kW
Answer: a
Clarification: Weight of car = m x g
= 200 x 10
= 2,000 N
Frictional force = (25/100) x (2,000)
= 500 N
Total force (F) = 2000 + 500
= 2,500 N
Velocity (v) = 36 km/h
= 36 x (1000/3600)
= 10 m/s
P = F x v
= 2500 x 10
= 25 kW.

10. A body of mass 3 kg starts from rest with a uniform acceleration of unknown magnitude. If the body has a velocity of 30 m/s in 6 seconds, what is the power consumed in 3 seconds?
a) 125 W
b) 225 W
c) 325 W
d) 425 W
Answer: b
Clarification: v = u + at
30 = a x 6
a = 5 m/s²
Force (F) = m x a
= 3 x 5
= 15 N
v’ = u + at’
v’ = 5 x 3
= 15 m/s
Power consumed in 3 seconds = F x v’
= 15 X 15
= 225 W.

11. The force required to tow a vehicle at constant velocity is directly proportional to the magnitude of velocity raised to the first power. It requires 1000 W to tow with a velocity of 10 m/s. How much power is required to tow at a velocity of 8 m/s?
a) 320 W
b) 640 W
c) 160 W
d) 80 W
Answer: b
Clarification: Since force is directly proportional to velocity, the power required will be directly proportional to the square of the velocity.
Power = Force x Velocity
Hence, power is a multiple of 8² and an integer such that;
1000/10² = P/8²
P = 640 W.

Physics,

Physics MCQs on “Acceleration due to Gravity of the Earth”.

1. The acceleration due to gravity on the surface of the earth is different at different points on the surface.
a) True
b) False
Answer: a
Clarification: Since the earth is not a perfect sphere and has many irregularities, the acceleration due to gravity is different at different points on the earth’s surface.

2. The acceleration due to gravity on the surface of the earth is _____
a) greater towards the equator and lesser towards the poles
b) lesser towards the equator and greater towards the poles
c) same at all points on the surface of the earth
d) same everywhere except at the poles
Answer: b
Clarification: The earth is not a perfect sphere. The radius of the earth at the equator is greater than that at the poles, hence, the acceleration due to gravity is lesser towards the equator and greater towards the poles.

3. The dimensions of acceleration due to gravity are _____
a) [M⁰L¹T^-2]
b) [M¹ L^-1T^-2]
c) [M^-1 L² T^-1]
d) [M⁰L^-1T²]
Answer: a
Clarification: The unit of acceleration is m/s².
m/s² = [L¹ T^-2]
= [M⁰ L¹ T^-2].

4. For an object on the surface of the earth, the magnitude of the acceleration due to the gravity of the earth it experiences depends also depends on the mass of that object.
a) True
b) False
Answer: b
Clarification: g = (G*M1)/R²;
M1 = Mass of the earth
The acceleration due to the gravity of the earth experienced by any object on the surface of the earth depends only on the mass of earth and the square of the distance between the object and the centre of the earth.

5. What would be the magnitude of the acceleration due to gravity on the surface of the earth if the radius of the earth were reduced by 20%?
a) 9.81 m/s²
b) 12.26 m/s²
c) 15.33 m/s²
d) 49.05 m/s²
Answer: c
Clarification: g = (G*M1)/R²;
M1 = Mass of the earth
R = Radius of the earth
We know; g = 9.81 m/s²
If the radius is reduced by 20% then the new radius is 80% of the original one;
New radius = 0.8R
Therefore, new acceleration;
g / (0.8 x 0.8) = 9.81 / 0.64
= 15.33 m/s².

6. What would be the magnitude of the acceleration due to gravity on the surface of the earth if the density of the earth increased by 3 times and the radius remained the same?
a) 9.81 m/s²
b) 12.26 m/s²
c) 15.33 m/s²
d) 29.43 m/s²
Answer: d
Clarification: g = (G*M1)/R²;
M1 = Mass of the earth
R = Radius of the earth
Since the radius is the same, the volume would remain constant.
Density = mass/volume
Since density is increased 3 times and volume is the same, this implies that mass is increased 3 times.
We know; g = 9.81 m/s²
New mass = 3 x M1
Therefore, new acceleration;
3 x (G*M1)/R² = 3x g
= 3 x 9.81
= 29.43 m/s².

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