Category: Engineering Physics Objective Questions

250+ TOP MCQs on Forces of Nature and Answers

Engineering Physics Multiple Choice Questions on “Forces of Nature”.

1. When we steer a bicycle round a bend with both our hands on the handlebars, we apply a ___________
a) Force
b) Torque
c) Couple
d) Energy
Answer: c
Clarification: A pair of equal and opposite forces acting on a body along two different lines of action constitutes a couple. A Couple has a turning effect, but no resultant force acts on a body. When we steer a bicycle round a bend with both our hands on the handlebars, we apple a couple.

2. Couple cannot produce a translational motion.
a) True
b) False
Answer: a
Clarification: A pair of equal and opposite forces acting on a body along two different lines of action constitutes a couple. A couple has a turning effect, but no resultant force acts on a body. Thus it cannot produce translational motion.

3. The areal velocity of a planet is constant. This is stated by ___________
a) Newton’s third law of motion
b) Kepler’s second law of planetary motion
c) Fourier’s law
d) Kirchhoff’s planetary law
Answer: b
Clarification: A planet revolves around the sun under the influence of a gravitational force which acts towards the sun. This means that the areal velocity of a planet is constant. The is Kepler’s second law of planetary motion which states that the line joining the planet to the sun sweeps out equal areas in equal intervals of time.

4. The internal torque on the system due to internal force is ___________
a) Increasing
b) Decreasing
c) Constant
d) Zero
Answer: d
Clarification: According to Newton’s third law, the internal torque on the system due to internal forces is zero because the forces between any two particles are equal and opposite and directed along the line joining the two particles. Hence the total torque is due to external forces only.

5. Which of the following can execute both translational and rotational motion under the influence of the external field?
a) Rigid body
b) Ideal body
c) Rotating body
d) A body which is stationary
Answer: a
Clarification: A rigid body is one for which the distance between different particles does not change, even though they move. Under the influence of an external force, a rigid body can execute rotational and translational motion.

6. For a body in equilibrium, the linear acceleration of its centre of mass would be zero and also the angular acceleration of the rigid body about any axis would be zero.
a) True
b) False
Answer: a
Clarification: A rigid body is in equilibrium if both the linear and angular momentum of the rigid body remains constant with time. Hence for a body on equilibrium, the linear acceleration of its centre of mass would be zero and also the angular acceleration of the rigid body about any axis would be zero.

7. A spring 40mm ling is stretched by the application of force. If 10N force is required to stretch the spring through 1mm, then the work done in stretching the spring through 40mm is?
a) 23J
b) 68J
c) 84J
d) 8J
Answer: d
Clarification: Spring constant, k = F/x = 10N/1mm = 10N/(10-3 m)= 104 N/m
Work done in stretching the spring through 40m,
W = 1/2 kx2=1/2×104×(40×10-3)2 = 8J.

8. Two bodies w of masses m and 4m are moving with equal kinetic energies. What is the ratio of their linear momenta?
a) 1:2
b) 1:1
c) 4:1
d) 1:4
Answer: a
Clarification: p = √2mE
For same kinetic energy, p1/p2 = √(m1/m2) = √(m/4m)=1:2.

9. Water falls from a height of 60m at the rate of 15kg.s to operate a turbine. The losses due to friction forces are 10% of energy. How much power is generated by the turbine? g=m/s2
a) 7kW
b) 10.2kW
c) 12.3kW
d) 8.1kW
Answer: d
Clarification: P = W/t=mgh/t=(15×10×60)/1 = 9000W = 9kW
Loss=10%
Useful power=90% of 9kW = (90×9)/100=8.1kW.

10. A shell explodes into four unequal parts. Which one of the following is conserved?
a) Potential energy
b) Kinetic energy
c) Momentum
d) Both potential and kinetic energy
Answer: c
Clarification: In the absence of external force, linear momentum is conserved. If a shell explodes into four equal parts, momentum is conserved.

250+ TOP MCQs on Thrust of a Liquid and Answers

Engineering Physics Questions and Answers for Campus interviews focuses on “Thrust of a Liquid”.

1. What will be the length of a mercury column in a barometer tube, when the atmospheric pressure is 75cm of mercury and the tube is inclined at an angle of 60° with the horizontal direction?
a) 8.66cm
b) 86.6cm
c) 866cm
d) 0.866cm
Answer: b
Clarification: If l is the length of a mercury column in the barometer tube, then
h/l=sin60
75/l=√3/2
l=(75×2)/√3=86.6cm.

2. The density of the atmosphere at sea level is 1.29kgm-3. Assume that is does not change with altitude. Then how high would the atmosphere extend? (g=9.8mm-2)
a) 0.5km
b) 0.8km
c) 8km
d) 80km
Answer: c
Clarification: Pa=hρg
h=Pa/ρg=(1.01×105)/(1.29×9.81)=7981m=8km(approx).

3. It is painful to walk barefooted on the ground covered with edged pebbles.
a) True
b) False
Answer: a
Clarification: While walking, when the entire weight of our body gets supported on the sharp edge of any pebble, it will exert a large pressure on our feet due to the reaction. This causes considerable pain on our feet.

4. A closed compartment containing gas in moving with some acceleration on horizontal direction. Neglect effect of gravity. Then the pressure in the compartment is ___________
a) Same everywhere
b) Lower in the front side
c) Lower in the rear side
d) Lower in the upper side
Answer: b
Clarification: As the compartment has acceleration in the forward direction, the gas molecules experience an acceleration in the backward direction. This lowers the pressure in the compartment in the front side.

5. Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.
Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.
a) Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1
b) Statement 1 is true, statement 2 is true. Statement 2 is not a correct explanation for statement 1
c) Statement 1 is true, statement 2 is false
d) Statement 1 is false, statement 2 is true
Answer: a
Clarification: Volume rate of flow, Q=Av = constant
As speed of upstream decreases, its area of cross-section increases. As the speed of downstream increases, its area of cross-section decreases.

6. If two soap bubbles of different radii are connected by a tube ___________
a) Air flow from the bigger bubble to the smaller bubble till the sizes become equal
b) Air flows from bigger bubble to the smaller bubble till the sizes are interchanged
c) Air flows from the smaller bubble to the bigger
d) There is no flow of air
Answer: c
Clarification: The excess pressure inside of a soap bubble is inversely proportional to its radius. So, the pressure is more inside the smaller bubble than the bigger bubble. When these two bubbles are connected by a tube, air will flow from the smaller bubble to the bigger bubble.

7. A 20cm long capillary tube is dipped in water. The water rises up to 8cm. If the entire arrangement is put in a freely falling elevator the length of the water column in the capillary tube will be ___________
a) 4cm
b) 20cm
c) 8cm
d) 10cm
Answer: b
Clarification: In the freely falling elevator, the entire arrangement is in a state of weightlessness, g = 0. So, water will rise (h = 2σcosθ/rρg ) to fill the entire 20cm length of the tube.

8. Assertion: A this stainless steel needle can lay floating on a still water surface.
Reason: Any objection floats, when the buoyancy force balances the weight of the object.
a) Both assertion and reason are true and the reason is the correct explanation of the assertion
b) Both assertion and reason are true but the reason is not the correct explanation of the assertion
c) Assertion is true but the reason is false
d) Assertion and reason is false
Answer: c
Clarification: The assertion is true but the reason is false. The needle floats when the upward tension on the needle balances its weight.

9. At a critical temperature, the surface tension of a liquid ___________
a) Is zero
b) Is infinity
c) Is the same as any other temperature
d) Cannot be determined
Answer: a
Clarification: At a critical temperature, the surface tension of a liquid becomes zero.

10. In a capillary tube experiment, a vertical, 30 cm long capillary tube is dipped in water. The water rises up to a height of 10 cm due to capillary action. If this experiment is conducted in a freely falling elevator, the length of the water column becomes ___________
a) 10 cm
b) 20 cm
c) 30 cm
d) Zero
Answer: c
Clarification: In the freely falling elevator, g = 0.
Water will rise (h=2σcosθ/rρg) to fill the entire 30 cm length of the tube.

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250+ TOP MCQs on Specific Heat and Thermodynamics and Answers

Engineering Physics Multiple Choice Questions on “Specific Heat and Thermodynamics”.

1. The internal energy change in a system that has absorbed 2kcal of heat and done 500 J of work is?
a) 6400 J
b) 5400 J
c) 7900 J
d) 8900 J
Answer: c
Clarification: As Q=∆U+W
∆U=Q-W=2×4.2×1000-500
∆U=8400-500=7900 J.

2. 110 J of heat is added to a gaseous system, whose internal energy increases by 40 J. Then the amount of external work done is?
a) 150 J
b) 70 J
c) 110 J
d) 40 J
Answer: b
Clarification: ∆Q=+110 J, ∆U=+4J
∆W=∆Q-∆U=110-40=70J.

3. The molar specific heat constant pressure of an ideal gas is 7R/2. The ratio of specific heat at constant pressure to that at constant volume is?
a) 9/7
b) 8/7
c) 7/5
d) 5/7
Answer: c
Clarification: Cp=7R/2
CV=CP-R=7R/2-R=5R/2
r=Cp/Cv = (7R/2)/(5R/2)=7/5.

4. The change internal energy in a cyclic process is ___________
a) Zero
b) Infinity
c) Constant
d) Unity
Answer: a
Clarification: The change in internal energy in a cyclic process is zero because the system returns to its initial state.

5. It is possible that the temperature of the body changes even without giving heat to it or taking heat from it.
a) True
b) False
Answer: a
Clarification: During an adiabatic compression, temperature increases and an adiabatic expansion, temperature decreases, although no heat is given to or taken from the system in these changes.

6. The mechanical energy can be completely converted into heat energy but the whole of the heat energy cannot be converted into mechanical energy.
a) True
b) False
Answer: a
Clarification: The whole of mechanical energy can be absorbed by the molecules of the system in the form of their kinetic energy. This kinetic energy gets converted into heat. But the whole of the heat energy cannot be converted into work as a part of it is always retained by the system as its internal energy.

7. Which statement is incorrect?
a) All reversible cycles have the same efficiency
b) Reversible cycle has more efficiency than an irreversible one
c) Carnot cycle is a reversible one
d) Carnot cycle has the maximum efficiency of the cycles
Answer: a
Clarification: Work done per cycle = Area of the loop representing the cycle
As different reversible cycles may have different loop areas, their efficiencies will also be different.

8. Which is an intensive property?
a) Volume
b) Mass
c) Refractive index
d) Weight
Answer: c
Clarification: An intensive property is that which does not depend on the quality of matter of the system. Refractive index is an intensive property. Volume, mass and weight are extensive properties.

9. The latent heat of vaporisation of water is 2,240 J. If the work done in the process of vaporisation of 1g is 168 J, then the increase in internal energy is?
a) 2408 J
b) 2240 J
c) 2072 J
d) 1904 J
Answer: c
Clarification: From the first law of thermodynamics,
dQ = mL = dU + dW
dU = m L – dW = 1×2240-168
dU = 2072 J.

10. If the amount of heat given to a system is 35 J and the amount of work done by the system is -15J and the amount of work done by the system is -15J, then the change in the internal energy of the system is?
a) -50J
b) 20J
c) 30
d) 50J
Answer: d
Clarification: ∆Q=∆U+∆W
35=∆U+15
∆U=35+15=50J.

11. Assertion: Reversible systems are difficult to find in the real world.
Reason: Most process is dissipative in nature.
a) Both assertion and reason are true and the reason is the correct explanation of the assertion
b) Both assertion and reason are true but the reason is not a correct explanation of the assertion
c) Assertion is true but the reason is false
d) Both assertion and reason are false
Answer: a
Clarification: Both the assertion and reason are true. The energy consumed is doing work against dissipative forces cannot be recovered.

12. The change in internal energy, when a gas is cooled from 927° to 27° is?
a) 100%
b) 300%
c) 200%
d) 75%
Answer: d
Clarification: U=nCv T
∆U/U×100=∆T/T×100
∆U/U×100=(1200-300)/1200×100=9/12×100=75%.

13. During adiabatic compression of a gas, its temperature ___________
a) Falls
b) Remains constant
c) Rises
d) Becomes zero
Answer: c
Clarification: The work done on the gas during the adiabatic process increases its internal energy and hence its temperature rises.

250+ TOP MCQs on Wave Motion – 2 and Answers

Engineering Physics online test focuses on “Wave Motion – 2”.

1. A particle in simple harmonic motion is described by the displacement function x(t)=Acos⁡(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s.
a) 1 cm
b) √2 cm
c) 2 cm
d) 2.5 cm
Answer: b
Clarification: v=ω√(A2-x2)
π=π√(A2-1)
A2-1=1 or A2=2
A=√2cm.

2. A particle executes simple harmonic motion, its time period is 16s. If it passes through the centre of oscillation, then its velocity is 2 m/s at time 2s. The amplitude will be ___________
a) 7.2m
b) 4cm
c) 6cm
d) 0.72m
Answer: a
Clarification: Here,
t=2s,v=2m/s,T=16s
v=Aωcosωt
2=A×2π/14×cos⁡(2π/16×2)
A=(16√2)/π=7.2m.

3. A body is executing the simple harmonic motion with an angular frequency of 2rad/sec. Velocity of the body at 20m displacement, when amplitude of motion is 60m, is ___________
a) 90 m/s
b) 118 m/s
c) 113 m/s
d) 131 m/s
Answer: c
Clarification: v=ω√(A2-y2)=2√(602-202)
v=80√2
v=113m/s.

4. A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is ___________
a) 10 cm/s
b) 12 cm/s
c) 16√16 cm/s
d) 16 cm/s
Answer: b
Clarification: v=2π/T×√(A2-y2)
v=2π/π √(102-82)
=2×6=12cm/s.

5. The magnitude of acceleration of particle executing simple harmonic motion at the position of maximum displacement is?
a) Zero
b) Minimum
c) Maximum
d) Infinity
Answer: c
Clarification: Acceleration in simple harmonic motion is
a=ω2 y
At ymax=A, amax=ω2 A
Acceleration is maximum at the position of maximum displacement.

6. The maximum velocity and maximum acceleration of a body moving in a simple harmonic motion are 2m/s and 4m/s2 respectively. What will be the angular velocity?
a) 4 rad/sec
b) 3 rad/sec
c) 2 rad/sec
d) 8 rad/sec
Answer: c
Clarification: vmax=ωA, amax=ω2 A
ω=amax/vmax = 4/2
ω=2rad/sec.

7. A particle executing simple harmonic motion has amplitude 0.01 and frequency 60Hz. The maximum acceleration of the particle is ___________
a) 144 π2 m/s2
b) 80 π2 m/s2
c) 120 π2 m/s2
d) 60 π2 m/s2
Answer: a
Clarification: amax=ω2 A=4π2 v2 A
=4π2×60×60×0.01=144 π2 m/s2.

8. A particle having potential energy 1/3 of the maximum value at a distance of 4 cm from mean position. Amplitude of motion is ___________
a) 4√3
b) 6/√2
c) 2/√6
d) 2√6
Answer: a
Clarification: Ep=1/3 E
1/2 ky2=1/3×1/2×kA2
A=√3 y=√3×4=4√3 cm.

9. A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
a) 0.51A
b) 0.71A
c) 0.61A
d) 0.81A
Answer: b
Clarification: Ek=Ep
1/2 k(A2-y2)=1/2×ky2
y=±A/√2
y=±0.71A.

10. To show that a simple pendulum executes simple harmonic motion, it is necessary to assure that ___________
a) Length of the pendulum is small
b) Amplitude of oscillation is small
c) Mass of the pendulum is small
d) Acceleration due to gravity is small
Answer: b
Clarification: Motion is simple harmonic only when the amplitude of oscillation is small because only then f is proportional to x.

11. Time period of a simple pendulum will be double if we ___________
a) Decrease the length 2 times
b) Decrease the length 4 times
c) Increase the length 2 times
d) Increase the length 4 times
Answer: d
Clarification: T=2π√(l/g) or T∝√l
When the length is increased four times, time period gets doubled.

12. The time period of a simple pendulum is 2 sec. If its length is increased by 4 times, then its period becomes ___________
a) 16 sec
b) 8 sec
c) 12 sec
d) 4 sec
Answer: d
Clarification: T∝√l, T‘∝√4l
T‘/T=2
T‘=2T=2×2=4 sec.

13. A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the increase in the time period of the pendulum of increased length is?
a) 10%
b) 30%
c) 21%
d) 50%
Answer: a
Clarification: As T is proportional to l, the percentage increase in a time period on increasing the length by 21%
=1/2×∆l/l×100=1/2×21=10.5%.

14. A hollow spherical pendulum is filled with mercury has time period T. If mercury is thrown out completely, then the new time period ___________
a) Increases
b) Decreases
c) Remains the same
d) First increases and then decreases
Answer: c
Clarification: Position of centre of gravity remains unaffected when mercury is thrown out. Hence effective length and time period remain the same.

15. A simple pendulum is vibrating in an evacuated chamber. It will oscillate with ___________
a) Constant amplitude
b) Increasing amplitude
c) Decreasing amplitude
d) First increasing amplitude and then decreasing amplitude
Answer: a
Clarification: In a vacuum, there is no loss of energy due to resistive forces. So, amplitude remains constant.

16. Two simple pendulum, whose lengths are 100cm and 121cm, are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again?
a) 11
b) 10
c) 21
d) 20
Answer: b
Clarification: T∝√l
For the two pendulums in same phase
nT1=(n+1) T2
n√(l1)=(n+1) √(l2)
n√121=(n+1)√100
n×11=(n+1)10
n=10.

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250+ TOP MCQs on Einstein’s Special Theory of Relativity and Answers

Engineering Physics Multiple Choice Questions on “Einstein’s Special Theory of Relativity”.

1. As an object approaches the speed of light, it’s mass becomes _____________
a) Zero
b) Double
c) Remains Same
d) Infinite

Answer: d
Clarification: We know, for an object in motion m=(frac{m_0}{sqrt{1-frac{V^2}{c^2}}})
Thus, as v = c, the quantity in denominator becomes zero. Hence the mass of the object becomes infinite.

2. If the sun radiates energy at the rate of 4 x 1026 Js-1, what is the rate at which its mass is decreasing?
a) 5.54 x 109 kgs-1
b) 4.44 x 109 kgs-1
c) 3.44 x 109 kgs-1
d) 2.44 x 109 kgs-1

Answer: b
Clarification: As we know from Einstein’s mass energy relation, E = mc2.
Therefore, ΔE= Δmc2
ΔE = 4 x 1026 Js-1, c = 3 x 108 ms-1.
Δm = 4 x 1026 Js-1/ 9 x 1016m2s-2
Δm = 4.44 x 109 kgs-1.

3. The orbit of mercury is changing slightly due to the sun’s gravity.
a) True
b) False

Answer: a
Clarification: According to Einstein’s theory of relativity, a massive object distorted space and time. Thus, due to the curvature of space-time around the massive sun, the orbit of mercury is shifting gradually over time.

4. According to Einstein’s Special Theory of Relativity, laws of physics can be formulated based on ____________
a) Inertial Frame of Reference
b) Non-Inertial Frame of Reference
c) Both Inertial and Non-Inertial Frame of Reference
d) Quantum State

Answer: a
Clarification: One of the postulates of Einstein’s Special Theory of Relativity states that all the inertial frames are equivalent to the formulation of laws of physics. Thus, it is suitable for all the inertial frames.

5. For Einstein’s relation, E2 – p2c2 = _____________
a) moc2
b) mo2c4
c) moc4
d) mo2c6

Answer: b
Clarification: We know, E = mc2 and momentum, p = mv
Now, E2 – p2c2 = m2c4 – m2v2c2
Now, we know, m = (frac{m_0}{sqrt{1-frac{V^2}{c^2}}})
Therefore, E2 – p2c2 = m2c4(1-v2/c2)
=c4m2(1-v2/c2)
Using, m2(1-v2/c2) = mo2 we get
E2 – p2c2 = mo2c4.

6. A frame of reference has four coordinates, x, y, z, and t is referred to as the_____________
a) Inertial frame of reference
b) Non-inertial frame of reference
c) Space-time reference
d) Four-dimensional plane

Answer: c
Clarification: Such a frame, having four coordinates of x, y, z and time t is called the space-time frame. It plays a major role in Einstein’s special theory of relativity.

7. A man, who weighs 60 kg on earth, weighs 61 kg on a rocket, as measured by an observer on earth. What is the speed of the rocket?
a) 2.5 X 108 m/s
b) 2.5 X 107 m/s
c) 5.5 X 107 m/s
d) 5.5 X 108 m/s

Answer: c
Clarification: Now, as we know m = (frac{m_0}{sqrt{1-frac{V^2}{c^2}}})
Here, m = 61 kg and m0 = 60 kg
Therefore, 61/60 = 1/({sqrt{1-frac{V^2}{c^2}}})
v2/c2 = 1 – (60/61)2 = 121/3600
v = 11 X c/60 = 5.5 X 107 m/s.

8. The momentum of a photon having energy 1.00 X 10-17 J is ____________
a) 2.33 X 10-26 kg m/s
b) 3.33 X 10-26 kg m/s
c) 4.33 X 10-26 kg m/s
d) 5.33 X 10-26 kg m/s

Answer: b
Clarification: Now, the rest mass of a photon is zero.
Therefore, it’s momentum, p = E/c
= 1.00 X 10-17/3 X 108
= 3.33 X 10-26 kg m/s.

250+ TOP MCQs on Finite Potential Well and Answers

Engineering Physics Multiple Choice Questions on “Finite Potential Well”.

1. In a finite Potential well, the potential energy outside the box is ____________
a) Zero
b) Infinite
c) Constant
d) Variable
Answer: c
Clarification: In a finite potential well, the potential energy of the particle outside the box is a finite constant unlike infinite potential well, where the potential energy outside the box was infinite.

2. The Schrodinger for the particle inside a finite potential well becomes ____________
a) x > 0
b) x < 0
c) 0 < X < L
d) x > L
Answer: c
Clarification: The particle cannot exist outside the box, as it cannot have infinite amount of energy. Thus, it’s wave function is between 0 and L, where L is the length of the side of the box.

3. When the particle strikes the wall of the well, it bounces off completely.
a) True
b) False
Answer: b
Clarification: Whenever the particle is incident on the wall of the potential well, there is a probability that the particle may move on to the next region even though it’s energy is less than the potential energy of the barrier.

4. The Energy of the particle is proportional to ____________
a) n
b) n-1
c) n2
d) n-2
Answer: c
Clarification: In a particle inside a box, the energy of the particle is directly proportional to the square of the quantum state in which the particle currently is n2.

5. For a particle inside a box of finite potential well, the particle is most stable at what position of x?
a) x > L
b) x < 0
c) 0 < x < L
d) Not stable in any state
Answer: c
Clarification: The particle is most stable when it is inside the box of finite potential well. In that case, the potential energy of the particle is zero.

6. When the Schrodinger equation is solved for E > Vo, the solutions will be __________
a) Non-oscillatory
b) Oscillatory Inside
c) Oscillatory Outside
d) Oscillatory inside as well as outside
Answer: d
Clarification: If we solve the time-independent Schrödinger equation for an energy E > Vo, the solutions will be oscillatory both inside and outside the well. Thus, the solution is never square integrable; that is, it is always a non-normalizable state.

7. Particle in a box of finite potential can never be at rest.
a) True
b) False
Answer: a
Clarification: If the particle in a box has zero energy, it will be at rest inside the well and it violates the Heisenberg’s Uncertainty Principle. Thus, the minimum energy possessed by a particle is not equal to zero.

8. What is the minimum Energy possessed by the particle in a box?
a) Zero
b) (frac{pi^2hbar^2}{2mL^2})
c) (frac{hbar^2}{2m})
d) (frac{pi^2hbar}{2m})
Answer: c
Clarification: The minimum energy possessed by a particle inside a box in a finite potential well is equal to (frac{hbar^2}{2m}). The particle can never be at rest, as it will violate Heisenberg’s Uncertainty Principle.

9. The wave function of a particle in a box is given by ____________
a) A sin(kx)
b) A cos(kx)
c) Asin(kx) + Bcos(kx)
d) A sin(kx) – B cos(kx)
Answer: c
Clarification: The wave function for the particle in a box is given by: Asin(kx)+Bcos(kx). The Energy possessed by the particle is given by: (frac{n^2hbar}{2m}).

10. What does the following figure shows?
engineering-physics-questions-answers-finite-potential-well-q10
a) Wave function for Infinite Potential Well
b) Wave function for Finite Potential Well
c) Probability Density function for Infinite Potential Well
d) Probability Density function for Finite Potential Well
Answer: d
Clarification: The given figure shows the probability densities of a particle in a finite potential well. The particle has a certain probability of being outside the well.