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Multiple choice question for engineering

Set 1

1. What does conductivity of metals depend upon?
a) The nature of the material
b) Number of free electrons
c) Resistance of the metal
d) Number of electrons

View Answer

Answer: b [Reason:] The conducting property of a solid is not a function of total number of electrons in the metal, but it is due to the number of valance electrons called free electrons.

2. The free electrons collide with the lattice elastically. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] The free electrons move randomly in all directions. The free electrons collide with each other and also with the lattice Elastically, without loss in energy.

3. What happens to the free electrons when electric field is applied?
a) They move randomly and collide with each other
b) They move in the direction of the field
c) They remain stable
d) They move in the direction opposite to that of the field

View Answer

Answer: d [Reason:] The free electrons move in the direction opposite to that of field direction. Since they are assumed to be a perfect gas as they obey classical kinetic theory of gases and the electron velocities in the metal obey the Maxwell-Boltzmann statistics.

4. Thermal conductivity is due to photons.
a) True
b) False

View Answer

Answer: a [Reason:] Thermal conductivity is due to both photons and free electrons and not just photons.

5. Which of the following theories cannot be explained by classical theory?
a) Electron theory
b) Lorentz theory
c) Photo-electric effect
d) Classical free electron theory

View Answer

Answer: c [Reason:] Classical theory states that all free electrons will absorb energy. This theory cannot explain photo-electric effect.

6. Which of the following theories can be adopted to rectify the drawbacks of classical theory?
a) Compton theory
b) Quantum theory
c) Band theory
d) Electron theory

View Answer

Answer: b [Reason:] In classical theory the properties of metals, such as electrical and thermal conductivities are well explained on the assumption that the electrons in the metal freely moves like the particles of a gas. Hence it can be used to rectify the drawbacks of classical theory.

7. What is the level that acts as a reference which separated the vacant and filled states at 0K?
a) Excited level
b) Ground level
c) Valance orbit
d) Fermi energy level

View Answer

Answer: d [Reason:] Fermi energy level is the maximum energy level up to which the electrons can be filled at 0K. Thus it acts as reference level which separated the vacant and filled states at 0K.

8. A uniform silver wire has a resistivity of 1.54×10-18 ohm/m at room temperature. For an electric field along the wire of 1 volt/cm. Compute the mobility, assuming that there are 5.8×1028 conduction electrons/m3.
a) 1.54 m2/Vs
b) 6.9973m2/Vs
c) 6.9973×10-3 m2/Vs
d) 0.69973m/s

View Answer

Answer: c [Reason:] Mobility of the electrons = 1/ƿne Mobility = 6.9973×10-3 m2/Vs

9. Calculate the drift velocity of the free electrons with mobility of 3.5×10-3 m2/Vs in copper for an electric field strength of 0.5 V/m.
a) 3.5 m/s
b) 1.75×103 m/s
c) 11.5 m/s
d) 1.75×10-3 m/s

View Answer

Answer: d [Reason:] Drift velocity = μE Drift velocity = 3.5×10-3×0.5 = 1.75×10-3m/s

10. The Fermi temperature of a metal is 24600K. Calculate the Fermi velocity.
a) 0.5m/s
b) 1.38m/s
c) 0.8633×106m/s
d) 9.11×10-3m/s

View Answer

Answer: c [Reason:] EF = KB TF= ½ mv(F)2 vF = √((2×KB×TF/m) vF = 0.8633×106m/s

Set 2

1. Three different objects of masses m1, m2 and m3 are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three different objects on reaching the ground will be in the ratio of
a) m1:m2:m3
b) 1:1:1
c) 1/m1:1/m2:1/m3
d) m1:2m2:3m3

View Answer

Answer: b [Reason:] v2-02=2gh v=√2gh This velocity does not depend on mass.

2. A truck and a car are moving with equal velocity. On applying brakes, both will stop after certain distance, then
a) Truck will cover less distance before stopping
b) Car will cover less distance before stopping
c) Both will cover equal distance
d) None of the mentioned

View Answer

Answer: b [Reason:] Being lighter than truck, the car has less kinetic energy. On applying brakes with the same force, the car will cover less distance before coming to rest.

3. If a ladder weighing 250N is placed against a smooth vertical wall having coefficient of friction between it and floor 0.3, then what is the maximum force of friction available at the point of contact between the ladder and the floor
a) 75N
b) 35N
c) 50N
d) 25N

View Answer

Answer: a [Reason:] f = μR = μmg f = 0.3×250 = 75N

4. A body of mass 2kg begins to slide down a plane inclined at an angle of 45° with the horizontal. The force of friction will be
a) 19.6sin 45°
b) 9.8sin 45°
c) 19.6cos 45°
d) 9.8cos 45°

View Answer

Answer: a [Reason:] When the body just slides down on inclined plane, f = mgsinθ = 2×9.8sin 45° f = 19.6sin 45°

5. A person sitting in an open car moving at constant velocity throws a ball vertically into air, the balls falls
a) Exactly in the hand of thrower
b) Outside the car
c) In the car ahead of the person
d) In the car behind the person

View Answer

Answer: a [Reason:] Both the person and the ball share the horizontal velocity of the car. Hence the ball falls exactly in the hand of the thrower.

6. A stone of mass m tied to one end of string of length l is rotated in a circle with the other end of the string as the centre. The speed of the stone is v. If the string breaks, the stone will
a) Move towards the centre
b) Move away from the centre
c) Move along tangent
d) Stop

View Answer

Answer: c [Reason:] When the string breaks, the centripetal force becomes zero. The stone moves along the tangent to the circular path. That is in the direction of velocity.

7. Two racing cars of masses m1 and m2 are moving in circle of radii r1 and r2 respectively. Heir speeds are such that each car makes a complete circle in same time t. The ratio of the angular speed of the first to the second car is
a) 1:1
b) r1:r2
c) m1:m2
d) m1 m2:r1 r2

View Answer

Answer: a [Reason:] w=2π/T For both cars, period of revolution T is same, so their angular speed w is also same.

8. A man weighs 80kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5m/s2. What would be the reading on the scale?
a) Zero
b) 400N
c) 800N
d) 1200N

View Answer

Answer: d [Reason:] As the lift accelerates upward, the apparent weight, R = m(g+a) = 80(10+5) = 1200N

9. Consider a car moving along a straight horizontal road with a speed of 72km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g=10m/s2)
a) 30m
b) 40m
c) 72m
d) 20m

View Answer

Answer: b [Reason:] Retardation, a= -μg = -0.5×10 =-5m/s2 u = 72km/h = 20m/s As v2-u2=2as s = 40m

10. A boy of 50kg is standing in a lift moving down with an acceleration 9.8m/s2. The apparent weight of the boy is
a) Zero
b) (50×9.8)N
c) 50N
d) 50/9.8N

View Answer

Answer: a [Reason:] Apparent weight = m(g-a) Apparent weight = 50(9.8-9.8) = 0

Set 3

1. Assertion: In pressure-temperature phase diagram of water, the slope of the melting curve is found to be negative [Reason:] Ice contracts on melting to water
a) Both assertion and reason are true and reason is the correct explanation of the assertion
b) Both assertion and reason are true but reason is not correct explanation of the assertion
c) Assertion it true but reason is false
d) Both assertion and reason are false

View Answer

Answer: b [Reason:] Both the assertion and reason are true but the reason is not the correct explanation of the assertion. The negative slope of the melting curve indicates that the melting point of ice decreases with the increase in pressure

2. At constant volume temperature is increases, then
a) Collisions on walls will be less
b) Number of collisions per unit time will increase
c) Collisions will be in straight line
d) Collisions will not change

View Answer

Answer: b [Reason:] As the temperature increases, the average molecular velocity increases. This increases collisions frequency.

3. Gases at low temperature and high pressure show large deviation from an ideal behaviour. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] At low temperature and high pressure, the intermolecular attractions become appreciable. Moreover, the volume occupied by the gas molecules cannot be neglected in comparison to the volume of the gas. Hence the real gases show large deviations from ideal gas.

4. When an automobile travels for a long distance, the pressure in the gas
a) Increases slightly
b) Decreases slightly
c) Remains the same
d) Decreases and then increases

View Answer

Answer: a [Reason:] Due to the friction between the tyres and the road, the tyres get heated. The temperature of air inside the tyres increases. Consequently, the air pressure in the tyres increases slight.

5. A vessel contains one mole of O2 gas at a temperature T. The pressure of the gas is P. An identical vessel containing one hole of He gas at a temperature 2T has a pressure of
a) P/8
b) P
c) 2P
d) 8P

View Answer

Answer:c [Reason:] PV = nRT or P=nRT/V For same n, R and V, P is proportional to T P2/P1 =T2/T1 Or P2/P=2T/T or P2=2P

6. The average translational kinetic energy of O2 molecules (relative molar mass 32) at a particular temperature is 0.048eV. The translational kinetic energy of N2 molecules (relative molar mass) in eV at the same temperature is
a) 0.0015
b) 0.003
c) 0.048
d) 0.768

View Answer

Answer: c [Reason:] Average translational kinetic energy of a molecule=3/2kT It depends on temperature and not on molecular mass. Hence average translational kinetic energy for N2 will also be 0.048 eV

7. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
a) 4RT
b) 9RT
c) 11RT
d) 15RT

View Answer

Answer: c [Reason:] For two moles of diatomic nitrogen with no vibrational mode, U1=2×5/2 RT=5RT For four moles of monoatomic argon, U2=4×3/2 RT=6RT Therefore, U=U1+U2=11RT

8. Surface the lake is at 2°C and depth of the lake is 20m. Find the temperature of the bottom of the lake
a) 2°C
b) 3°C
c) 4°C
d) 552°C

View Answer

Answer: d [Reason:] P1/T1 =P2/T2 (1×105)/(273+2)=(hρg+1×105)/T2 (1×105)/275=(20×103×10+1×105)/T2 T2=275×3=825K=552℃

9. The degree of freedom in case of a monoatomic gas is
a) 1
b) 3
c) 5
d) 2

View Answer

Answer: b [Reason:] A monoatomic gas has 3 degrees of freedom due to translator motion

10. The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is
a) (1/2) kT
b) (3/2) kT
c) (3/2) RT
d) (1/2) RT

View Answer

Answer: d [Reason:] Mean kinetic energy per mole pf a gas per degree of freedom = (1/2) RT

Set 4

1. In a vessel, the gas at a pressure P. If the mass of all molecules is halved and their speed is doubled, then the resultant pressure will be
a) 4P
b) 2P
c) P
d) P/2

View Answer

Answer: c [Reason:] According to kinetic theory of gases, the pressure exerted by a gas, P=1/3×mn/V v2 P/P=2 P=2P

2. The absolute zero is the temperature, at which
a) All substances exist in solid state
b) Water freezes
c) Molecular motion ceases
d) Solid melts

View Answer

Answer: c [Reason:] The absolute zero is the temperature at which all molecular motion ceases.

3. When we heat a gas-sample from 27°C to 327°C, then the initial average kinetic energy of the molecules was E. What will b the average kinetic energy?
a) 327E
b) 300E
c) 2E
d) √2E

View Answer

Answer: c [Reason:] E/E=(273+327)/(273+27)=600/300 E=2E

4. Temperature less than absolute zero is not possible. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] According to kinetic interpretation of temperature, absolute temperature is proportional to the average kinetic energy of molecules. As the heat is removed, the temperature falls and velocity of molecules decreases. At absolute zero, the kinetic energy becomes zero. As the kinetic energy cannot be negative so no further decrease in kinetic energy is possible. Hence temperature cannot be decreases below absolute zero.

5. A given quantity of an ideal gas at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is
a) 2/3 P
b) P
c) 3/2 P
d) 2P

View Answer

Answer: a [Reason:] At a given temperature, P∆V+V∆P=0 P=V∆P/∆V=(∆P/∆V)/V=Isothermal bulk modulus

6. Statement: The ration Cp/Cv is more for helium gas than that for hydrogen gas [Reason:] Atomic mass of helium is more than that of hydrogen
a) Both statement and reason are true and reason is the correct explanation of the statement
b) Both statement and reason are true but reason is not correct explanation of the statement
c) Statement it true but reason is false
d) Both statement and reason are false

View Answer

Answer: a [Reason:] Both the statement and reason are true but the reason is not a correct explanation of the statement For monoatomic He, γ=Cp/Cv =5/3 For diatomic H2, γ=Cp/Cv =7/5

7. The relation PV=RT can describe behaviour of the real gas at
a) High temperature and high density
b) High temperature and low density
c) Low temperature and low density
d) Low temperature and high density

View Answer

Answer: b [Reason:] At high temperature and low density, real gas behaves like an ideal gas. Then the intermolecular attractions and actual volume of gas molecules become negligible.

8. An absolute zero is the temperature at which
a) Water solidifies
b) All gases become liquid
c) rms velocity becomes zero
d) Solids melt

View Answer

Answer: c [Reason:] According to kinetic theory of gases, absolute zero is that temperature at which all molecular motion ceases. Hence at T=0K, rms velocity becomes zero.

9. The temperature of a given mass is increased from 27°C to 327°C. The rms velocity of the molecules increases
a) √2 times
b) 2 times
c) 2√2 times
d) 4 times

View Answer

Answer: a [Reason:] vrms is proportional √T v2/v1 =√(T2/T1)=√((273+327)/(273+27))=√(600/300)=√2 v2=√2 v1

10. The ratio of the vapour densities of two gases at a given temperature is 9:8. The ratio of the rms velocities of their molecules is
a) 3:2√2
b) 2√2:3
c) 9:8
d) 8:9

View Answer

Answer: b [Reason:] At a given temperature, (vrms (1))/(vrms (2) )=√(8/9)=2√2:3

Set 5

1. Statement 1- The length of the day is slowly increasing
Statement 2- The dominant effect causing a slow down in the rotation of the earth is the gravitational pull of other planets in the solar system.
a) Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
b) Statement 1 and statement 2 are correct but statement 2 is not the correct explanation of statement 1
c) Statement 1 is true but statement 2 is false
d) Both statement 1 and 2 are false

View Answer

Answer: d [Reason:] Both the statements are false. The length of the day depends on the speed of rotation of the earth about its axis. As it remains constant, the length of the day is not changing. Further, the earth is not slowing down due to gravitational pull of other planets in the solar system.

2. If the earth stops rotating, the value of g at the equator
a) Increases
b) Decreases
c) No effect
d) First increases and then decreases

View Answer

Answer: a [Reason:] At the equator, ge=g-Rω2 When ω=0, ge=g The value of g increases if the earth stops rotating.

3. A satellite revolves very near to the earth surface. Its speed should be around
a) 5km/s
b) 8km/s
c) 2km/s
d) 11km/s

View Answer

Answer: b [Reason:] Near the earth’s surface, v0=√gR=7.2km/s Approximately 8km/s

4. The value of escape velocity on a certain planet is 2km/s. Then the value of orbital speed for a satellite orbiting close to its surface is
a) 12km/s
b) 1km/s
c) √2km/s
d) 2√2km/s

View Answer

Answer: c [Reason:] Near earth’s surface, v0=ve/√2=2/√2=√2 km/s

5. If gravitational constant is decreasing with time, what will remain unchanged in case of a satellite orbiting around the earth?
a) Time period
b) Orbiting radius
c) Areal velocity
d) Angular velocity

View Answer

Answer: c [Reason:] The central gravitational force decreases which does not exert any torque. Hence the angular momentum and consequently areal velocity remains unchanged.

6. An elephant and an ant are to be projected out of earth into space. What is the velocity needed to do so?
a) Elephant needs to be projected with a higher velocity
b) Ant should be projected with a higher velocity
c) Both should be projected with the same velocity
d) Elephant cannot be projected to space

View Answer

Answer: c [Reason:] The escape velocity ve=√2gR, does not depend upon the mass of the projected body. Thus we need the same velocity to project an elephant and an ant into space.

7. An astronaut, while revolving in a circular orbit happens to throw a spoon outside. What will happen to the spoon?
a) It will reach the earth surface
b) It will escape the universe
c) It will remain stationary
d) It will continue to move in a circular orbit

View Answer

Answer: d [Reason:] While revolving in a circular orbit, if a spoon in thrown outside, the spoon will continue to rotate in the same circular orbit, and chase the astronaut. It will never reach the surface of the earth.

8. It is possible to place an artificial satellite in an orbit, such that it is always visible over New Delhi. True or false?
a) True
b) False

View Answer

Answer: b [Reason:] A satellite remains always visible only if it revolves in the equatorial plane with a period of revolution equal to that of the earth. New Delhi does not lie in the region of equatorial plane. Hence it is not possible to place an artificial satellite in an orbit such that it is always visible from New Delhi.

9. A satellite does not need ant fuel to circle around the earth. True or false?
a) True
b) False

View Answer

Answer: a [Reason:] The gravitational force between satellite and earth provides the centripetal force required by the satellite to move in a circular orbit.

10. A satellite orbiting the earth with speed v0. To make the satellite escape, what should be the minimum percentage increase in its velocity?
a) 41.5%
b) 44.1%
c) 41.4%
d) 40%

View Answer

Answer: c [Reason:] Required percentage increase in velocity = (ve-v0)/v0 ×100 (√2-1)×100=41.4%

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