Sindh Mdcat Exclusive Course Physics Work Energy Power — Solved MCQs

Question 1

An automobile is moving forwards with uniform velocity due to the force exerted by its engine. If the force is doubled with the velocity remaining constant, what happens to its total power?

Accepted Answer

C. It is doubled — Power = force * velocityWith velocity kept constant, Power is directly proportional to force. Therefore, doubling force would double power.

Question 2

A mass of 20 kg is lifted from the floor to a height of 2m in 4.9 sec. Calculate the power in Watts.

Accepted Answer

B. 80 W — Firstly, we’ll find the increase in the gravitational potential energy as we have the object traveling vertically and we are unaware of the force by which the object is being lifted. It could be equivalent to the weight of the object or greater. The height is increased by 2 meters and by using the formula 'mgh', where g is the gravitational field strength, we can find the increase in GPE. The GPE increased by 392.4 J in 4.9 seconds and the formula for power is 'energy over time' so, we divide 392.4 J by 4.9 seconds giving us 80W.

Question 3

Which of the following statement shows that no work is done?

Accepted Answer

E. C and D — When you lift weights, you apply a force (the gravitational force) to move the weights against gravity. Work is done because there is a force applied (lifting) and a displacement (the weights move vertically) in the direction of the force.

Question 4

Coal and petrol are stores of chemical energy when are burnt, chemical energy is converted into heat energy i.e chemical energy = heat energy + _:

Accepted Answer

B. Losses — When coal or petrol is burned, the chemical energy stored in the fuel is primarily converted into heat energy. However, not all of this energy conversion is perfectly efficient; some energy is inevitably lost to the surroundings as waste heat, which is why 'Losses' is the correct answer. Electrical energy is not directly produced from combustion without additional equipment. Gains are incorrect because energy is not gained but redistributed with losses. Wind energy is unrelated to burning fuels. 'Nothing' is incorrect because energy transformations usually accompany some degree of energy loss.

Question 5

A neutron travels a distance of 24 m in a time interval of 3.6 x 10-4 s. Assuming its speed was constant, find its kinetic energy. Take 1.7 x 10-27 kg as the mass of the neutron:

Accepted Answer

A. 37.2 x 10-19 joule — The following is the solution:speed of neutron = distance travelled by neutron / time taken by neutron for covered that distance .speed of neutron = 24/3.6 × 10‐4 m/sspeed of neutron ( V)= 6.6 × 10⁴ m/sno ,kinetic energy = 1/2mv²K.E = 1/2 × 1.7 × 10-27 ×( 6.6 × 10⁴)² J= 1/2 × 1.7 × (6.6)² × 10-19 j=37.2 × 10-19 jso,K.E = 37.2 × 10-19.Hence the correct answer is A which is closest to the correct answer

Question 6

Which one is the unit of energy:

Accepted Answer

D. All of these — All of these are units of energy. The most commonly used unit is Joules, however erg is also a unit of energy and it is equal to 10-7 Joules. Energy is defined as Power x time so the units are kWh (kilo Watt hour).

Question 7

Energy consumed by 60 watt bulb in 2 minutes is equal to:

Accepted Answer

A. 7.2 kilo joules — Power=Energy/time Energy=power ×time 60 watt×120 sec =7200 J =7.2 kilo J.

Question 8

An overhead tank of capacity 1000 litres has to be filled in 1/2 an hour using a water pump. The tank is kept at a height 10m above ground and the water level is 10m below ground. The opening of the inlet pipe inside the tank is 1.11cm2. Assuming the efficiency of the motor to be 60%, the electric power used is (Neglect viscosity):

Accepted Answer

D. 193 W — Determine Total Height and Mass of Water: The total height (H) the water needs to be lifted is the sum of the height of the tank above ground and the depth of the water level below ground.H=10 m+10 m=20 mThe mass (m) of the water is calculated from its volume (1000 litres = 1 m3) and density (1000 kg/m3).m=1 m3×1000 kg/m3=1000 kgCalculate Water Velocity: The time to fill the tank is 30 minutes, which is 1800 seconds. The flow rate (Q) is the volume divided by time.Q=1m3/1800 s≈0.000556 m3/sThe area (A) of the inlet pipe is 1.11 cm², which converts to 0.000111 m². The velocity (v) is the flow rate divided by the area.v=Q/A=0.000556 m³s-1 /0.000111 m²≈5.005 m/sCalculate Total Work Done: The total work is the sum of the potential energy (for lifting the water) and kinetic energy (for the water's final speed).Potential Energy (PE) = mgh=1000 kg×9.8 m/s²×20 m=196,000 JKinetic Energy (KE) = 1/2mv²=0.5×1000 kg×(5.005 m/s)²≈12,525 JTotal Work (W) = PE+KE=196,000 J+12,525 J=208,525 JDetermine Electric Power: The ideal power (Pideal​) is the total work divided by the time taken.Pideal​​=W/t=208,525 J/1800 s≈115.85 WGiven the motor's efficiency (η) is 60% (or 0.60), the electric power (Pelectric​) is the ideal power divided by the efficiency.Pelectric=​Pideal​​/η=115.85 W/0.60≈193.08 W

Question 9

Kilowatt-hour is a unit of:

Accepted Answer

A. Electric Energy — The kilowatt-hour (SI symbol: kW⋅h or kW h; commonly written as kWh) is a unit of energy equal to 3600 kilojoules (3.6 megajoules). The kilowatt-hour is commonly used as a billing unit for energy delivered to consumers by electric utilities.

Question 10

The sum of all forms of molecular energies (kinetic and potential) of a substance is termed as:

Accepted Answer

C. Internal energy — Elastic energy is a form of potential energy. Absolute energy does not exist. By definition, Internal energy is the sum of total kinetic and potential energy as defined in 1st law of thermodynamics.

Question 11

The rate at which work is being done is called:

Accepted Answer

A. Power — The rate at which work is done is called power. It is expressed as the amount of work per unit of time.

Question 12

The K.E. of a body of mass 2kg and momentum of 2Ns is:

Accepted Answer

A. 1 J — Given:mass(m)=2kg, momentum(p)=2Nswe know; p=mvv=p/m​v=1ms-1Kinetic energy (K.E)= 1/2(mv2)K.E=1/2(pv) K.E=1/2x(2x1)K.E=1J

Question 13

A particle of mass m=9 x 10–31 kg moving towards the wall of a vessel at a velocity of v=600 ms-1 strikes it at an angle of 60° to the normal and rebounds at the same angle at the same speed. The impulse of the force experienced by the wall during the impact is:

Accepted Answer

C. 5.4x10-28 Ns — Mass m = 9 × 10⁻³¹ kgVelocity v = 600 m/sAngle θ = 60°Change in momentum Δp (normal to wall)= −2mv cosθ= −2 × 9 × 10⁻³¹ × 600 × cos 60°= −2 × 9 × 10⁻³¹ × 600 × 0.5= −5.4 × 10⁻²⁸ kg·m/sImpulse on wall = −Δp = 5.4 × 10⁻²⁸ N·s

Question 14

A bullet of mass m and velocity a is fired into a large block of wood of mass M. The final velocity of the system is:

Accepted Answer

D. Option D — The following is the solution:

Question 15

A cord is used to lower vertically a block of mass M a distance d at a constant downward acceleration of g/4. Then the work done by the cord on the block is :

Accepted Answer

D. –3Mgd / 4 — When the block moves vertically downward with acceleration g/4 the tension in the cord is:T=M(g-g/4)=3Mg/4Work done by cord = F.S cosθAs tension is acting upwards and the block is moving downwards the angle will be 180Work=(3Mg/4)X(d)Xcos(180)Work Done= -3Mgd/4

Question 16

A particle moves along the x-axis from x = 0 to x = 5m under the influence of a force given byF = 7–2x+3x2. The work done in the process is:

Accepted Answer

D. 135 J — W = ∫ F . dx = ∫ from 0 to 5 (7 – 2x + 3x²) dx= [ 7x – (2x²)/2 + (3x³)/3 ] from 0 to 5 = 35 – 25 + 125 = 135 J

Question 17

A particle moves in a circle of radius r under the action of a centripetal force equal to -k/r2 r, where k is a constant. The total energy of the particle is:

Accepted Answer

C. Option C — As the particle moves in a circle mv²/r = k/r² ∴ K.E. = 1/2 mv² = k/2rNow potential energy U = – ∫ F dr = – k/r∴ total energy = – k/2r

Question 18

A block of mass m is stationary with respect to a wedge of mass M moving with uniform speed v on a horizontal surface. Find the work done by the friction force on the block in t seconds:

Accepted Answer

B. Option B — This is the following solution:

Question 19

When the output power equals to one-half of the input power, efficiency of the transformer becomes:

Accepted Answer

C. 50% — Following is the formula for efficiency. As we can plug in values of input and output, we will get 50%.

Question 20

1 kilowatt hour =

Accepted Answer

B. 3.6x106 J — The commercial unit of electrical energy is a kilowatt-hour (kWh). The energy generated by a kilowatt power source in 1 hour is 1 kWh.1kWh = 1kW × 1hour1kWh = 1000W (Joules/seconds) × (60 × 60) seconds1kWh = 3.6 × 106 J

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