A one-ton car moves with a constant velocity of 15 ms^–1 on a rough horizontal road. The total resistance to the motion of the car is 12% of the weight of the car. The power required to keep the car moving with the same constant velocity of 15ms^–1 is[Take g = 10 ms^–2]

9 kW
18 kW
24 kW
36 kW

Solution

F = ¹²⁄₁₀₀ × 1000 10N = 1200 N

P = Fv = 1200 N × 15 ms^–1= 18 kW

The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5c. The change in its energy will be

0.026 MeV
0.051 MeV
0.08 MeV
0.105 MeV

Solution

A nucleus ruptures into two nuclear parts which have their velocity ratio equal to 2 : 1. The ratio of their respective nuclear sizes (nuclear radii )is

1 : 2
1:√2
1 : 2^1/3
1 : 8

Solution

Ten litre of water per second is lifted from a well through 10m and delivered with a velocity of 10 ms^-1. If g = 10 ms–2 , then the power of the motor is

1 kW
1.5 kW
2 kW
2.5 kW

Solution

Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over s = 0 to 20 m will be

250 J
200 J
150 J
10 J

Solution

Loss in K.E = Area under the curve

A body of mass m is suspended from a mass less spring of natural length ι. It stretches the spring through a vertical distance y.The potential energy of the stretched spring is

mg(ι + y)
¹⁄₂ mg(ι + y)
¹⁄₂ mgy
mgy

Solution

A moving body with a mass m₁ and velocity u strikes a stationary body of mass m₂. The masses m₁ and m₂ should be in the ratio m₁/m₂ so as to decrease the velocity of the first body to 2u/3 and giving a velocity of u to m₂ assuming a perfectly elastic impact.Then the ratio m₁/m₂ is

5
1/ 5
1/25
25.

Solution

A body is dropped from a height of 20m and rebounds to a height 10m. The loss of energy is

Solution

Since the new height gained is half ,therefore there is 50% loss of energy.

In the non-relativistic regime, if the momentum, is in crease by 100% , the percentage increase in kinetic energy is

Solution

A body of mass m moving with velocity v makes a head on elastic collision with another body of mass 2m which in initially at rest. The loss of kinetic energy of the colliding body (mass m ) is

¹⁄₂ of its initial kinetic energy
¹⁄₉ of its initial kinetic energy
⁸⁄₉ of its initial kinetic energy
¹⁄₄ of its initial kinetic energy

Solution

Fraction of energy transferred =\(\frac{4\times 2}{(1+2)^{2}}=\frac{8}{9}\)

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