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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]
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
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
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
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
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
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
A body is dropped from a height of 20m and rebounds to a height 10m. The loss of energy is
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
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
Fraction of energy transferred =\(\frac{4\times 2}{(1+2)^{2}}=\frac{8}{9}\)