Johnny and his sister Jane race up a hill. Johnny weighs twice as much as jane and takes twice as long as jane to reach the top . Compared to Jane

Johnny did more work and delivered more power.
Johnny did more work and delivered the same amount of power.
Johnny did more work and delivered less power
Johnny did less work and johnny delivered less power.

Solution

The work is done against gravity so it is equal to the change in potential energy. W = E_p= mgh

For a fixed height, work is proportional to weight lifted.Since Johnny weighs twice as much as Jane he works twice as hard to get up the hill.

Power is work done per unit time. For Johnny this is W/Δt. Jane did half the work in half the time, (1/2 W)/(1/2 Δt)= W/t which is the same power delivered by Johnny.

A sphere of mass 8m collides elastically (in one dimension)with a block of mass 2m.If the initial energy of sphere is E.What is the final energy of sphere?

0.8 E
0.36 E
0.08 E
0.64 E

Solution

A particle of mass m₁ moving with velocity v collides with a mass m₂ at rest, then they get embedded. Just after collision, velocity of the system

increases
decreases
remains constant
becomes zero

Solution

A neutron with velocity v strikes a stationary deuterium atom, its K.E. changes by a factor of

¹⁵⁄₁₆
¹⁄₂
²⁄₁
None of these

Solution

Let mass of neutron =m

then mass of deuterium = 2m

[∵ it has double nuclides thus has neutron].

Let initial velocity of neutron = v and final velocities of neutron and deuterium are v₁ and v₂ respectively.

In figure, a carriage P is pulled up from A to B. The relevant coefficient of friction is 0.40. The work done will be

10 kJ
23 kJ
25 kJ
28 kJ

Solution

Work done against gravity

W_g= 50 × 10 × 30 = 15 kJ

Work done against friction

W_f=μ mg cos × s = 0.4 × 50 × 10 × ⁴⁄₅ × 50 =8 kJ

Total work done = W_g + W_f = 15 kJ + 8kJ = 23 kJ

A crane is used to lift 1000 kg of coal from a mine 100 m deep. The time taken by the crane is 1 hour. The efficiency of the crane is 80%. If g = 10 ms^-2, then the power of the crane is

10⁴ W
10⁵ W
\(\frac{10^{4}}{36\times 8}W\)
\(\frac{10^{5}}{36\times 8}W\)

Solution

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms^-1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?(a)2(b)4(c)1(d)3

Solution

A body accelerates uniformly from rest to a velocity of 1 ms^-1 in 15 seconds. The kinetic energy of the body will be ⁹⁄₂ J when ‘t’ is equal to [Take mass of body as 1 kg]

Solution

A machine, which is 75% efficient, uses 12 J of energy in lifting up a 1kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in m/s)

\(\sqrt{24}\)
\(\sqrt{12}\)
\(\sqrt{18}\)
√9

Solution

A body of mass 5 kg initially at rest explodes into 3 fragments with mass ratio 3 : 1 : 1. Two of fragments each of mass ‘m’ are found to move with a speed 60 m/s in mutually perpendicular directions. The velocity of third fragment is

60√2
20√3
10√2
20√2

Solution

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