NEET physics

Question 1

Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to

  • 1R

  • 1R

  • R

  • 1R32

Solution

A.

1R

Kinetic energy of the satellite

KE = 12mvo2       ...... 1Where vo =GMRNow putting the value of vo is eqn1KE =12mGMR22       =12mGMR KE 1R

So the kinetic energy of the satellite is inversely proportional 

to R. Therefore R increases then KE decreases.

Question 2

A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?

  • 2 s

  • 4 s

  • 1 s

  • 6 s

Solution

B.

4 s

When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

ν g ralso ν = rω =r2πTHence, 2πrTr gT2πrr g=2πrg=2π49.8=2×3.14×0.6389=4.0 secso Tmax =4 sec

Question 3

The maximum range of a gun on horizontal terrain is 16 km if g=10m/s2. What must be the muzzle velocity of the shell?

  • 200 m/s

  • 100 m/s

  • 400 m/s

  • 300 m/s

Solution

C.

400 m/s

The velocity with which a bullet or shell leaves the muzzle of a gun.
 
We  know that in projection of the particle for maximum range, θ = 45o
 
Now maximum rangeR =μ2sin 2θgR=μ2 sin 90og        sin 90o =1μ =Rg               .....(1)Given :- Rmax = 16 km= 16×103 mg = 10 m/s2Now from equation(1)μ =16×103×10=400 m/s
 
 
Question 4

The length, breadth and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm. The volume of the block according to the idea of significant figures should be

  • 1 × 102 cm3

  • 2 × 102 cm3

  • 1.76 × 102 cm3

  • None of these

Solution

B.

2 × 102 cm3

Using relation for volume

Given:- length of a block  =  12 cm

          breadth of a block  =  6 cm

        thickness of a block  =  2.45cm

V =length  × breadth  × thickness

  =12  × 6  × 2.45  = 176.4

 =1.764 ×102 cm3

The minimum number of significant figures is in thickness, hence the volume will contain only one significant figure.

Therefore V= 2 ×102 cm3 

Question 5

Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

  • 1 kg m2

  • 0.1 kg m2

  • 2 kg m2

  • 0.2 kg m2

Solution

B.

0.1 kg m2

The moment of inertia of given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by :

MI of disc + MI of particle

Since the mass of the disc is negligible

∴ MI of the system = MI of particle

= 5 × 2 × (0.1)2

= 0.1 Kg m2