JEE advance 2013 - A particle of mass 0.2 kg

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. The initial speed (in m / s ) of the particle is zero, the speed ( in m / s ) after 5 sec is_ _ _ _ 

{ JEE advance 2013 }

The magnetic induction at center of coil

A Closely wound flat circular coil of 50 turns wound flat circular coil of 50 turns has a diameter of 10 cm carries a current of 4 ampere. Determine the magnetic induction at center of coil.

Solution:

In LCR series circuit, an alternating e.m.f. ‘ e ’ and current ‘ i ’ is given by equation e = 100 sin ( 100 t ) and i = 100 sin [ 100 t + (π/3) ] mA The average power dissipated in the circuit will be _ _ _ _ _ _ _ _ _ ( MHT - CET 2014 )

a ) 100 W              

b ) 10 W        

c ) 5 W                  

d ) 2.5 W

What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?

Solution :

Given :

a)      q ₁ = 2 × 10 ^{– 7} C

b)      q ₂ = 3 × 10 ^{– 7} C

c)       r   = 30 cm = 30 X 10 ^{– 3} = 3 X 10 ^{– 2}

d)      ε ₀ = 8.854 × 10 ^{– 12} C ² N ^–1 m^{– 2}

Formula : 

The force between two small charged spheres can be given as

F = [ 1 / ( 4 π ε₀ ) ] [ ( q₁ q₂) / ( r² ) ]

Therefore,

F = [ 1 / ( 4 X 3.14 X 8.854 × 10^–12 ) ] { ( 2 × 10^–7 X 3 × 10^–7) / [ ( 3 X 10^{– 2})² ] }

F = [ 1/(1.11 × 10^–10 ) ] [ ( 6 × 10^–14 ) / ( 9 X 10 ^–4 ) ]

F = [ 9 X 10 ⁹ ) ] [ ( 6 × 10^–10 ) / ( 9 ) ]

F = 60 N

The force between two small charged spheres having charges of 2 × 10^–7C and 3 × 10^–7C placed 30 cm apart in air is 60 newton 

The angle of minimum deviation of a glass prism is π – 2 A , where A is angle of prism. The angle of prism is _ _ _ _ _ _ _ _ _ _ _ _ .

a )      2 tan μ

    b )      2 tan^-1 μ   

    c )       2 cot^-1 μ

    d )      2 cot μ

A ray of monochromatic light is incident at angle ‘i’ on the surface of glass slab. If the angle of refraction ‘r’ is twice the angle of incidence, the angle of incidence is_ _ _ _ _ _ _ _ .

a)      i =  sin ^{- 1} ( μ / 2 )

b)      i = 2 cos ^{- 1} ( μ / 2 )

c)       i =  cos ^{- 1} ( μ / 2 )

d)      i = 2 sin ^{- 1} ( μ / 2 )

A planet of mass m moves in the inverse square central force field of the Sun of mass M . If the semi-major and semi-minor axes of the orbit are a and b , respectively, the total energy of the planet is:

          (a)    - ( G M m ) / ( a + b )

          (b)   - ( G M m ) [ ( 1 / a ) + ( 1 / b ) ]

          (c)    – [ ( G M m ) / a ] [ ( 1 / b ) - ( 1 / a ) ]

          (d)   - ( G M m ) / [ ( a – b ) / ( a + b )² ]