NEET physics

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Question
CBSEENPH11020703

The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2(R2> R1) is 

  • GmM space open parentheses fraction numerator 1 over denominator straight R subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight R subscript 2 superscript 2 end fraction close parentheses
  • GmM space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses
  • 2 GmM space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses
  • 1 half GmM space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

Solution

D.

1 half GmM space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

The kinetic energy changing the orbit of satellite
WiredFaculty

Sponsor Area

Question
CBSEENPH11020704

The speed of the projectile at its maximum height is half of its initial speed. The angle of projection is

  • 60o

  • 15o

  • 30o

  • 45o

Solution

A.

60o

The speed of projectile at its maximum height
v' = vo cos θ
vo/2 = vo cos θ
cos θ = 1/2
θ = 60o

Question
CBSEENPH11020705

From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

  • 40MR2/9

  • MR2

  • 4 MR2

  • 4MR2/9

Solution

A.

40MR2/9

The moment of inertia of the remaining disc about axis perpendicular to the plane of the disc and passing through its centre.

WiredFaculty

Question
CBSEENPH11020706

A particle moves in the x-y plane according to rule x = a sin ωt and y = a cos ωt. The particle follows

  • an elliptical path

  • a circular path 

  • a parabolic path

  • a straight line path inclined equally to x and y- axes

Solution

B.

a circular path 

WiredFaculty
This is the equation of a circle, so the particle follows a circular path.

Question
CBSEENPH11020707

A closely wound solenoid of 2000 turns and area of cross -section 1.5 x 10-4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 x 10-2 T making an angle of 30o with axis of  the solenoid. The torque on the solenoid will be

  • 3 x 10-3 N-m

  • 1.5 x 10-3 N-m

  • 1.5 x 10-2 N-m

  • 3 x 10-2 N-m

Solution

A.

3 x 10-3 N-m

Given, N = 2000,
A = 1.5 x 10-4 m2
i = 2.0 A B = 5 x 10-2 T
and
θ = 30o
Torque = NiBA sin θ
2000 x 2 x 5 x 10-2 x 1.5 x 10-14 x sin 30o
= 2000 x 50 x 10-6 x (1/2)
= 1.5 x 10-2 Nm