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
KE space equals space open parentheses fraction numerator GMm over denominator 2 straight R subscript 1 end fraction close parentheses space minus open parentheses negative fraction numerator GMm over denominator 2 straight R subscript 1 end fraction close parentheses

KE space equals space GMm over 2 open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets

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.

straight I space equals space straight I subscript 1 minus straight I subscript italic 2
space equals fraction numerator 9 MR squared over denominator 2 end fraction minus MR squared over 18
equals space fraction numerator 81 space MR squared minus MR squared over denominator 18 end fraction
equals space fraction numerator 40 space MR squared over denominator 9 end fraction

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 

straight x space equals space straight a space sin space ωt space rightwards double arrow space straight x over straight a space equals space sin space ωt
straight x space equals space straight a space cos space ωt space rightwards double arrow space straight x over straight a squared space equals space cos space ωt
therefore space straight x squared over straight a squared space plus straight y squared over straight b squared space equals space 1
or space
straight x squared space plus space straight y squared space equals space straight a squared
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