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NEET Practice Questions (MCQs) with Answers & Solutions

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If the gravitational force between two objects were proportional to 1/R (and not as $ 1/R^2$ )where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

A. $ {1 \over R^2 } $ B. $ R^0$ C. $ R^1$ D. 1/R

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Explanation

Gravitational force provides the required centripetal force for orbiting the satellite $ { m \upsilon^2 \over R } = { K \over R } $ because $ F \alpha { 1 \over R } $ $ \therefore \upsilon \alpha R^o $

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A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of the satellite is M , its total energy is

A. $ - { 1 \over 2 } MV^2 $ B. $ { 1 \over 2 } MV^2 $ C. $ { 3 \over 2 } MV^2 $ D. $ MV^2 $

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Explanation

Total energy = -K.E. = $ {-1/2 m \upsilon^2 } $

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A satellite with K.E. $E_k$ is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outerspace ?

A. $ E_k$ B. $ 2 E_k$ C. $ 1/2 E_k$ D. $ 3 E_k$

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Explanation

B.E. = -K.E. And it this amount of energy(Ek) given to satellite it will escape into outer space

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Potential energy of a satellite having mass m and rotating at a height
of $6.4 x 10^6$ m from the surface of earth

A. 0.5 mg Re B. -mgRe C. -2mgRe D. 4mgRe

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Explanation

$ P.E = { GMm \over r } = - { GMm \over Re+h } = -{ GMm \over 2 Re } = - {g Re^2 m \over 2 Re } = - { 1 \over 2 } mg Re = 0.5 mg Re $

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When a satellite going round the earth in a circular obrit of radius r and speed v loses some of its energy, then r and v changes as

A. r and v both will increase B. r and v both will decease C. r will decrease and v will increase D. r will increase and v will decrease

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Explanation

$ B.E = { - GMm \over r } $ if B.E. decreases the r also decreases and V increases as $ v \alpha { 1 \over r } $

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A person sitting in a chair in a satellite feels weightless because

A. the earth does not attract the objects in a satellite B. the normal force by the chair on the person balances the earthâ€™s attraction C. the normal force is zero D. the person in satellite is not accelerated

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Explanation

$ {V_B \over V_A } = \sqrt { \gamma_A \over \gamma_B } = \sqrt { 4R \over R } = 2 $

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Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectively If the speed of satellite A is 3v, then speed of satellite B is

A. 3v/2 B. 4v/2 C. 6v D. 12v

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Explanation

For satellites in circular orbits, the orbital speed is inversely proportional to the square root of the orbital radius. If satellite A has a speed of 3v at a radius of 4R, then the speed of satellite B, which has a radius R, can be calculated as follows:

$$v_B = v_A imes rac{ ext{sqrt}(R_A)}{ ext{sqrt}(R_B)} = 3v imes rac{ ext{sqrt}(4R)}{ ext{sqrt}(R)} = 3v imes 2 = 6v.$$

Therefore, the speed of satellite B is 6v.

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A satellite moves in a circle around the earth, the radius of this circlr is equal to one half of the radius of the moonâ€™s orbit the satellite completes one revolution inâ€¦â€¦â€¦... lunar month

A. 1/2 B. 2/3 C. $ 2 ^ { - 3/2 } $ D. $ 12 ^ { 3/2 } $

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Explanation

time period of revolution of moon around the earth = 1 lunar month $ {T_e \over T_m } = \left( {r_e \over r_m } \right)^{3/2} = {1 \over 2 } ^ {3/2} = 2^{-2/3} $

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The additional K.E. 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 radius R2 (R2 > R1) is

A. $ GMm \left( { 1 \over R_1^2 } - { 1 \over R_2^2 } \right) $ B. $ GMm \left( { 1 \over R_1 } - { 1 \over R_2 } \right) $ C. $ 2GMm \left( { 1 \over R_1 } - { 1 \over R_2 } \right) $ D. $ {1 \over 2 } GMm \left( { 1 \over R_1 } - { 1 \over R_2} \right) $

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Explanation

$ - { GMm \over 2 R_1} + K.E = -{GMm \over 2R_2} $ $ K.E = { GMm \over 2 } \left( {1 \over R_1} - {1 \over R_2 } \right) $

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Rockets are launched in eastward direction to take advantage of

A. the clear sky on eastem side B. the thinner atmosphere on this side C. earthâ€™s rotation D. earthâ€™s tilt

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Explanation

Because Earth rotation from west to east direction

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