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Physics NEET Questions — Free MCQ Practice

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The ratio of escape velocity at earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of earth is

A.

$1 : 2 \sqrt{2}$

B.

$1 : 4$

C.

$1 : \sqrt{2}$

D.

$1 : 2$

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Explanation

(a) Since, the escape velocity of earth can be given as

$(v_{e})$ = $\sqrt{2 g R} = R \sqrt{\frac{8}{3} πG ρ}$ $[ρ = d e n s i t y o f e a r t h]$

$\Rightarrow v_{e} = R \sqrt{\frac{8}{3} πGρ}$ ...(i)

As it is given that the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be

$v_{p} = 2 R \sqrt{\frac{8}{3} πG 2 ρ}$ ...(ii)

Divide, eq, (i) by eq.(ii), we get

$\frac{v_{e}}{v_{p}} = \frac{R \sqrt{\frac{8}{3} πGρ}}{2 R \sqrt{\frac{8}{3} πG (2 ρ)}} \Rightarrow \frac{v_{e}}{v_{p}} = \frac{1}{2 \sqrt{2}}$

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Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between the sun and planet i.e. T²=Kr^3, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitation force of attraction between them is F=GMm/r², here G is gravitational constant. The relation between G and K is described as

A.

GK=4π²

B.

GMK=4π²

C.

K=G

D.

K=l/G

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Explanation

The gravitational force of attraction between the planet and sun provide the centripetal force
i.e. $\frac{G M m}{r^{2}}$ =mv²/r =>v= $\sqrt{\frac{G M}{r}}$

The time period of planet will be

T=2πr/v =>T²= $\frac{4 π^{2} r^{2}}{Gm / r}$ = $\frac{4 π^{2} r^{3}}{GM}$ ...(i)

Also from Kepler's third law

T²=Kr³...(ii)

From Eqs. (i) and (ii), we get
$\frac{4 π^{2} r^{3}}{GM}$ =Kr³=>GMK=4π²

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Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

A.

2.5 R

B.

4.5 R

C.

7.5 R

D.

1.5 R

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Explanation

The collision distance between two spherical bodies of masses M and 5M with initial separation 12R is 7.5R for the smaller body. This can be derived using Newton's law of gravitation and principles of conservation of energy and momentum.

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A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25 x 10⁶ m above the surface of earth. If earth’s radius is 6.38x10⁶ m and g=9.8ms^-1, then the orbital speed of the satellite is

A.

7.76kms^-1

B.

8.56kms^-1

C.

9.13kms^-1

D.

6.67kms^-1

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A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Then,

A.

the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

B.

the total mechanical energy of S varies periodically with time

C.

the linear momentum of S remains constant in magnitude

D.

the acceleration of S is always directed towards the centre of the earth

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Explanation

As we know that, force on satellite is only gravitational force which will always be towards the centre of earth Thus, the acceleration of S is always directed towards the centre of the earth

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A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass=5.98X 10²⁴ kg) have to be compressed to be a black hole?

A.

10^-9m

B.

10^-6 m

C.

10^-2m

D.

100 m

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Explanation

Problem Solving Strategy For the black hole, the escape speed is more than c (speed of light). We should compare the escape speed with the c (Note that the escape speed should be at least just greater than c)

$v_{e} = \sqrt{2 G M / R}' [R' - > N e w r a d i u s o f t h e e a r t h] c = \sqrt{2 G M / R'} [V_{e} \approx c]$

=> c²=2GM/R'

R'=2GM/c²= $\frac{2 x 6.67 x 10^{- 11} x 6 x 10^{24}}{9 \times 10^{16}}$ =8.89x10^-3=0.889x10^-2=10^-2 m

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Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1m,2m,4m,8m, respectively from the origin. The resulting gravitational potential due to this system at the origin will be

A.

-G

B.

-8/3G

C.

-4/3G

D.

-4G

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Explanation

(d)The resulting gravitational potential,

$V = - 2 G [1 / 1 + 1 / 2 + 1 / 4 + 1 / 8 + . . .] V = - 2 G [1 + 1 / 2 + 1 / 2^{2} + 1 / 2^{3} . . .]$

V=-2G(1-1/2)^-1

V=-2G/(1-1/2)

=-2G/(1/2)

=-4G

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The height at which the weight of a body becomes 1/16th, its weight on the surface of earth(radius R), is

A.

5 R

B.

15 R

C.

3 R

D.

4 R

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Explanation

According to question $\frac{G M m}{{(R + h)}^{2}} = \frac{1}{16} \frac{G M m}{R^{2}}$

$\frac{1}{{(R + h)}^{2}} = \frac{1}{16 R^{2}}$

$o r \frac{R}{R + h} = \frac{1}{4}$

$\frac{R + h}{R} = 4$

h=3R

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A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth.The time period of another satellite in hours at a height of 2R from the surface of the earth is

A.

5

B.

10

C.

6 $\sqrt{2}$

D.

6/ $\sqrt{2}$

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Explanation

From Keplar third's law

$T^{2} \propto r^{3}$

Hence, $T_{1}^{2} \propto r_{1}^{3}$

and $T_{2}^{2} \propto r_{2}^{3}$

So, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{r_{2}^{3}}{r_{1}^{3}}$

$= \frac{{(3 R)}^{3}}{{(6 R)}^{3}}$

or $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{1}{8}$

$T_{2}^{2} = \frac{1}{8} T_{1}^{2}$

$T_{2} = \frac{24}{2 \sqrt{2}} = 6 \sqrt{2} h$

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If $v_{e}$ is escape velocity and $v_{0}$ is orbital

velocity of a satellite for orbit close to the

Earth's surface, then these are related by

A.

$v_{o} = \sqrt{2 v_{e}}$

B.

$v_{o} = v_{e}$

C.

$v_{e} = \sqrt{2 v_{o}}$

D.

$v_{e} = \sqrt{2} v_{o}$

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Explanation

Orbital velocity of satellite

$v_{o} = \sqrt{\frac{GM}{R}}$ ....(i)

and escape velocity of satellite

$v_{e} = \sqrt{\frac{2 GM}{R}}$ ...(ii)

Hence, $v_{e} = \sqrt{2} v_{o}$

Physics MCQs for NEET — Practice Questions with Answers

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