Test: Gravitation (October 4) - NEET MCQ

# Test: Gravitation (October 4) - NEET MCQ

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## 15 Questions MCQ Test Daily Test for NEET Preparation - Test: Gravitation (October 4)

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Test: Gravitation (October 4) - Question 1

### A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of V.

Detailed Solution for Test: Gravitation (October 4) - Question 1

*Multiple options can be correct
Test: Gravitation (October 4) - Question 2

### An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change

Detailed Solution for Test: Gravitation (October 4) - Question 2

Let’s see the formulae for each quantity given there,
Gravitational potential energy  =−GMm/r​
angular velocity =(GM/r3​)0.5
linear orbital velocity  =(GM/r​)0.5
centripetal acceleration  =GM/ r2
We can observe that, the quantities in the options B,C,D are all indirectly proportional to r
i.e. ω, v, a α(1/rn)​ where n is a distinct positive integer in each quantity.
So, with increase in r all these quantities decrease.
Now, (-potential energy) α(1/r)​ i.e. negative of potential energy decreases with increase in r. Therefore, potential increases with increase in orbital radius.

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*Multiple options can be correct
Test: Gravitation (October 4) - Question 3

### For a satellite to orbit around the earth, which of the following must be true ?

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Test: Gravitation (October 4) - Question 4

The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse.

If U is the potential energy and K kinetic energy then |U| > |K| at

Detailed Solution for Test: Gravitation (October 4) - Question 4

If a planet is orbiting the sun then |U|> |K| else it will leave the system because it won’t be under the influence of the sun’s gravity.

*Multiple options can be correct
Test: Gravitation (October 4) - Question 5

In side a hollow spherical shell

Detailed Solution for Test: Gravitation (October 4) - Question 5

The gravitational field inside a uniform spherical shell is 0 from gauss law for gravitation since no mass is enclosed in any Gaussian surface.
Since gravitational potential is given by φ=−∫ gdr, hence, φ=constant since g=0.
Since the gravitational field is 0 everywhere, it is apparently the same everywhere.

*Multiple options can be correct
Test: Gravitation (October 4) - Question 6

Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start under their mutual gravitational attraction

Detailed Solution for Test: Gravitation (October 4) - Question 6

The total energy of the system remains constant.

Test: Gravitation (October 4) - Question 7

The orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse.

If t1 is the time taken by the planet to travel along ACB and t2 the time along BDA, then

Detailed Solution for Test: Gravitation (October 4) - Question 7

Here, the angular momentum is conserved i.e. mvr is constant. Where r is the distance from centre of the sun to centre of the planet.
Now, consider one point each from both the mentioned paths. (Shown in fig.)
Applying conservation of angular momentum for these points we get
mv1​r1​=mv2​r2​. Simplifying this, v1/​v2​​=r2​/r1​​<1
Time period are given by
t1​=L/ v1​, t2​= L​/ v2
Comparing them by taking the ratio
t1/t2​​=(L​/ v1)×(v2/L)
So, t1/t2=v2​/v1​​>1.Thus t1​>t2

*Multiple options can be correct
Test: Gravitation (October 4) - Question 8

When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then

Detailed Solution for Test: Gravitation (October 4) - Question 8

Due to air drag some mechanical energy of satellite will converted into heat energy, there will be loss of ME of satellite, so radius of orbit will decrease and satellite will follow a spiral path towards earth.
If r is decreases,
P.E.=−GMm​/r, P.E. is also decreases.
K.E=GMm​/2r K.E increasing, increased speed in spite of decrease in M.E. But rate of P.E decreases is more than the rate at which M.E decreases.
Angular momentum L=mrv=mr√GM​r=m√GMr​⇒L∝√r​.
So, angular momentum decreases as r decreases.

*Multiple options can be correct
Test: Gravitation (October 4) - Question 9

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

Detailed Solution for Test: Gravitation (October 4) - Question 9

As the gravitational force on satellites due to earth acts always towards the centre of earth, thus acceleration of S is always directed towards the centre of the earth.
Also, as there is no external force so according to conservation of energy, total mechanical energy of S is constant always.
Also, as in the absence of external torque L is constant in magnitude and direction.
Thus, mrv=constant ⟹v varies as r changes
Hence, p=mv is not constant.

Test: Gravitation (October 4) - Question 10

Two masses m1 & m2 are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of d is

Test: Gravitation (October 4) - Question 11

A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

Test: Gravitation (October 4) - Question 12

Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Select the correct statement.

Detailed Solution for Test: Gravitation (October 4) - Question 12

Kinetic Energy of the Satellite =GMm​/2r
Potential Energy of the satellite =−GMm​/r
Total Energy of the Satellite =−GMm​/2r
Hence, Kinetic Energy Cure will be above 0 , hence X
Potential energy will be the lowest curve , hence Y
Total Energy is the curve is below 0 , but higher magnitude than Potential Energy  . Hence, Z

Test: Gravitation (October 4) - Question 13

Two uniform spherical stars made of same material have radii R and 2R. Mass of the smaller planet is m. They start moving from rest towards each other from a large distance under mutual force of gravity. The collision between the stars is inelastic with coefficient of restitution 1/2.

Kinetic energy of the system just after the collision is

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Test: Gravitation (October 4) - Question 14

A mass is at the center of a square, with four masses at the corners as shown.

(A)  (B)

(C)  (D)

Rank the choices according to the magnitude of the gravitational force on the center mass.

Detailed Solution for Test: Gravitation (October 4) - Question 14

Gravitational force by mass M on another body of m at a distance of r is given by F=GMmr2
For given figure Right diagonal is x axis and left y axis in figure A forces due to 5M will cancel out each other so effective force will be due to only 3M an M. In figure B 2M forces will cancel out each other and effective force will be due to 3M and M.
Hence total force on M in both the cases will be the same.
Similarly, in figure C, 5M mass will cancel each other out and in Figure D, 2M will cancel each other out and effective force will be due to 3M and M on the mass of 2M. So, force on c and d will be the same.
Force on C will be more than B.

Test: Gravitation (October 4) - Question 15

The escape velocity for a planet is ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

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