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101.

At what height above the surface of the earth, the acceleration due to the gravity of the earth becomes 5% of that of the surface?(a) h = 0.5 R(b) h = 1.5 R(c) h = 2.5 R(d) h = 3.5 RThe question was asked in my homework.The origin of the question is Acceleration due to Gravity below and above the Surface of Earth topic in portion Gravitation of Physics – Class 11

Answer»

Right ANSWER is (d) H = 3.5 R

To explain I would SAY: Acceleration due to gravity on the surface of the earth;

g = (G*M1)/R^2

Acceleration due to gravity above the surface of the earth;

g’ = (G*M1)/(R + h)^2

g’ = (5/100) x g

(G*M1)/(R + h)^2 = (5/100) x (G*M1)/R^2

R^2 x 100 = (R + h)^2 x 5

Taking square root on both sides;

R x 10 = (R + h) x 2.24

7.76 x R = 2.24 x h

h = 3.5 x R.
Discussion
102.

Acceleration due to gravity increases as we move away from the surface of the earth radially towards the sky.(a) True(b) FalseThis question was posed to me at a job interview.Origin of the question is Acceleration due to Gravity below and above the Surface of Earth topic in division Gravitation of Physics – Class 11

Answer»

Correct ANSWER is (b) False

Explanation: The volume or mass enclosed by the radius vector from the centre of the earth remains the same as we move HIGHER from the earth’s SURFACE. However, the increase in radius decreases the acceleration due to gravity. The mathematical relationship is as follows;

G = (G*M1)/R^2;

M1 = Mass of the earth

R = Radius vector from the centre of the earth.
Discussion
103.

What would be the magnitude of the acceleration due to gravity on the surface of the earth if the density of the earth increased by 3 times and the radius remained the same?(a) 9.81 m/s^2(b) 12.26 m/s^2(c) 15.33 m/s^2(d) 29.43 m/s^2The question was posed to me in an internship interview.Query is from Acceleration due to Gravity of the Earth in portion Gravitation of Physics – Class 11

Answer»

The CORRECT choice is (d) 29.43 m/s^2

The explanation is: g = (G*M1)/R^2;

M1 = Mass of the earth

R = Radius of the earth

Since the radius is the same, the volume would remain constant.

Density = mass/volume

Since density is increased 3 times and volume is the same, this implies that mass is increased 3 times.

We know; g = 9.81 m/s^2

New mass = 3 X M1

Therefore, new ACCELERATION;

3 x (G*M1)/R^2 = 3X g

= 3 x 9.81

= 29.43 m/s^2.
Discussion
104.

Gravitational force is _____(a) an imaginary force(b) a long-range force(c) a short-range force(d) the strongest fundamental forceI had been asked this question during an interview.My doubt is from Universal Law of Gravitation in division Gravitation of Physics – Class 11

Answer»

The correct option is (b) a long-range force

Easy explanation: Gravitational force is a long-range force which is INVERSELY proportional to the square of the distance. The STRONG nuclear force is the STRONGEST fundamental force and is a short-range force.
Discussion
105.

Which of the following is the variation of acceleration due to gravity at a depth “d” below the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume the density of the earth to be constant)(a) g = (G*M1)/(R – d)(b) g = [(G*M1) x density]/d(c) g = (G x M1/R^3) / (R – d)(d) g = (G x M1/R^3) x (R – d)The question was posed to me in an interview.Query is from Acceleration due to Gravity below and above the Surface of Earth topic in section Gravitation of Physics – Class 11

Answer»

The CORRECT option is (d) g = (G x M1/R^3) x (R – d)

EASY explanation: g = (G*M)/r^2;

M = MASS contained in an enclosed volume

r = distance from centre of the earth to the DEPTH “d”

Therefore; r = R – d; where, R = Radius of the earth

M = (density) x (volume)

 = (density) x [(4/3) x (pi) x (r^3)]

 = (density x 4 x pi x r^3) / 3

 = (M1 x r^3) / R^3

Therefore;

g = (G x density x 4 x pi x r^3) / (3 x r^2)

 = (G x density x 4 x pi x r^1) / 3

Therefore;

g = [(G x density x 4 x pi) / 3] x (R – d)

 = (G x M1/R^3) x (R – d).
Discussion
106.

What apparatus did Henry Cavendish use in his experiment to determine the gravitational constant?(a) 1 bar, 1 small sphere and 1 large sphere(b) 1 bar, 2 small spheres and 2 large spheres(c) 2 bar, 1 small sphere and 2 large spheres(d) 2 bar, 2 small spheres and 1 large sphereI had been asked this question in a national level competition.This interesting question is from The Gravitational Constant in chapter Gravitation of Physics – Class 11

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Correct ANSWER is (b) 1 BAR, 2 small spheres and 2 large spheres

Explanation: Henry Cavendish used 1 bar, 2 small spheres and 2 large spheres in his EXPERIMENT to determine the gravitational CONSTANT. By measuring the TORQUE and angle of deflection produced due to the proximity of large spheres to the small ones, Henry Cavendish was able to determine the universal gravitational constant.
Discussion
107.

Kepler’s laws of planetary motion replaced circular orbits with _____(a) elliptical orbits(b) parabolic orbits(c) conical orbits(d) hyperbolic orbitsThe question was asked during an online exam.Query is from Gravitation topic in portion Gravitation of Physics – Class 11

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The CORRECT answer is (a) elliptical orbits

Easy explanation: From the first law of KEPLER’s laws of planetary motion, we can infer that the orbit of a planet is an ellipse with the SUN at ONE of the FOCI.
Discussion
108.

What does Kepler’s law of period relate?(a) Time period and semi-minor axis(b) Time period and eccentricity(c) Time period and semi-major axis(d) Time period and area swept by the planetThe question was posed to me during an online exam.This question is from Gravitation in chapter Gravitation of Physics – Class 11

Answer»

The correct option is (c) Time period and semi-major axis

The BEST explanation: According to KEPLER’s LAW of periods, the square of time period f revolution of a planet is directly proportional to the cube of the semi-major axis of the planet’s ELLIPTICAL orbit.
Discussion
109.

From Kepler’s law of orbit, we can infer that the sun is located _____ of the planet’s orbit.(a) at the centre(b) at one of the foci(c) at both foci(d) anywhere along the semi-minor axisThis question was posed to me by my college professor while I was bunking the class.I'd like to ask this question from Gravitation topic in portion Gravitation of Physics – Class 11

Answer»

Right answer is (B) at one of the FOCI

For explanation: According to Kepler’s law of orbit, every planet revolves around the sun in an ELLIPTICAL orbit and the sun is at one of the foci.
Discussion
110.

The acceleration due to gravity on the surface of the earth is _____(a) greater towards the equator and lesser towards the poles(b) lesser towards the equator and greater towards the poles(c) same at all points on the surface of the earth(d) same everywhere except at the polesThis question was addressed to me in an interview for job.Asked question is from Acceleration due to Gravity of the Earth topic in division Gravitation of Physics – Class 11

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Correct choice is (b) LESSER towards the equator and greater towards the poles

To elaborate: The earth is not a perfect sphere. The radius of the earth at the equator is greater than that at the poles, HENCE, the ACCELERATION DUE to gravity is lesser towards the equator and greater towards the poles.
Discussion
111.

Acceleration due to gravity increases as we go towards the centre of the earth.(a) True(b) FalseThe question was asked in an online interview.I'm obligated to ask this question of Acceleration due to Gravity below and above the Surface of Earth topic in portion Gravitation of Physics – Class 11

Answer» CORRECT option is (b) False

The best I can explain: Acceleration due to gravity decreases as we GO towards the centre of the earth because the volume enclosed by the RADIUS vector from the centre of the earth decreases as DEPTH increases. This reduction in volume leads to a reduction in enclosed mass and hence, the acceleration due to gravity is reduced.
Discussion
112.

For an object on the surface of the earth, the magnitude of the acceleration due to the gravity of the earth it experiences depends also depends on the mass of that object.(a) True(b) FalseThis question was posed to me during an interview.This interesting question is from Acceleration due to Gravity of the Earth in chapter Gravitation of Physics – Class 11

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Correct choice is (b) False

The BEST I can explain: g = (G*M1)/R^2;

M1 = MASS of the earth

The acceleration due to the gravity of the earth experienced by any OBJECT on the surface of the earth depends only on the mass of earth and the square of the distance between the object and the centre of the earth.
Discussion
113.

The dimensions of acceleration due to gravity are _____(a) [M^0L^1T^-2](b) [M^1 L^-1T^-2](c) [M^-1 L^2 T^-1](d) [M^0L^-1T^2]The question was posed to me in semester exam.My question comes from Acceleration due to Gravity of the Earth in chapter Gravitation of Physics – Class 11

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The CORRECT CHOICE is (a) [M^0L^1T^-2]

Easy EXPLANATION: The unit of ACCELERATION is m/s^2.

m/s^2 = [L^1 T^-2]

= [M^0 L^1 T^-2].
Discussion
114.

The universal law of gravitation becomes more inapplicable as the size and distance between objects decreases.(a) True(b) FalseI had been asked this question in examination.I'd like to ask this question from Universal Law of Gravitation topic in division Gravitation of Physics – Class 11

Answer» CORRECT answer is (a) True

The explanation is: As the size and distance between OBJECTS decrease, nuclear forces become stronger and the law of gravitation cannot be APPLIED. This calls for a new branch of PHYSICS KNOWN as “quantum physics”.
Discussion
115.

The elliptical orbits of planets were indicated by calculations of the orbit of which astronomical body?(a) Mercury(b) Earth(c) Earth’s moon(d) MarsI got this question in an interview.I'm obligated to ask this question of Gravitation in portion Gravitation of Physics – Class 11

Answer»

The correct answer is (d) MARS

The best explanation: By observing the MOTION of Mars in the sky, Kepler inferred that the planets have ELLIPTICAL orbits around the sun. Kepler DISCOVERED that a simple ellipse would clearly define the orbit of Mars and eliminate many complexities. It would ALSO eliminate the need for epicycles.
Discussion
116.

What will be the value of acceleration due to gravity on the surface of the earth if the radius of the earth suddenly decreases to 60% of its present value, keeping the mass of the earth unchanged?(a) 9.81 m/s^2(b) 5.89 m/s^2(c) 16.35 m/s^2(d) 27.25 m/s^2This question was addressed to me in my homework.My question is from Universal Law of Gravitation in portion Gravitation of Physics – Class 11

Answer»

The correct option is (d) 27.25 m/s^2

For explanation I WOULD say: From Newton’s LAW of gravitation, we have;

g = (G*M1)/R^2

Since the radius is REDUCED to 60% of its original value;

The NEW radius R’ = 0.6 x R

Therefore;

g’ = (G*M1)/(0.6*R) ^2

g’ = g/(0.6) ^2

= 9.81 / (0.6 x 0.6)

g’ = 27.25 m/s^2.
Discussion
117.

Gravitational force is the strongest fundamental force.(a) True(b) FalseThe question was posed to me during an interview.My question is taken from Universal Law of Gravitation in division Gravitation of Physics – Class 11

Answer»

The CORRECT CHOICE is (b) False

For explanation: Gravitational FORCE is the weakest fundamental force. The strong NUCLEAR force is the strongest fundamental force of nature.
Discussion
118.

Let the radius of the earth be R. Now, assume that the earth shrunk by 20% but the mass is the same. What would be the new value of acceleration due to gravity at a distance R from the centre of the earth if the value at the same distance in the previous case was “g’”?(a) g’(b) 2g’(c) 3g’(d) 4g’I had been asked this question by my college professor while I was bunking the class.The above asked question is from Acceleration due to Gravity below and above the Surface of Earth topic in portion Gravitation of Physics – Class 11

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The correct option is (a) g’

Easy explanation: The value of acceleration due to gravity before SHRINKING at a distance R from the centre of the earth, i.e., on the surface of the earth is;

g’ = (G*M)/R^2

Where; M = MASS of the earth

Now, the earth shrunk by 20%, HOWEVER, the mass remains the same. This IMPLIES an increase in density.

The value of acceleration due to gravity from an object at a point outside the object is DEPENDENT only on the distance between the centre of gravity of the object and the distance between the point and the object. It is independent of density.

Hence, the new value of acceleration due to gravity at the same distance will remain unchanged, i.e., g’.
Discussion
119.

What would be the magnitude of the acceleration due to gravity on the surface of the earth if the radius of the earth were reduced by 20%?(a) 9.81 m/s^2(b) 12.26 m/s^2(c) 15.33 m/s^2(d) 49.05 m/s^2The question was posed to me by my school principal while I was bunking the class.My question comes from Acceleration due to Gravity of the Earth topic in division Gravitation of Physics – Class 11

Answer»

Correct ANSWER is (c) 15.33 m/s^2

Easy explanation: g = (G*M1)/R^2;

M1 = Mass of the earth

R = RADIUS of the earth

We KNOW; g = 9.81 m/s^2

If the radius is reduced by 20% then the new radius is 80% of the original one;

New radius = 0.8R

Therefore, new acceleration;

g / (0.8 x 0.8) = 9.81 / 0.64

= 15.33 m/s^2.
Discussion
120.

The value of acceleration due to gravity of earth at the equator is less than that of the poles due to _____(a) shape and rotation of the earth(b) mass of the sun(c) mass of the earth(d) mass of the moonThis question was addressed to me in an internship interview.My question is taken from Universal Law of Gravitation in chapter Gravitation of Physics – Class 11

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The CORRECT choice is (a) shape and rotation of the earth

The explanation: The gravitational FORCE is a central force. Acceleration due to GRAVITY has DIFFERENT values for different POINTS on the earth’s surface.
Discussion
121.

What is the constant of proportionality in Kepler’s law of periods known as?(a) Universal gravitational constant(b) Escape velocity(c) There is no constant of proportionality(d) Cannot be determinedI got this question in an interview.Origin of the question is Gravitation in division Gravitation of Physics – Class 11

Answer»

The correct option is (c) There is no constant of proportionality

Best explanation: There is no particular constant of proportionality for KEPLER’s LAW of periods. The law of periods only relates the proportionality of the square of the time PERIOD of revolution of a planet to the CUBE of the semi-major axis of the orbit of the planet.
Discussion