Physics Questions and Answers – Acceleration due to Gravity of the Earth

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This set of Physics MCQs for NEET Exam focuses on “Acceleration due to Gravity of the Earth”.

1. The acceleration due to gravity on the surface of the earth is different at different points on the surface.
a) True
b) False
View Answer

Answer: a
Explanation: Since the earth is not a perfect sphere and has many irregularities, the acceleration due to gravity is different at different points on the earth’s surface.
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2. 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 poles
View Answer

Answer: b
Explanation: 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.

3. The dimensions of acceleration due to gravity are _____
a) [M0L1T-2]
b) [M1 L-1T-2]
c) [M-1 L2 T-1]
d) [M0L-1T2]
View Answer

Answer: a
Explanation: The unit of acceleration is m/s2.
m/s2 = [L1 T-2]
= [M0 L1 T-2].
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4. 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) False
View Answer

Answer: b
Explanation: g = (G*M1)/R2;
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.

5. 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/s2
b) 12.26 m/s2
c) 15.33 m/s2
d) 49.05 m/s2
View Answer

Answer: c
Explanation: g = (G*M1)/R2;
M1 = Mass of the earth
R = Radius of the earth
We know; g = 9.81 m/s2
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/s2.
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6. 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/s2
b) 12.26 m/s2
c) 15.33 m/s2
d) 29.43 m/s2
View Answer

Answer: d
Explanation: g = (G*M1)/R2;
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/s2
New mass = 3 x M1
Therefore, new acceleration;
3 x (G*M1)/R2 = 3x g
= 3 x 9.81
= 29.43 m/s2.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter