Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth

Short Answer

Expert verified
The weight of a body is maximum at the poles of the earth (B) due to higher gravitational acceleration caused by the Earth's shape (oblate spheroid).

Step by step solution

01

(1) Moon Vs Earth

The gravity of the moon is 1/6th that on the earth, so the weight of a body would be less on the moon compared to anywhere on earth.
02

(2) Weight at Poles Vs the Equator

The earth's shape (oblate spheroid) causes the weight of a body to vary slightly at different locations. Since the earth is bulging at the equator and flattened at the poles, the weight will be higher at poles due to higher gravitational acceleration.
03

(3) Weight at Centre of Earth

As we approach the center of the earth, the force exerted within the interior is balanced out by the forces acting from other sides, making gravitational acceleration, and hence the weight, zero at the center of the earth. By comparing these locations, we find that the weight of a body is maximum at the poles of the earth. The correct answer is: (B) poles of earth

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The square of the period of revolution of a planet is proportional to the cube of its distance from the sun. Reason: Sun's gravitational field is inversely proportional to the square of its distance from the planet (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free