if you want to remove an article from website contact us from top.

# the value of universal gravitational constant g was first experimentally determined by

Category :

### Mohammed

Guys, does anyone know the answer?

get the value of universal gravitational constant g was first experimentally determined by from screen.

## The value of universal gravitational constant 'G' was first experimentally determined by :

The value of universal gravitational constant

Question ′ G ′

was first experimentally determined by :

A

B

C

D

## Kelvin

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

Cavendish experiment . The Cavendish experiment , performed in 1797–1798 by British scientist Henry Cavendish, was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield accurate value for the gravitational constant

Hence,

Solve any question of Gravitation with:-

Patterns of problems

>

14 0

स्रोत : www.toppr.com

## The value of gravitational constant G was first experimentally class 11 physics CBSE

The value of gravitational constant G was first experimentally determined by A Galileo B Newton C Cavendish D Kelvin

The value of gravitational constant G was first experimentally determined by:

A) Galileo B) Newton C) Cavendish D) Kelvin Answer Verified 223.8k+ views 2 likes

Hint: The value of gravitational constant, G first determined was

6.75× 10 −11 N m 2 k g −2 6.75×10−11Nm2kg−2 .

After nearly a century in 1798, Lord Henry Cavendish came to determine the value for that constant G by the experiment of torsional balance.

He attached the two small lead spheres in each side of the rod and the rod was suspended by a thin wire. When the rod is twisted, the torsion of the wire begins to exert the torsional force which is proportional to the angle of rotation. Then the more twisting leads the system to restore itself backward more. By this, Cavendish calculated the relationship between the angle of rotation to the amount of torsional force.

And then he brought two large spheres near to the small lead spheres connected to the rod. The large spheres exert the gravitational force on the small spheres which causes the rod to twist. Once the torsional force equals the gravitational force, the twisting stops. The rod comes to rest. Using this, Cavendish calculated the force of attraction. Then finally he determined the value of G.

Cavendish determined G as

6.75× 10 −11 N m 2 k g −2 6.75×10−11Nm2kg−2

. Now the accepted value is

6.67× 10 −11 N m 2 k g −2 6.67×10−11Nm2kg−2 ∴ ∴

The correct option is option (C).

(i)Sir Issac Newton in 1687, in his book philosophiae Naturalis Principia Mathematica discovered the term Gravitational constant for the first time while explaining the gravitational force.

(ii)Newton states that the gravitational force between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

∴ F grav α m 1 m 2 d 2 ∴Fgravαm1m2d2

. To remove this proportionality he introduced the constant G.

F grav =G m 1 m 2 d 2 Fgrav=Gm1m2d2

(iii) After nearly a century in 1798, Lord Henry Cavendish came to determine the value for that constant G by the experiment of torsional balance.

(vi) Cavendish determined G as

6.75× 10 −11 N m 2 k g −2 6.75×10−11Nm2kg−2

. Now the accepted value is

6.67× 10 −11 N m 2 k g −2 6.67×10−11Nm2kg−2 .

Note:

The gravitational force is the weakest of all forces in the universe. It has an appreciable value only if it is between the two objects having large masses.

स्रोत : www.vedantu.com

## [Solved] The value of ‘G’, the Universal Gravitational Co

The correct answer is Cavendish. Key Points Cavendish experiment. The Cavendish experiment, performed in 1797–1798 by British scientist Hen

Home Physics Gravitation The gravitational constant

## The value of ‘G’, the Universal Gravitational Constant, was measured experimentally by:

This question was previously asked in

Kerala SET Paper 1: Held on July 2018

View all Kerala SET Papers >

Newton Cavendish Copernicus Kepler

Option 2 : Cavendish

Crack with

India's Super Teachers

FREE

Demo Classes Available*

Free Tests

View all Free tests >

FREE

Kerala SET Paper 1: Held on 10th January 2020

0.5 K Users

120 Questions 120 Marks 120 Mins

Start Now

## Detailed Solution

Key PointsCavendish experiment.

The Cavendish experiment, performed in 1797–1798 by British scientist Henry Cavendish.

It was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield an accurate value for the gravitational constant.

Cavendish was the first to establish the value of the universal gravitational constant G through his experiment.

The apparatus used by him included two small spheres made of lead attached to the ends of a bar of length L.

The bar is suspended from rigid support by a fine wire.

He noticed that, when two spheres were brought simultaneously on the opposite sides of the small spheres, the big spheres attract the nearby small ones by an equal and opposite gravitational force (F). This produced torque of magnitude FL.

Due to this torque, the suspended wire gets twisted until the restoring torque of the wire equals the gravitational torque.

If θ was the angle of twist of the wire, the restoring torque = τθ (Where τ is the restoring couple per unit angle of twist)

On equating the equations,

FL = τθ θ ⇒GMmd2L=τθ

Where m is the mass of the small sphere, and M is the mass of the big sphere.

Observation of θ thus enables one to calculate G from this equation.

The Cavendish experiment determines the magnitude of gravitational attraction between two bodies.

It is based on the principle that a suspended wire resists torsion with a force that is due to gravity.

Hence, it is based on force-torque equilibrium.

Share on Whatsapp

Last updated on Sep 22, 2022

Kerala SET has released the notification for January 2023. The candidates can apply from 1st October to 20th October 2022. Candidates who qualify this exam can apply for teaching posts at Higher Secondary Schools in the state. The Kerala SET exam consists of two papers, i.e. Paper I which comprises subjects like General Knowledge and Aptitude. Paper I is common for all candidates, and Paper-II which comprises tests based on the subject of specialization of the candidate at the Post Graduate (PG) Level. Know the Kerala SET cut off details here.

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Trusted by 3.2 Crore+ Students

‹‹ Previous Ques Next Ques ››

## More The gravitational constant Questions

Q1. The motion of the moon around the earth is due to ________-Q2. The value of 'g' is ________-Q3. The value of 'g' is less than the actual value of 'g' in ________.Q4. An object weighs 60 N when measured on the surface of the earth. Its weight on the surface of the Moon will be _______.Q5. The value of the universal gravitational constant in CGS system of units is equal to:Q6. The universal gravitational constant numerically equals-Q7. The value of the gravitational constant in the cgs unit is-Q8. The spheres used in Cavendish's experiment were made out of-Q9. Cavendish's experiment is based on the principle of-

## More Gravitation Questions

Q1. An object weighs 9 N on the surface of the Earth. What would be its weight, when measured on the surface of a planet where the acceleration due to gravity is 9 times that on the surface of the Earth?Q2. What happens to the gravitational force between two objects if the mass of one object is doubled and the distance between them is also doubled?Q3. Which one of the following statements with respect to Global Positioning System (GPS) is not correct?Q4. Which of these phenomena cannot be explained on the basis of Gravitational force?Q5. Read the following paragraph and fill in the blanks with the correct option Every object is attracted by earth with a force which depends on the A of the object and acceleration due to gravity. The B of the object is equal to the force with which it is attracted towards earth. However, on the surface of moon C changes but D remains same.

स्रोत : testbook.com

Do you want to see answer or more ?
Mohammed 2 month ago

Guys, does anyone know the answer?