Question
Easy
A uniform ring of mass 'm' is lying at a distance $\sqrt{3}R$ from the center of the sphere of mass 'M' just over the sphere (where 'R' is the radius of the ring as well as of the sphere), then the magnitude of the gravitational force between them is:
1
$\frac{\sqrt{3}GMm}{R^{2}}$
2
$\frac{\sqrt{3}}{8}\frac{GMm}{R^{2}}$
3
$\frac{GMm}{8R^{2}}$
4
$\frac{G}{2}\frac{Mm}{R^{2}}$
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 2
Chapter: Gravitation
Topic: Mass and Weight
Correct Answer
Option B
Explanation
To determine the gravitational force between a uniform ring and a sphere, we need to consider the gravitational interaction between the two objects. The given problem specifies that the ring is at a distance of \(\sqrt{3}R\) from the center of the sphere, where \(R\) is the radius of both the ring and the sphere. ### Explanation for Option 2: The gravitational force between two masses is given by Newton's lawтАжRead More
Similar Questions from REET Exam - Paper 1 - Year 2018
Question 1
Easy
Source :
HTET 2022
Which of the Planet has the largest number of satellites ?
Chapter :
Gravitation
Topic :
Motion under Gravity
Question 2
Easy
Source :
HTET 2023
Range of the gravitational force is :
Chapter :
Gravitation
Topic :
Mass and Weight
Question 3
Easy
Source :
HTET 2021
When vapour pressure of a liquid is equal to the atmospheric pressure, what happens to the liquid ?
Chapter :
Fluid Mechanics
Topic :
Pressure in Fluids
Question 4
Easy
Source :
HTET 2022
In which of the following conditions, the distance between the molecules of hydrogen gas would increase, filled in a container ? (i) Some hydrogen gasтАж
Chapter :
Fluid Mechanics
Topic :
Pressure in Fluids