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DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET PDF Download

Introductory Exercise 10.1

Q.1. Calculate the change in the value of g at altitude 45°. Take radius of earth = 6.37 × 103 km
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
= g -0.0168 m/s2
Dg = g-g' = 0.0168 m/s

Q.2. Determine the speed with which the earth would have to rotate on its axis, so that a person on the equator would weigh 3/5th as much as at present. Take R = 6400 km
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.3. At what height from the surface of earth will the value of g be reduced by 36% from the value at the surface? R = 6400 km
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.4. The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square of r, the acceleration of the body A is a. If the gravitational force follows an inverse fourth power law, then what will be the acceleration of body A?
Sol.
 

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.5. Find the force of attraction on a particle of mass m placed at the centre of a semicircular wire of length L and mass M.
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Introductory Exercise 10.2

Q.1. A particle of mass m is placed at the centre of a uniform spherical shell of same mass and radius R. Find the gravitational potential at a distance R/2 from the centre.
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
V = V+ Vs = DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.2. A particle of mass 20 g experiences a gravitational force of 4.0 N along positive x-direction. Find the gravitational field at that point.
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
200 N/kg i

Q.3. The gravitational potential due to a mass distribution is V = 3x2y + y3z. Find the gravitational field.
Sol.

DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.4. Gravitational potential at x = 2 m is decreasing at a rate of 10 J/kg-m along the positive x-direction. It implies that the magnitude of gravitational field at x = 2 m is also 10 N/kg. Is this statement true or false?

Sol. Only the variation is given along x-axis, nothing is about y and z axis, so, the statement is false.

Q.5. The gravitational potential in a region is given by, V = 20(x+ y) J/kg. Find the magnitude of the gravitational force on a particle of mass 0.5 kg placed at the origin.

Sol. DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

Q.6. The gravitational field in a region is given by
DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET
Find the work done by the gravitational field when a particle of mass 1 kg is moved on the line 3y + 2x = 5 from (1 m, 1 m) to (-2 m, 3 m).

Sol.

 DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET

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FAQs on DC Pandey Solutions: Gravitation - 1 - Physics Class 11 - NEET

1. What is the formula for gravitational force?
Ans. The formula for gravitational force between two objects is given by Newton's law of universal gravitation, which states that the force is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them. Mathematically, it can be represented as: F = G * (m1 * m2) / r^2 Where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers.
2. How does the distance between two objects affect the gravitational force between them?
Ans. The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that as the distance between the objects increases, the gravitational force decreases. Conversely, as the distance decreases, the gravitational force increases. Therefore, the closer the objects are to each other, the stronger the gravitational force between them.
3. What is the significance of the gravitational constant in the formula for gravitational force?
Ans. The gravitational constant (G) is a fundamental constant in physics that determines the strength of the gravitational force between two objects. It is a universal constant and has the same value everywhere in the universe. The value of G is approximately 6.674 × 10^-11 N(m/kg)^2. The significance of the gravitational constant is that it allows us to calculate the gravitational force accurately based on the masses and distances involved.
4. What is the difference between mass and weight in the context of gravity?
Ans. Mass and weight are often used interchangeably, but they have different meanings in the context of gravity. Mass is a measure of the amount of matter in an object and remains constant regardless of the gravitational field. On the other hand, weight is the force exerted on an object due to gravity and depends on the mass of the object and the strength of the gravitational field. Weight can vary depending on the location, as the strength of the gravitational field differs from one place to another.
5. How does the gravitational force affect the motion of celestial bodies in the universe?
Ans. The gravitational force plays a crucial role in determining the motion of celestial bodies in the universe. It is responsible for keeping planets in orbit around the sun, moons around planets, and stars in galaxies. The gravitational force between celestial bodies causes them to attract each other, resulting in the formation and maintenance of stable orbital paths. It is this force that keeps the celestial bodies in motion and prevents them from flying off into space.
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