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A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is (Given : gravitational potential at the centre of earth = -3/2 GM/R)
- a)√R
- b)√gR
- c)√2.5gR
- d)√7.1gR
Correct answer is option 'B'. Can you explain this answer?
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A diametrical tunnel is dug across the earth. A ball is dropped into ...
Loss in potential energy = Gain in kinetic energy
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A diametrical tunnel is dug across the earth. A ball is dropped into ...
Given information:
- Gravitational potential at the center of the Earth = -3/2 GM/R
To find: Velocity of the ball when it reaches the center of the Earth
- Let's assume the radius of the Earth as R and the mass of the Earth as M.
- The gravitational potential energy at any point inside the Earth can be given by the formula U = -GMm/r, where G is the gravitational constant, m is the mass of the object, and r is the distance from the center of the Earth.
- As the ball is dropped from one side of the tunnel, it falls freely under the influence of gravity. Therefore, it will gain kinetic energy and lose potential energy.
- At the surface of the Earth, the potential energy of the ball is zero. As it falls towards the center, the potential energy decreases.
- At the center of the Earth, the potential energy is minimum and equal to -3/2 GM/R.
- According to the law of conservation of mechanical energy, the total mechanical energy of the ball remains constant throughout its motion.
- Therefore, at the surface of the Earth, the total mechanical energy of the ball is equal to its kinetic energy.
- At the center of the Earth, the total mechanical energy is equal to the kinetic energy plus the potential energy, which is -3/2 GM/R.
- Using the equation for kinetic energy (K.E. = 1/2 mv^2), we can equate the total mechanical energy at the surface to the kinetic energy at the center of the Earth.
- 1/2 mv^2 = -3/2 GM/R
- Simplifying the equation, we get v^2 = -3GM/R
- Taking the square root of both sides, we get v = √(3GM/R)
- Since g = GM/R, where g is the acceleration due to gravity, we can rewrite the equation as v = √(3gR)
- Therefore, the velocity of the ball when it reaches the center of the Earth is √(3gR), which is represented by option B.
- Gravitational potential at the center of the Earth = -3/2 GM/R
To find: Velocity of the ball when it reaches the center of the Earth
- Let's assume the radius of the Earth as R and the mass of the Earth as M.
- The gravitational potential energy at any point inside the Earth can be given by the formula U = -GMm/r, where G is the gravitational constant, m is the mass of the object, and r is the distance from the center of the Earth.
- As the ball is dropped from one side of the tunnel, it falls freely under the influence of gravity. Therefore, it will gain kinetic energy and lose potential energy.
- At the surface of the Earth, the potential energy of the ball is zero. As it falls towards the center, the potential energy decreases.
- At the center of the Earth, the potential energy is minimum and equal to -3/2 GM/R.
- According to the law of conservation of mechanical energy, the total mechanical energy of the ball remains constant throughout its motion.
- Therefore, at the surface of the Earth, the total mechanical energy of the ball is equal to its kinetic energy.
- At the center of the Earth, the total mechanical energy is equal to the kinetic energy plus the potential energy, which is -3/2 GM/R.
- Using the equation for kinetic energy (K.E. = 1/2 mv^2), we can equate the total mechanical energy at the surface to the kinetic energy at the center of the Earth.
- 1/2 mv^2 = -3/2 GM/R
- Simplifying the equation, we get v^2 = -3GM/R
- Taking the square root of both sides, we get v = √(3GM/R)
- Since g = GM/R, where g is the acceleration due to gravity, we can rewrite the equation as v = √(3gR)
- Therefore, the velocity of the ball when it reaches the center of the Earth is √(3gR), which is represented by option B.
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A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is (Given : gravitational potential at the centre of earth = -3/2 GM/R)a)√Rb)√gRc)√2.5gRd)√7.1gRCorrect answer is option 'B'. Can you explain this answer?
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A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is (Given : gravitational potential at the centre of earth = -3/2 GM/R)a)√Rb)√gRc)√2.5gRd)√7.1gRCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is (Given : gravitational potential at the centre of earth = -3/2 GM/R)a)√Rb)√gRc)√2.5gRd)√7.1gRCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is (Given : gravitational potential at the centre of earth = -3/2 GM/R)a)√Rb)√gRc)√2.5gRd)√7.1gRCorrect answer is option 'B'. Can you explain this answer?.
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