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A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.?
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A solid sphere of uniform density and radius R applies a gravitational...
Analysis:
To solve this problem, we need to consider the gravitational force exerted by the solid sphere and the sphere with the cavity on the particle placed at point P. Let's break down the problem step by step.
Step 1: Gravitational force by the solid sphere
The solid sphere exerts a gravitational force of attraction, F1, on the particle placed at point P, which is at a distance of 2R from the center O of the sphere. The force of gravity between the particle and the solid sphere can be calculated using Newton's law of universal gravitation:
F1 = G * (m * M) / r^2
Where:
- F1 is the gravitational force between the particle and the solid sphere
- G is the gravitational constant
- m is the mass of the particle
- M is the mass of the solid sphere
- r is the distance between the particle and the center of the solid sphere (2R in this case)
Step 2: Gravitational force by the sphere with the cavity
Now, a spherical cavity of radius R/2 is made in the solid sphere, touching one of the points on the sphere's edge. The new sphere with the cavity still exerts a gravitational force, F2, on the particle placed at point P. Since the cavity does not contain any mass, the mass of the sphere with the cavity is now M - m_cavity, where m_cavity is the mass of the cavity.
Step 3: Calculating the ratio of F1/F2
To find the ratio of F1/F2, we need to calculate the forces F1 and F2 and then take their ratio.
F2 = G * (m * (M - m_cavity)) / r^2
The ratio of F1/F2 can be calculated as:
F1/F2 = (G * (m * M) / r^2) / (G * (m * (M - m_cavity)) / r^2)
By simplifying the equation, we can cancel out the gravitational constant, mass, and distance terms:
F1/F2 = (M / (M - m_cavity))
Conclusion:
The ratio of F1/F2 is (M / (M - m_cavity)). Therefore, the ratio depends on the mass of the solid sphere and the mass of the cavity. As the mass of the cavity increases, the ratio decreases.
To solve this problem, we need to consider the gravitational force exerted by the solid sphere and the sphere with the cavity on the particle placed at point P. Let's break down the problem step by step.
Step 1: Gravitational force by the solid sphere
The solid sphere exerts a gravitational force of attraction, F1, on the particle placed at point P, which is at a distance of 2R from the center O of the sphere. The force of gravity between the particle and the solid sphere can be calculated using Newton's law of universal gravitation:
F1 = G * (m * M) / r^2
Where:
- F1 is the gravitational force between the particle and the solid sphere
- G is the gravitational constant
- m is the mass of the particle
- M is the mass of the solid sphere
- r is the distance between the particle and the center of the solid sphere (2R in this case)
Step 2: Gravitational force by the sphere with the cavity
Now, a spherical cavity of radius R/2 is made in the solid sphere, touching one of the points on the sphere's edge. The new sphere with the cavity still exerts a gravitational force, F2, on the particle placed at point P. Since the cavity does not contain any mass, the mass of the sphere with the cavity is now M - m_cavity, where m_cavity is the mass of the cavity.
Step 3: Calculating the ratio of F1/F2
To find the ratio of F1/F2, we need to calculate the forces F1 and F2 and then take their ratio.
F2 = G * (m * (M - m_cavity)) / r^2
The ratio of F1/F2 can be calculated as:
F1/F2 = (G * (m * M) / r^2) / (G * (m * (M - m_cavity)) / r^2)
By simplifying the equation, we can cancel out the gravitational constant, mass, and distance terms:
F1/F2 = (M / (M - m_cavity))
Conclusion:
The ratio of F1/F2 is (M / (M - m_cavity)). Therefore, the ratio depends on the mass of the solid sphere and the mass of the cavity. As the mass of the cavity increases, the ratio decreases.
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A solid sphere of uniform density and radius R applies a gravitational...
Does it is 9/7 please check it is correct ask me i explain
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A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.?
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A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.?.
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P,distance 2R from centre O of the sphere.A spherical cavity of radius R/2 is now made in the sphere touching one of the point on sphere edge.the sphere with cavity now applies an grav forceF2 on same particle placed at P.the ratio of F1/F2 will be.?.
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