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Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–
- a)7.5 R
- b)4.5 R
- c)2.5 R
- d)1.5 R
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are ...
Ratio of masses is 1:5.
Ratio of acceleration is 5:1.
Ratio of distances covered is 5:1.
Distance to be covered =(12R−R−2R)=9R
Distance moved by smaller sphere
Ratio of acceleration is 5:1.
Ratio of distances covered is 5:1.
Distance to be covered =(12R−R−2R)=9R
Distance moved by smaller sphere
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Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are ...
Let's first calculate the gravitational force between the two bodies. Using the formula for gravitational force:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two bodies, and r is the distance between their centers.
Plugging in the values, we get:
F = 6.67 x 10^-11 * (M * 5M) / (12R)^2
F = 27.78 * 10^-11 * M^2 / R^2
Now, using Newton's second law of motion, we can find the acceleration of the smaller body towards the larger body:
F = ma
a = F / m
a = 27.78 * 10^-11 * M / (2R) (mass of smaller body is M)
Integrating this acceleration with respect to time, we can find the velocity of the smaller body as it approaches the larger body:
v = at
v = 27.78 * 10^-11 * M / (2R) * t
The distance covered by the smaller body just before collision is equal to the distance between the two bodies when their surfaces touch. At this point, the sum of their radii is equal to the initial separation between their centers:
R + 2R = 12R
3R = 12R
r = 4R
So, the distance covered by the smaller body is:
distance = initial separation - final separation
distance = 12R - 4R
distance = 8R
To find the time it takes for the bodies to collide, we can use the formula for the distance traveled by an object under constant acceleration:
d = (1/2)at^2
Plugging in the values, we get:
4R = (1/2) * (27.78 * 10^-11 * M / (2R)) * t^2
t^2 = (8R^2) / (27.78 * 10^-11 * M)
t = sqrt((8R^2) / (27.78 * 10^-11 * M))
Substituting the value of t in the expression for distance, we get:
distance = 27.78 * 10^-11 * M / (2R) * sqrt((8R^2) / (27.78 * 10^-11 * M))
distance = sqrt(8) * R
Therefore, the distance covered by the smaller body just before collision is equal to sqrt(8) times the radius of the smaller body, or approximately 2.83 times the radius of the smaller body.
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two bodies, and r is the distance between their centers.
Plugging in the values, we get:
F = 6.67 x 10^-11 * (M * 5M) / (12R)^2
F = 27.78 * 10^-11 * M^2 / R^2
Now, using Newton's second law of motion, we can find the acceleration of the smaller body towards the larger body:
F = ma
a = F / m
a = 27.78 * 10^-11 * M / (2R) (mass of smaller body is M)
Integrating this acceleration with respect to time, we can find the velocity of the smaller body as it approaches the larger body:
v = at
v = 27.78 * 10^-11 * M / (2R) * t
The distance covered by the smaller body just before collision is equal to the distance between the two bodies when their surfaces touch. At this point, the sum of their radii is equal to the initial separation between their centers:
R + 2R = 12R
3R = 12R
r = 4R
So, the distance covered by the smaller body is:
distance = initial separation - final separation
distance = 12R - 4R
distance = 8R
To find the time it takes for the bodies to collide, we can use the formula for the distance traveled by an object under constant acceleration:
d = (1/2)at^2
Plugging in the values, we get:
4R = (1/2) * (27.78 * 10^-11 * M / (2R)) * t^2
t^2 = (8R^2) / (27.78 * 10^-11 * M)
t = sqrt((8R^2) / (27.78 * 10^-11 * M))
Substituting the value of t in the expression for distance, we get:
distance = 27.78 * 10^-11 * M / (2R) * sqrt((8R^2) / (27.78 * 10^-11 * M))
distance = sqrt(8) * R
Therefore, the distance covered by the smaller body just before collision is equal to sqrt(8) times the radius of the smaller body, or approximately 2.83 times the radius of the smaller body.
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Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer?
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Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer?.
Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two spherical bodies of mass Mand 5Mand radii Rand 2Rrespectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is–a)7.5 Rb)4.5 Rc)2.5 Rd)1.5 RCorrect answer is option 'A'. Can you explain this answer?.
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