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The height at which the weight of a body becomes 1/16th its weight on the surface of earth (radius R), is
  • a)
    5 R
  • b)
    15 R
  • c)
    3 R
  • d)
    4 R
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The height at which the weight of a body becomes 1/16th its weight on...
Acceleration due to gravity at a height h from the surface of earth is
where g is the acceleration due to gravity at the surface of earth and R is the radius of earth.
Multiplying by m (mass of the body) on both sides in (i), we get
∴ Weight of body at height h ,W′ = mg′
Weight of body at surface of earth, W = mg
According to question,W′ = 1/16W
or h/R = 3 or h = 3R
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Community Answer
The height at which the weight of a body becomes 1/16th its weight on...
Concept:
When a body is at a height above the surface of the earth, the weight of the body decreases due to the decrease in gravitational force acting on it.

Solution:
To find the height at which the weight of a body becomes 1/16th its weight on the surface of earth, we can use the concept of gravitational force and the inverse square law.
- Let the height above the surface of the earth be 'h'.
- The weight of the body at height 'h' can be calculated using the formula: W = (G * m * M) / (R + h)^2, where G is the gravitational constant, m is the mass of the body, M is the mass of the earth, and R is the radius of the earth.
- Given that the weight at height 'h' is 1/16th its weight on the surface of the earth, we can write: W = (1/16) * (m * g), where g is the acceleration due to gravity on the surface of the earth.
- Equating the two expressions for weight, we get: (G * m * M) / (R + h)^2 = (1/16) * (m * g).
- Simplifying the above equation, we get: h = 3R.
Therefore, the height at which the weight of the body becomes 1/16th its weight on the surface of earth is 3 times the radius of the earth, which is option 'C'.
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