MHT CET · Physics · Gravitation
A body weights 500 Non the surface of earth. At what distance below the surface of earth it weight \(250 \mathrm{~N}\) ?
(Radius of earth, \(R=6400 \mathrm{~km}\) )
- A \(800 \mathrm{~km}\)
- B 32000 km
- C \(1600 \mathrm{~km}\)
- D \(6400 \mathrm{~km}\)
Answer & Solution
Correct Answer
(B) 32000 km
Step-by-step Solution
Detailed explanation
Value of acceleration due to gravity decreases on going below the surface of the earth. Weight \((w)\) of a body is defined as product of mass \((m)\) and acceleration due to gravity \(w=m g\).
The value of g below the surface of earth at a distance \(h\) is given by,
\(\begin{aligned} & g^{\prime}=g\left(1-\frac{h}{R}\right) \\ & \Rightarrow W^{\prime}=W\left(1-\frac{h}{R}\right)\end{aligned}\)
Given, \(W^{\prime}=250 \mathrm{~N}\) and \(W=500\)
\(\begin{aligned} & 250=500\left(1-\frac{h}{R}\right) \\ & \Rightarrow h=\frac{R}{2}=32000 \mathrm{~km}\end{aligned}\)
The value of g below the surface of earth at a distance \(h\) is given by,
\(\begin{aligned} & g^{\prime}=g\left(1-\frac{h}{R}\right) \\ & \Rightarrow W^{\prime}=W\left(1-\frac{h}{R}\right)\end{aligned}\)
Given, \(W^{\prime}=250 \mathrm{~N}\) and \(W=500\)
\(\begin{aligned} & 250=500\left(1-\frac{h}{R}\right) \\ & \Rightarrow h=\frac{R}{2}=32000 \mathrm{~km}\end{aligned}\)
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