MHT CET · Physics · Gravitation
The value of gravitational acceleration \(g\) at a height \(h\) above the earth's surface is \(\frac{g}{4}\) then (\(R =\) radius of earth)
- A
- B
- C
- D
Answer & Solution
Correct Answer
(A)
Step-by-step Solution
Detailed explanation
\(g^{\prime}=g\left(\frac{R}{R+h}\right)^2\)
When \(g^{\prime}=\frac{g}{4}\) then.
\(\frac{g}{4}=g \times\left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{1}{2}=\frac{R}{R+h}\)
\(\therefore 2 R=R+h \Rightarrow R=h\)
When \(g^{\prime}=\frac{g}{4}\) then.
\(\frac{g}{4}=g \times\left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{1}{2}=\frac{R}{R+h}\)
\(\therefore 2 R=R+h \Rightarrow R=h\)
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