MHT CET · Physics · Gravitation
A boy weighs 72 N on the surface of earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be
- A 32 N
- B 48 N
- C 96 N
- D 162 N
Answer & Solution
Correct Answer
(A) 32 N
Step-by-step Solution
Detailed explanation
At height \(\left(\mathrm{h}=\frac{\mathrm{R}}{\mathrm{n}}\right)\), the value of acceleration due to gravity is given by, \(\mathrm{g}_{\mathrm{h}}=\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2 \mathrm{~g}\).
For \(\mathrm{n}=2\),
\(\begin{aligned}
& \mathrm{g}_{\mathrm{h}}=\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2 \mathrm{~g}=\left(\frac{2}{2+1}\right)^2 \mathrm{~g}=\frac{4}{9} \mathrm{~g} \\
& \mathrm{~W}_{\mathrm{h}}=\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \times \mathrm{mg}=\frac{4}{9} \times 72=32 \mathrm{~N}
\end{aligned}\)
For \(\mathrm{n}=2\),
\(\begin{aligned}
& \mathrm{g}_{\mathrm{h}}=\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)^2 \mathrm{~g}=\left(\frac{2}{2+1}\right)^2 \mathrm{~g}=\frac{4}{9} \mathrm{~g} \\
& \mathrm{~W}_{\mathrm{h}}=\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \times \mathrm{mg}=\frac{4}{9} \times 72=32 \mathrm{~N}
\end{aligned}\)
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