MHT CET · Physics · Oscillations
The acceleration due to gravity on the planet \(A\) is 9 times the acceleration due to gravity on planet \(B\). A man jumps to a height of \(2 \mathrm{~m}\) on the surface of \(A\). What is the height of jump by the same person on the planet \(B\) ?
- A \(6 \mathrm{~m}\)
- B \(\frac{2}{3} \mathrm{~m}\)
- C \(2 / 9 \mathrm{~m}\)
- D \(18 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(18 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
It is given that, acceleration due to gravity on planet \(A\) is 9 times the acceleration due to gravity on planet \(B\) ie,
\(
g_{A}=9 g_{B}
\)
From third equation of motion
\(
v^{2}=2 g h
\)
At planet \(A, \quad h_{A}=\frac{v^{2}}{2 g_{A}}\)
At planet \(B, \quad h_{B}=\frac{v^{2}}{2 g_{B}}\)
Dividing Eq. (ii) by Eq. (iii), we have
\(
\frac{h_{A}}{h_{B}}=\frac{g_{B}}{g_{A}}
\)
From Eq. (i), \(g_{A}=9 g_{B}\)
\(
\frac{h_{A}}{h_{B}}=\frac{g_{B}}{9 g_{B}}=\frac{1}{9}
\)
or \(h_{B}=9 h_{A}=9 \times 2=18 \mathrm{~m}\left(\because h_{A}=2 \mathrm{~m}\right)\)
\(
g_{A}=9 g_{B}
\)
From third equation of motion
\(
v^{2}=2 g h
\)
At planet \(A, \quad h_{A}=\frac{v^{2}}{2 g_{A}}\)
At planet \(B, \quad h_{B}=\frac{v^{2}}{2 g_{B}}\)
Dividing Eq. (ii) by Eq. (iii), we have
\(
\frac{h_{A}}{h_{B}}=\frac{g_{B}}{g_{A}}
\)
From Eq. (i), \(g_{A}=9 g_{B}\)
\(
\frac{h_{A}}{h_{B}}=\frac{g_{B}}{9 g_{B}}=\frac{1}{9}
\)
or \(h_{B}=9 h_{A}=9 \times 2=18 \mathrm{~m}\left(\because h_{A}=2 \mathrm{~m}\right)\)
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