KCET · Physics · Dual Nature of Matter
The value of acceleration due to gravity at adepth of \( 1600 \mathrm{~km} \) is equal to:
[Radius of earth \( =6400 \mathrm{~km} \) ]
- A \( 9.8 \mathrm{~ms}^{-2} \)
- B \( 4.9 \mathrm{~ms}^{-2} \)
- C \( 19.6 \mathrm{~ms}^{-2} \)
- D \( 7.35 \mathrm{~ms}^{-2} \)
Answer & Solution
Correct Answer
(D) \( 7.35 \mathrm{~ms}^{-2} \)
Step-by-step Solution
Detailed explanation
Acceleration due to gravity at a depth \( d \) and radius \( R \) isgiven as
\[
\begin{array}{l}
g=g\left[1-\frac{d}{R}\right]=g\left(1-\frac{1600}{6400}\right)=g\left(1-\frac{1}{4}\right)=g \times \frac{3}{4} \\
\Rightarrow g=9.8 \times \frac{3}{4}=7.35 m_{S}^{-2}
\end{array}
\]
Therefore, the acceleration due to gravity at a depth of \( 1600 \mathrm{~km} \) is equal to \( 7.35 \mathrm{~ms}^{-2} \)
\[
\begin{array}{l}
g=g\left[1-\frac{d}{R}\right]=g\left(1-\frac{1600}{6400}\right)=g\left(1-\frac{1}{4}\right)=g \times \frac{3}{4} \\
\Rightarrow g=9.8 \times \frac{3}{4}=7.35 m_{S}^{-2}
\end{array}
\]
Therefore, the acceleration due to gravity at a depth of \( 1600 \mathrm{~km} \) is equal to \( 7.35 \mathrm{~ms}^{-2} \)
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