Q128 / 180 · KCET 2017 · Physics
The value of acceleration due to gravity at adepth of \( 1600 \mathrm{~km} \) is equal to:
[Radius of earth \( =6400 \mathrm{~km} \) ]
[Radius of earth \( =6400 \mathrm{~km} \) ]
- A\( 9.8 \mathrm{~ms}^{-2} \)
- B\( 4.9 \mathrm{~ms}^{-2} \)
- C\( 19.6 \mathrm{~ms}^{-2} \)
- Correct answerD\( 7.35 \mathrm{~ms}^{-2} \)
Solution
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 \( 7.35 \mathrm{~ms}^{-2} \)
Hence, the correct answer is option: 'D'.