Q105 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
The acceleration due to gravity at a height 'h' above the surface of earth is '$g_h$'. At the depth 90 km below the earth's surface the acceleration due to gravity is also '$g_h$'. The value of 'h' is
- A$180$ km
- B$120$ km
- C$90$ km
- Correct answerD$45$ km
Solution
The acceleration due to gravity at a depth $d$ is given by:
$g_d = g\left(1 - \dfrac{d}{R}\right)$
The acceleration due to gravity at a height $h$ (where $h \ll R$) is given by:
$g_h = g\left(1 - \dfrac{2h}{R}\right)$
Given that the acceleration due to gravity at height $h$ is equal to that at depth $d = 90$ km:
$g_h = g_d$
$g\left(1 - \dfrac{2h}{R}\right) = g\left(1 - \dfrac{d}{R}\right)$
$\dfrac{2h}{R} = \dfrac{d}{R}$
$2h = d$
$h = \dfrac{d}{2}$
Substituting $d = 90$ km:
$h = \dfrac{90}{2} = 45$ km
Answer: $45$ km
$g_d = g\left(1 - \dfrac{d}{R}\right)$
The acceleration due to gravity at a height $h$ (where $h \ll R$) is given by:
$g_h = g\left(1 - \dfrac{2h}{R}\right)$
Given that the acceleration due to gravity at height $h$ is equal to that at depth $d = 90$ km:
$g_h = g_d$
$g\left(1 - \dfrac{2h}{R}\right) = g\left(1 - \dfrac{d}{R}\right)$
$\dfrac{2h}{R} = \dfrac{d}{R}$
$2h = d$
$h = \dfrac{d}{2}$
Substituting $d = 90$ km:
$h = \dfrac{90}{2} = 45$ km
Answer: $45$ km
Hence, the correct answer is option: 'D'.