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A small body of super dense material, whose mass is twice the mass of Earth but whose size is very small compared to the size of Earth, starts from rest at a height H << R above the Earth’s surface and reaches the surface in time t. Then t is equal to:
A) √(2H/g)
B) √(H/g)
C) √(2H/3g)
D) √(4H/3g)

Aniket Singh , 11 Months ago
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anser 1 Answers
Askiitians Tutor Team

Given the problem, we have a small body with a mass twice that of Earth and a very small size. The body is dropped from a height \( H \) (where \( H \) is much smaller than the Earth's radius \( R \)), and we need to find the time \( t \) it takes to reach the Earth's surface.

Since the body's mass is much greater than the Earth's, it will accelerate towards the Earth primarily due to the Earth's gravitational pull, which can be approximated by the standard acceleration due to gravity \( g \) near the Earth's surface.

For an object falling freely under gravity from a height \( H \) near the Earth's surface, the time \( t \) taken to reach the ground is given by:

\[
H = \frac{1}{2} g t^2
\]

Solving for \( t \):

\[
t = \sqrt{\frac{2H}{g}}
\]

So, the correct answer is:

**A. \(\sqrt{\dfrac{2H}{g}}\)**

Last Activity: 11 Months ago
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