1
AP EAPCET 2024 - 21th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The time period of revolution of a satellite $T$ around the carth depends on the radius of the circular orbit $R$. mass of the earth $M$ and universal gravitational constant $G$. The expression for $T$, using dimensional analysis is ( $K$ is constant of proportionality)
A
$K \sqrt{\frac{R^2}{G M}}$
B
$K \sqrt{\frac{R}{G M}}$
C
$K \sqrt{\frac{R^3}{G M}}$
D
$K \sqrt{\frac{R^3}{G M^2}}$
2
AP EAPCET 2024 - 21th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If the time period of revolution of a satellite is $T$, the its kinetic energy is proportional to
A
$T^{-1}$
B
$T^{-2}$
C
$T^{-3}$
D
$T^{-2 / 3}$
3
AP EAPCET 2024 - 20th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
What is the height from the surface of earth, where acceleration due to gravity will be $1 / 4$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$
A
6400 km
B
3200 km
C
1600 km
D
640 km
4
AP EAPCET 2024 - 20th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5 \mathrm{~ms}^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth= 6400 km )
A
$1.11 \mathrm{~ms}^{-2}$
B
$1.5 \mathrm{~ms}^{-2}$
C
$2.22 \mathrm{~ms}^{-2}$
D
$1.25 \mathrm{~ms}^{-2}$
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