1
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Time period of simple pendulum on earth's surface is '$$\mathrm{T}$$'. Its time period becomes '$$\mathrm{xT}$$' when taken to a height $$\mathrm{R}$$ (equal to earth's radius) above the earth's surface. Then the value of '$$x$$' will be

A
4
B
2
C
$$\frac{1}{2}$$
D
$$\frac{1}{4}$$
2
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The height at which the weight of the body becomes $$\left(\frac{1}{9}\right)^{\text {th }}$$ its weight on the surface of earth is $$(\mathrm{R}=$$ radius of earth)

A
8R
B
4R
C
3R
D
2R
3
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Consider a light planet revolving around a massive star in a circular orbit of radius '$$r$$' with time period '$$T$$'. If the gravitational force of attraction between the planet and the star is proportional to $$\mathrm{r}^{\frac{7}{2}}$$, then $$\mathrm{T}^2$$ is proportional to

A
$$r^{9 / 2}$$
B
$$r^{7 / 2}$$
C
$$r^{5 / 2}$$
D
$$r^{3 / 2}$$
4
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The radius of the orbit of a geostationary satellite is (mean radius of earth is '$$R$$', angular velocity about own axis is '$$\omega$$' and acceleration due to gravity on earth's surface is '$$g$$')

A
$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$$
B
$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}$$
C
$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}$$
D
$$\frac{\mathrm{gR}^2}{\omega^2}$$
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