MHT CET 2023 12th May Morning Shift | Gravitation Question 44 | Physics

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

$$\frac{1}{2}$$

$$\frac{1}{4}$$

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

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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)

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

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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

$$r^{9 / 2}$$

$$r^{7 / 2}$$

$$r^{5 / 2}$$

$$r^{3 / 2}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

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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 '$$g$$')

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

$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}$$

$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}$$

$$\frac{\mathrm{gR}^2}{\omega^2}$$

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