MHT CET 2023 14th May Evening Shift | Gravitation Question 33 | Physics

Earth is assumed to be a sphere of radius R. If '$$\mathrm{g}_\phi$$' is value of effective acceleration due to gravity at latitude $$30^{\circ}$$ and '$$g$$' is the value at equator, then the value of $$\left|g-g_\phi\right|$$ is ($$\omega$$ is angular velocity of rotation of earth, $$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$$ )

$$\frac{1}{4} \omega^2 \mathrm{R}$$

$$\frac{3}{4} \omega^2 \mathrm{R}$$

$$\omega^2 \mathrm{R}$$

$$\frac{1}{2} \omega^2 \mathrm{R}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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A body (mass $$\mathrm{m}$$ ) starts its motion from rest from a point distant $$R_0\left(R_0>R\right)$$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $$\mathrm{G}=$$ universal constant of gravitation, $$\mathrm{M}=$$ mass of earth, $$\mathrm{R}$$ = radius of earth)

$$2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$$

$$\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$$

$$\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$$

$$2 \mathrm{GM}\left[\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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Considering earth to be a sphere of radius '$$R$$' having uniform density '$$\rho$$', then value of acceleration due to gravity '$$g$$' in terms of $$R, \rho$$ and $$\mathrm{G}$$ is

$$g=\sqrt{\frac{3 \pi R}{\rho G}}$$

$$\mathrm{g}=\sqrt{\frac{4}{3} \pi \rho \mathrm{GR}}$$

$$\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}$$

$$g=\frac{G M}{\rho R^2}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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The value of acceleration due to gravity at a depth '$$d$$' from the surface of earth and at an altitude '$$h$$' from the surface of earth are in the ratio

$$1: 1$$

$$\frac{R-2 h}{R-d}$$

$$\frac{R-d}{R-2 h}$$

$$\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}-\mathrm{h}}$$

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