MHT CET 2023 14th May Morning Shift | MHT CET Year Wise Previous Years Questions

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 equation of the wave is $$\mathrm{Y}=10 \sin \left(\frac{2 \pi \mathrm{t}}{30}+\alpha\right)$$ If the displacement is $$5 \mathrm{~cm}$$ at $$\mathrm{t}=0$$ then the total phase at $$\mathrm{t}=7.5 \mathrm{~s}$$ will be $$\left(\sin 30^{\circ}=0.5\right)$$

$$\frac{\pi}{3} \mathrm{~rad}$$

$$\frac{\pi}{2} \mathrm{~rad}$$

$$\frac{2 \pi}{5} \mathrm{~rad}$$

$$\frac{2 \pi}{3} \mathrm{~rad}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The bob of simple pendulum of length '$$L$$' is released from a position of small angular displacement $$\theta$$. Its linear displacement at time '$$\mathrm{t}$$' is ( $$\mathrm{g}=$$ acceleration due to gravity)

$$L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \cos \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

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