1
MHT CET 2023 12th May Evening Shift
+1
-0

A simple pendulum is oscillating with frequency '$$F$$' on the surface of the earth. It is taken to a depth $$\frac{\mathrm{R}}{3}$$ below the surface of earth. ( $$\mathrm{R}=$$ radius of earth). The frequency of oscillation at depth $$\mathrm{R} / 3$$ is

A
$$\frac{2 \mathrm{~F}}{3}$$
B
$$\frac{\mathrm{F}}{\sqrt{1.5}}$$
C
F
D
$$\frac{\mathrm{F}}{3}$$
2
MHT CET 2023 12th May Evening Shift
+1
-0

The depth at which acceleration due to gravity becomes $$\frac{\mathrm{g}}{2 \mathrm{n}}$$ is $$(\mathrm{R}=$$ radius of earth, $$\mathrm{g}=$$ acceleration due to gravity on earth's surface, $$\mathrm{n}$$ is integer)

A
$$\frac{\mathrm{R}(1-2 \mathrm{n})}{\mathrm{n}}$$
B
$$\frac{\mathrm{R}(1-\mathrm{n})}{2 \mathrm{n}}$$
C
$$\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}$$
D
$$\frac{\mathrm{R}(2 \mathrm{n}-1)}{2 \mathrm{n}}$$
3
MHT CET 2023 12th May Morning Shift
+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
$$\frac{1}{4}$$
4
MHT CET 2023 12th May Morning Shift
+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
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