MHT CET 2023 14th May Morning Shift | Circular Motion Question 18 | Physics

A string of length '$$L$$' fixed at one end carries a body of mass '$$\mathrm{m}$$' at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The string makes angle '$$\theta$$' with the vertical. The angular frequency of the body is '$$\omega$$'. The tension in the string is

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

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

$$\frac{\omega^2}{\mathrm{~mL}}$$

$$\frac{\mathrm{m} \omega^2}{\mathrm{~L}}$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

-0

A stone is projected at angle $$\theta$$ with velocity $$u$$. If it executes nearly a circular motion at its maximum point for short time, then the radius of the circular path will be ( $$g=$$ acceleration due to gravity)

$$\frac{u^2}{g}$$

$$\frac{u^2 \cos ^2 \theta}{g}$$

$$\frac{u^2 \sin ^2 \theta}{g}$$

$$\frac{u^2 \cos ^2 \theta}{2 g}$$

MHT CET 2023 12th May Evening Shift

MCQ (Single Correct Answer)

-0

A particle is moving in a circle with uniform speed '$$v$$'. In moving from a point to another diametrically opposite point

the momentum changes by $$\mathrm{mv}$$

the momentum changes by $$2 \mathrm{~mv}$$

the kinetic energy changes by $$\frac{1}{2} \mathrm{mv}^2$$

the kinetic energy changes by $$\mathrm{m v^2}$$

MHT CET 2023 12th May Evening Shift

MCQ (Single Correct Answer)

-0

A body of mass '$$\mathrm{m}$$' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ( $$\mathrm{g}$$ = gravitational acceleration)

zero

$$\mathrm{mg}$$

$$3 \mathrm{~mg}$$

$$6 \mathrm{~mg}$$

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