MHT CET 2020 19th October Evening Shift | Circular Motion Question 45 | Physics

A child starts running from rest along a circular track of radius $r$ with constant tangential acceleration a. After time $t$ he feels that slipping of shoes on the ground has started. The coefficient of friction between shoes and the ground is

[g = acceleration due to gravity]

$\frac{\left[a^4 t^4+a^2 r^2\right]^{\frac{1}{2}}}{g r}$

$\frac{\left[a^4 t^4+a^2 r^2\right]}{r g}$

$\frac{\left[a^2 t^2+a^4 r^4\right]}{r g}$

$\frac{\left[a^4 t^4-a^2 r^2\right]^{\frac{1}{2}}}{r g}$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

A body is moving along a circular track of radius 100 m with velocity $20 \mathrm{~m} / \mathrm{s}$. Its tangential acceleration is $3 \mathrm{~m} / \mathrm{s}^2$, then its resultant acceleration will be

$5 \mathrm{~m} / \mathrm{s}^2$

$4 \mathrm{~m} / \mathrm{s}^2$

$2 \mathrm{~m} / \mathrm{s}^2$

$3 \mathrm{~m} / \mathrm{s}^2$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

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A particle starting from rest moves along the circumference of a circle of radius $$r$$ with angular acceleration $$\alpha$$. The magnitude of the average velocity, in the time it completes the small angular displacement $$\theta$$ is

$$r\left(\frac{2}{\alpha \theta}\right)^2$$

$$r\left(\frac{\alpha \theta}{2}\right)^2$$

$$\rho\left(\frac{\alpha \theta}{2}\right)$$

$$r\left(\frac{\alpha \theta}{2}\right)^{\frac{1}{2}}$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

A particle of mass $$m$$ is performing UCM along a circle of radius $$r$$. The relation between centripetal acceleration $$a$$ and kinetic energy $$E$$ is given by

$$a=\frac{2 E}{m r}$$

$$a=2 E m$$

$$a=\frac{E}{m r}$$

$$a=\left(\frac{2 E}{m r}\right)^2$$

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