MHT CET 2019 2nd May Evening Shift | Circular Motion Question 38 | Physics

In U.C.M., when time interval $\delta t \rightarrow 0$, the angle between change in velocity ( $\delta \mathbf{v}$ ) and linear velocity $(\boldsymbol{v})$ will be

$0^\circ$

$90^\circ$

$180^\circ$

$45^\circ$

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

A particle is performing U.C.M. along the circumference of a circle of diameter 50 cm with frequency 2 Hz . The acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is

$2 \pi^2$

$8 \pi^2$

$\pi^2$

$4 \pi^2$

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

A stone of mass 1 kg is tied to a string 2 m long and it's rotated at constant speed of $40 \mathrm{~ms}^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g=10 \mathrm{~ms}^{-2}$]

$\frac{81}{79}$

$\frac{79}{81}$

$\frac{19}{12}$

$\frac{12}{19}$

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The real force ' $F$ ' acting on a particle of mass $m$ ' performing circular motion acts along the radius of circle ' $r$ ' and is directed towards the centre of circle. The square root of magnitude of such force is ( $T=$ periodic time)

$\frac{2 \pi}{T} \sqrt{m r}$

$\frac{T m r}{4 \pi}$

$\frac{2 \pi T}{\sqrt{m r}}$

$\frac{T^2 m r}{4 \pi}$

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