1
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude $30^{\circ}$ is

$$\left[\begin{array}{l} \sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \\ \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2 \end{array}\right]$$

A
$\frac{\mathrm{V}}{\sqrt{2}}$
B
$\frac{\mathrm{V}}{2}$
C
$\frac{\sqrt{3}}{2} \mathrm{v}$
D
$\mathrm{V}$
2
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two objects of masses ' $m_1$ ' and ' $m_2$ ' are moving in the circles of radii ' $r_1$ ' and ' $r_2$ ' respectively. Their respective angular speeds ' $\omega_1$ ' and ' $\omega_2$ ' are such that they both complete one revolution in the same time ' $t$ '. The ratio of linear speed of ' $m_2$ ' to that of ' $m_1$ ' is

A
$\omega_1: \omega_2$
B
$\mathrm{T}_2: \mathrm{T}_1$
C
$\mathrm{m}_1: \mathrm{m}_2$
D
$\mathrm{r_2: r_1}$
3
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A body performing uniform circular motion of radius ' $R$ ' has frequency ' $n$ '. Its centripetal acceleration per unit radius is proportional to $(n)^x$. The value of $x$ is

A
1
B
2
C
$-$1
D
$-$2
4
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

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 time it completes the small angular displacement ' $\theta$ ' is

A
$\frac{r^2}{2 \alpha \theta}$
B
$\frac{\mathrm{r}}{2 \alpha \theta}$
C
$\frac{\mathrm{r} \alpha \theta}{2}$
D
$\frac{\mathrm{r}}{\sqrt{2}} \sqrt{\alpha \theta}$
MHT CET Subjects
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