MHT CET 2024 3rd May Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

A gardening pipe having an internal radius ' $R$ ' is connected to a water sprinkler having ' $n$ ' holes each of radius ' $r$ '. The water in the pipe has a speed ' $v$ '. The speed of water leaving the sprinkler is

$\left(\frac{\mathrm{R}^2}{\mathrm{r}^2}\right) \mathrm{nV}$

$\frac{\mathrm{R}^2 \mathrm{v}}{\mathrm{nr}^2}$

$\left(\frac{n r^2}{R^2}\right) V$

$\left(\frac{\mathrm{nR}^2}{\mathrm{r}^2}\right) \mathrm{V}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

A magnetic needle of magnetic moment $6 \times 10^{-2} \mathrm{Am}^2$ and moment of inertia $9.6 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^2$ performs simple harmonic motion in a magnetic field of 0.01 T . Time taken to complete 10 oscillations is [Take $\pi=3 \cdot 14$]

0.2512 s

$2.512 \mathrm{~s}$

25.12 s

251.2 s

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

In case of rotational dynamics, which one of the following statements is correct?

[$\vec{\omega}=$ angular velocity, $\overrightarrow{\mathrm{v}}=$ linear velocity

$\overrightarrow{\mathbf{r}}=$ radius vector, $\vec{\alpha}=$ angular acceleration

$\overrightarrow{\mathrm{a}}=$ linear acceleration, $\overrightarrow{\mathrm{L}}=$ angular momentum

$\overrightarrow{\mathrm{p}}=$ linear momentum, $\bar{\tau}=$ torque,

$\overrightarrow{\mathrm{f}}=$ centripetal force]

$\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{r}} \times \vec{\omega}, \overrightarrow{\boldsymbol{\alpha}}=\overrightarrow{\mathbf{r}} \times \vec{a}, \vec{L}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}, \vec{\tau}=\overrightarrow{\mathrm{f}} \times \overrightarrow{\mathrm{r}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

A convex lens of focal length ' $f$ ' $m$ forms a real, inverted image twice in size of the object. The object distance from the lens in metre is

0.5 f

0.66 f

1.5 f

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