MHT CET 2025 21st April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( $h=$ Planck's constant, $\mathrm{I}=$ moment of inertia of diatomic molecule)

$\frac{9 h^2}{8 \pi^2 I}$

$\frac{3 \mathrm{~h}^2}{8 \pi^2 \mathrm{I}}$

$\frac{6 h^2}{8 \pi I}$

$\frac{12 h^2}{7 \pi^2 I}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A body is rotating about its own axis. Its rotational kinetic energy is ' x ' and its angular momentum is ' $y$ '. Hence its moment of inertia about its own axis is

$\frac{x^2}{2 y}$

$\frac{y^2}{2 x}$

$\frac{x}{2 y}$

$\frac{y}{2 x}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A ray of light travelling through rarer medium is incident at a very small angle ' $i$ ' on a glass slab and after refraction its velocity is reduced by $20 \%$. The angle of deviation is

$\frac{\mathrm{i}}{5}$

$\frac{\mathrm{i}}{8}$

$\frac{\mathrm{i}}{4}$

$\frac{\mathrm{i}}{2}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text {th }}$ second of its motion to the total distance travelled in n second is

$\frac{2}{n^2}-\frac{1}{n}$

$\frac{2}{n}-\frac{1}{n^2}$

$\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}^2}$

$\frac{2}{\mathrm{n}^2}+\frac{1}{\mathrm{n}}$

PREVIOUS NEXT