MHT CET 2025 23rd April Evening Shift | Motion Question 1 | Physics

A body when projected at an angle ' $\theta$ ' with the horizontal reaches a maximum height ' $H$ '. The time of flight of the body will be ( $\mathrm{g}=$ acceleration due to gravity)

$\frac{1}{2} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

$\sqrt{\frac{g}{2 H}}$

$2 \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

$\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

The acceleration (a) - time (T) graph for the body starting from rest is given below. The maximum speed of the body is

MHT CET 2025 23rd April Morning Shift Physics - Motion Question 2 English

$40 \mathrm{~m} / \mathrm{s}$

$80 \mathrm{~m} / \mathrm{s}$

$160 \mathrm{~m} / \mathrm{s}$

$200 \mathrm{~m} / \mathrm{s}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

At any time ' $t$ ', the co-ordinates of moving particle are $x=a t^2$ and $y=b t^2$. The speed of the particle is

$2 t \sqrt{a^2+b^2}$

$2 t \sqrt{a^2-b^2}$

$2 \mathrm{t}(\mathrm{a}+\mathrm{b})$

$\frac{2 t}{\sqrt{a^2+b^2}}$

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

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

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

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

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

MHT CET Subjects