MHT CET 2025 25th April Evening Shift | Motion Question 2 | Physics

Two girls are standing at the ends ' $A$ ' and ' $B$ ' of a ground where $\mathrm{AB}=\mathrm{b}$. The girl at ' B ' starts running in a direction perpendicular to ' $A B$ ' with velocity ' $\mathrm{V}_1$ '. The girl at ' A ' starts running simultaneously with velocity ' $\mathrm{V}_2$ ' and in shortest distance meets the other girl in time ' $t$ '. The value of ' $t$ ' is

$\frac{b}{\sqrt{V_1{ }^2+V_2{ }^2}}$

$\frac{b}{V_1+V_2}$

$\frac{b}{V_2-V_1}$

$\frac{b}{\sqrt{V_2{ }^2-V_1{ }^2}}$

MHT CET 2025 25th April Morning Shift

MCQ (Single Correct Answer)

-0

Two boys are standing at points A and B on ground, where distance $\mathrm{AB}=\mathrm{x}$. The boy at B stars running perpendicular to $A B$ with velocity $\mathrm{v}_1$. The boy at A starts running simultaneously with velocity v and meets the other boy in time $t$. The value of $t$ is

$\left[\frac{x}{v-v_1}\right]^{1 / 2}$

$\left[\frac{x}{v_1-v}\right]^{1 / 2}$

$\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}$

$\left[\frac{x^2}{v_1^2-v^2}\right]^{1 / 2}$

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

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 5 English

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

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

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

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

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