MHT CET 2024 15th May Evening Shift | Center of Mass and Collision Question 5 | Physics

1000 small balls, each weighing 1 gram, strike one square cm of area per second with a velocity $50 \mathrm{~m} / \mathrm{s}$ in a normal direction and rebound with the same velocity. The value of pressure on the surface will be

$10^7 \mathrm{~N} / \mathrm{m}^2$

$10^6 \mathrm{~N} / \mathrm{m}^2$

$5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

$2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

A moving body with mass ' $\mathrm{m}_1$ ' strikes a stationary mass ' $\mathrm{m}_2$ '. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of first by (1.5) times the velocity after the collision?

$1: 25$

$1: 5$

$5: 1$

$25: 1$

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

A metal rod of weight ' $W$ ' is supported by two parallel knife-edges A and B . The rod is in equilibrium in horizontal position. The distance ' between two knife-edges is ' $r$ '. The centre of mass of the rod is at a distance ' $x$ ' from $A$. The normal reaction on A is

$\frac{\mathrm{W} \cdot \mathrm{r}}{\mathrm{x}}$

$\frac{\mathrm{W} \cdot \mathrm{x}}{\mathrm{r}}$

$\mathrm{\frac{W \cdot(r-x)}{x}}$

$\mathrm{\frac{W \cdot(r-x)}{r}}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

In the system of two particles of masses ' $\mathrm{m}_1$ ' and ' $\mathrm{m}_2$ ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, towards the centre of mass.

$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, away from the centre of mass.

$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, towards the centre of mass.

$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, away from the centre of mass.

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