Electromagnetism
Current Electricity
MCQ (Single Correct Answer)
Moving Charges and Magnetism
MCQ (Single Correct Answer)
Magnetism and Matter
MCQ (Single Correct Answer)
Electromagnetic Waves
MCQ (Single Correct Answer)
Electromagnetic Induction
MCQ (Single Correct Answer)
Alternating Current
MCQ (Single Correct Answer)
Modern Physics
Dual Nature of Radiation
MCQ (Single Correct Answer)
Semiconductor Devices and Logic Gates
MCQ (Single Correct Answer)
Communication Systems
MCQ (Single Correct Answer)
1
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-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?

A
$1: 25$
B
$1: 5$
C
$5: 1$
D
$25: 1$
2
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-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

A
$\frac{\mathrm{W} \cdot \mathrm{r}}{\mathrm{x}}$
B
$\frac{\mathrm{W} \cdot \mathrm{x}}{\mathrm{r}}$
C
$\mathrm{\frac{W \cdot(r-x)}{x}}$
D
$\mathrm{\frac{W \cdot(r-x)}{r}}$
3
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-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

A
$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, towards the centre of mass.
B
$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, away from the centre of mass.
C
$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, towards the centre of mass.
D
$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, away from the centre of mass.
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In projectile motion two particles of masses $\mathrm{m}_1$ and $m_2$ have velocities $\vec{V}_1$, and $\vec{V}_2$ respectively at time $t=0$. Their velocities become $\overline{V_1^{\prime}}$ and $\overrightarrow{V_2^{\prime}}$ at time 2 t while still moving in air. The value of $\left[\left(m_1 \overrightarrow{V_1^{\prime}}+m_2 \overrightarrow{V_2^{\prime}}\right)-\left(m_1 \vec{V}_1+m_2 \vec{V}_2\right)\right]$ is ( $\mathrm{g}=$ acceleration due to gravity)

A
zero
B
$\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{gt}$
C
$\left(m_1+m_2\right) g t$
D
$2\left(m_1+m_2\right) g t$
MHT CET Subjects