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

In the circuit shown in the figure, $P \neq R$. The reading of the galvanometer remains the same with switch ' $S$ ' open or closed. Then

MHT CET 2025 21st April Evening Shift Physics - Current Electricity Question 1 English

$\mathrm{I}_{\mathrm{Q}}=\mathrm{I}_{\mathrm{G}}$

$\quad \mathrm{I}_{\mathrm{Q}}=\mathrm{I}_{\mathrm{R}}$

$\mathrm{I}_{\mathrm{R}}=\mathrm{I}_{\mathrm{G}}$

$\mathrm{I}_{\mathrm{P}}=\mathrm{I}_{\mathrm{G}}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A unit vector in the direction of resultant vector of $\vec{A}=-2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{B}=\hat{i}+2 \hat{j}-4 \hat{k}$ is

$\frac{-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{35}}$

$\frac{\hat{i}+2 \hat{j}-4 \hat{k}}{\sqrt{35}}$

$\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{35}}$

$\frac{-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}}{\sqrt{35}}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

$\frac{\mathrm{I} \omega^2}{16}$

$\frac{\mathrm{I} \omega^2}{8}$

$\frac{\mathrm{I} \omega^2}{4}$

$\frac{\mathrm{I} \omega^2}{2}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

The two coherent sources produce interference with intensity ratio ' $b$ '. In the interference pattern, the ratio $\frac{I_{\text {max }}+I_{\text {min }}}{I_{\text {max }}-I_{\text {min }}}$ will be

$\frac{1+b}{\sqrt{b}}$

$\frac{1+b}{2 \sqrt{b}}$

$\frac{2 \sqrt{b}}{1+b}$

$\frac{2 \sqrt{b}}{(1+b)^2}$

PREVIOUS NEXT