1
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

According to Bohr's first postulate the kinetic energy $\frac{1}{2} m v^2$ of the electron in C.G.S. system is ( $\mathrm{m}=$ mass of electron, v is its velocity, r is the radius of the stationary orbit around the nucleus with charge Ze )

A

$\frac{\mathrm{Ze}}{\mathrm{r}^2}$

B

$\frac{\mathrm{Ze}^2}{\mathrm{r}}$

C

$\frac{\mathrm{Ze}^2}{2 \mathrm{r}}$

D

$\frac{\mathrm{Ze}^2}{2 \mathrm{r}^2}$

2
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

An air capacitor of plate area ' $A$ ' and separation between the plates is ' $d$ ' has a capacity ' $C$ '. Two dielectric slabs are inserted between its plates in two different manners as shown in figure. The capacitance of a capacitor is

MHT CET (PCB) 2025 9th April Evening Shift Physics - Capacitor Question 1 English
A

$\frac{\varepsilon_0 A}{d-\frac{t_1}{k_1}+\frac{t_2}{k_2}}$

B

$\frac{\varepsilon_0 A}{d+t_1+t_2+\frac{t_1}{k_1}+\frac{t_2}{k_2}}$

C

$\frac{\varepsilon_0 A}{d-t_1-t_2+\frac{t_1}{k_1}+\frac{t_2}{k_2}}$

D

$\frac{\varepsilon_0 A}{d+t_1+t_2-\frac{t_1}{k_1}-\frac{t_2}{k_2}}$

3
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

The output Y of a given logic circuit is (For inputs $\mathrm{A}, \mathrm{B}$ and C )

MHT CET (PCB) 2025 9th April Evening Shift Physics - Semiconductor Devices and Logic Gates Question 1 English
A

$\quad \mathrm{Y}=\overline{\mathrm{A}} \cdot(\overline{\mathrm{B}+\mathrm{C}})$

B

$\mathrm{Y}=(\overline{\mathrm{A} \cdot \mathrm{B}})+(\overline{\mathrm{B} \cdot \mathrm{C}})$

C

$ \mathrm{Y}=(\overline{\mathrm{A}+\mathrm{B}}) \cdot(\overline{\mathrm{B}+\mathrm{C}})$

D

$Y=(A+B) \cdot C$