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

If the electric flux entering and leaving an enclosed surface area are ' $\phi_1$ ' and ' $\phi_2$ ' respectively, the electric charge inside the surface will be ( $\varepsilon_0=$ permittivity of free space)

A

$\quad \varepsilon_0\left(\phi_1-\phi_2\right)$

B

$\quad \varepsilon_0\left(\phi_2-\phi_1\right)$

C

$\frac{\left(\phi_1+\phi_2\right)}{2}$

D

$\frac{\left(\phi_1-\phi_2\right)}{2}$

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

A simple pendulum has mass 2 gram and charge $2 \mu \mathrm{C}$. In a uniform horizontal electric field of intensity $1000 \mathrm{~V} / \mathrm{m}$, pendulum is at rest. At equilibrium, the angle made by the pendulum with the vertical is ( $\mathrm{g}=$ acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )

A

$\tan ^{-1}(0.2)$

B

$\tan ^{-1}(0.1)$

C

$\tan ^{-1}(0.4)$

D

$\tan ^{-1}(0.5)$

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

The charges are arranged at the four corners of square $A B C D$ of side ' $d$ ' as shown in figure. The work required to put this arrangement together is given by

MHT CET (PCB) 2025 9th April Morning Shift Physics - Electrostatics Question 2 English
A

$\frac{-9 q^2}{4 \pi \varepsilon_0 d}$

B

$\frac{-7 q^2}{4 \pi \varepsilon_0 d}$

C

$\frac{-q^2}{4 \pi \varepsilon_0 d}[9-2 \sqrt{2}]$

D

$\frac{-q^2}{4 \pi \varepsilon_0 d}[9+2 \sqrt{2}]$

4
MHT CET (PCB) 2024 22th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

Three charges $Q,-2 q$ and $-2 q$ are placed at the vertices of an isosceles right-angled triangle as shown in figure. The net electrostatic potential energy is zero if $Q$ is equal to

MHT CET (PCB) 2024 22th April Evening Shift Physics - Electrostatics Question 6 English

A
$\frac{\mathrm{q}}{2}$
B
$\sqrt{2} q$
C
$\frac{q}{\sqrt{2}}$
D
$\frac{q}{2 \sqrt{2}}$

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