Electricity
Current Electricity
MCQ (Single Correct Answer)
Moving Charges and Magnetism
MCQ (Single Correct Answer)
Magnetism and Matter
MCQ (Single Correct Answer)
Electromagnetic Induction
MCQ (Single Correct Answer)
Electromagnetic Waves
MCQ (Single Correct Answer)
Modern Physics
Semiconductor Electronics
MCQ (Single Correct Answer)
1
NEET 2024
MCQ (Single Correct Answer)
+4
-1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential (V) at any axial point, at $$2 \mathrm{~m}$$ distance $$(r)$$ from the centre of the dipole of dipole moment vector $$\vec{P}$$ of magnitude, $$4 \times 10^{-6} \mathrm{C} \mathrm{m}$$, is $$\pm 9 \times 10^3 \mathrm{~V}$$.

(Take $$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$$ units)

Reason R: $$V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$$, where $$r$$ is the distance of any axial point, situated at $$2 \mathrm{~m}$$ from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true and R is NOT the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.
2
NEET 2023 Manipur
MCQ (Single Correct Answer)
+4
-1

According to Gauss law of electrostatics, electric flux through a closed surface depends on :

A
the area of the surface
B
the quantity of charges enclosed by the surface
C
the shape of the surface
D
the volume enclosed by the surface
3
NEET 2023 Manipur
MCQ (Single Correct Answer)
+4
-1

A charge $$\mathrm{Q} ~\mu \mathrm{C}$$ is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit) :

A
$$\frac{Q}{\epsilon_0} \times 10^{-6}$$
B
$$\frac{2 \mathrm{Q}}{3 \epsilon_0} \times 10^{-3}$$
C
$$\frac{\mathrm{Q}}{6 \epsilon_0} \times 10^{-3}$$
D
$$\frac{\mathrm{Q}}{6 \epsilon_0} \times 10^{-6}$$
4
NEET 2023 Manipur
MCQ (Single Correct Answer)
+4
-1

If a conducting sphere of radius $$\mathrm{R}$$ is charged. Then the electric field at a distance $$\mathrm{r}(\mathrm{r} > \mathrm{R})$$ from the centre of the sphere would be, $$(\mathrm{V}=$$ potential on the surface of the sphere)

A
$$\frac{r V}{R^2}$$
B
$$\frac{R^2 V}{r^3}$$
C
$$\frac{R V}{r^2}$$
D
$$\frac{\mathrm{V}}{\mathrm{r}}$$
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