1
NEET 2024
MCQ (Single Correct Answer)
+4
-1
Change Language

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
Change Language

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
Change Language

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
Change Language

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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