1
JEE Main 2023 (Online) 13th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Two charges each of magnitude $$0.01 ~\mathrm{C}$$ and separated by a distance of $$0.4 \mathrm{~mm}$$ constitute an electric dipole. If the dipole is placed in an uniform electric field '$$\vec{E}$$' of 10 dyne/C making $$30^{\circ}$$ angle with $$\vec{E}$$, the magnitude of torque acting on dipole is:

A
$$4 \cdot 0 \times 10^{-10} ~\mathrm{Nm}$$
B
$$1.5 \times 10^{-9} ~\mathrm{Nm}$$
C
$$1.0 \times 10^{-8} ~\mathrm{Nm}$$
D
$$2.0 \times 10^{-10} ~\mathrm{Nm}$$
2
JEE Main 2023 (Online) 12th April Morning Shift
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 : If an electric dipole of dipole moment $$30 \times 10^{-5} ~\mathrm{C} ~\mathrm{m}$$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.

Reason R : Electric dipole consists of two equal and opposite charges.

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

A
A is true but R is false
B
A is false but R is true
C
Both A and R are true but R is NOT the correct explanation of A
D
Both A and R are true and R is the correct explanation of A
3
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor

A
move in the straight line paths in the same direction
B
move with the uniform velocity throughout from lower potential to higher potential
C
drift from higher potential to lower potential.
D
move in the curved paths from lower potential to higher potential
4
JEE Main 2023 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Electric potential at a point '$$\mathrm{P}$$' due to a point charge of $$5 \times 10^{-9} \mathrm{C}$$ is $$50 \mathrm{~V}$$. The distance of '$$\mathrm{P}$$' from the point charge is:

(Assume, $$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$$ )

A
0.9 cm
B
90 cm
C
3 cm
D
9 cm
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