1
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electrostatic potential due to an electric dipole at a distance '$$r$$' varies as :

A
$$\frac{1}{r^3}$$
B
$$\frac{1}{\mathrm{r}}$$
C
$$\frac{1}{r^2}$$
D
r
2
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A particle is placed at the point $$A$$ of a frictionless track $$A B C$$ as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is :

(Take $$g=10 \mathrm{~m} / \mathrm{s}^2$$).

JEE Main 2024 (Online) 30th January Morning Shift Physics - Work Power & Energy Question 11 English

A
$$2 \sqrt{10} \mathrm{~m} / \mathrm{s}$$
B
$$10 \mathrm{~m} / \mathrm{s}$$
C
$$\sqrt{10} \mathrm{~m} / \mathrm{s}$$
D
$$20 \mathrm{~m} / \mathrm{s}$$
3
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are $$\mathrm{C}_{\mathrm{A}}$$ and $$\mathrm{C}_{\mathrm{B}}$$. The molar heat capacity at constant pressure and constant volume are represented by $$\mathrm{C_P}$$ and $$\mathrm{C_V}$$, respectively. Choose the correct statement.

JEE Main 2024 (Online) 30th January Morning Shift Physics - Heat and Thermodynamics Question 23 English

A
$$\mathrm{C_P>C_B>C_A>C_V}$$
B
$$\mathrm{C}_{\mathrm{P}}>\mathrm{C}_{\mathrm{V}}>\mathrm{C}_{\mathrm{A}}=\mathrm{C}_{\mathrm{B}}$$
C
$$\mathrm{C}_{\mathrm{A}}=0$$ and $$\mathrm{C}_{\mathrm{B}}=\infty$$
D
$$\mathrm{C_A=\infty, C_B=0}$$
4
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electric field of an electromagnetic wave in free space is represented as $$\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}$$. The corresponding magnetic induction vector will be :

A
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}$$
B
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}$$
C
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}$$
D
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}$$
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