1
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit, rms value of current $\left(I_{\mathrm{rms}}\right)$ through the resistor $R$ is:

JEE Main 2023 (Online) 30th January Evening Shift Physics - Alternating Current Question 44 English
A
$ 2 \sqrt{2} \mathrm{~A}$
B
$2 \mathrm{~A}$
C
$\frac{1}{2} \mathrm{~A}$
D
$20 \mathrm{~A}$
2
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is

A
$4.5^{\circ}$
B
$7.8^{\circ}$
C
$1.3^{\circ}$
D
$6^{\circ}$
3
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A: The nuclear density of nuclides ${ }_{5}^{10} \mathrm{~B},{ }_{3}^{6} \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne}$ and ${ }_{83}^{209} \mathrm{Bi}$ can be arranged as $\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}}^{\mathrm{N}}>\rho_{\mathrm{Ne}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}$

Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R=R_{0} A^{1 / 3}$, where $R_{0}$ is a constant.

In the light of the above statements, choose the correct answer from the options given below
A
${Both ~\mathbf{A}}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B
Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$
C
$\mathbf{A}$ is false but $\mathbf{R}$ is true
D
$\mathbf{A}$ is true but $\mathbf{R}$ is false
4
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency. At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is:
A
$\frac{1}{40 \pi} \frac{W}{m^2}$
B
$\frac{1}{10 \pi} \frac{W}{m^2}$
C
$\frac{1}{20 \pi} \frac{W}{m^2}$
D
$\frac{1}{2 \pi} \frac{W}{m^2}$
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