1
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An alternating current is given by $\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}$. The r.m.s current will be

A
$\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}$
B
$\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}$
C
$\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}$
D
$\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}$
2
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For an experimental expression $y=\frac{32.3 \times 1125}{27.4}$, where all the digits are significant. Then to report the value of $y$ we should write

A
$y=1326.186$
B
$y=1326.2$
C
$y=1326.19$
D
$y=1330$
3
JEE Main 2025 (Online) 24th January Morning Shift
Numerical
+4
-1
Change Language

A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Your input ____
4
JEE Main 2025 (Online) 24th January Morning Shift
Numerical
+4
-1
Change Language

A current of 5 A exists in a square loop of side $\frac{1}{\sqrt{2}} \mathrm{~m}$. Then the magnitude of the magnetic field $B$ at the centre of the square loop will be $p \times 10^{-6} \mathrm{~T}$. where, value of p is ________ $\left[\right.$ Take $\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}$ ].

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