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JEE Main 2026 (Online) 23rd January Morning Shift
Numerical
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A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T . The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is

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

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2
JEE Main 2026 (Online) 23rd January Morning Shift
Numerical
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The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1=2, K_2=3$ and $K_3=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .

JEE Main 2026 (Online) 23rd January Morning Shift Physics - Capacitor Question 10 English
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3
JEE Main 2026 (Online) 23rd January Morning Shift
Numerical
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The equation of the electric field of an electromagnetic wave propagating through free space is given by : $E=\sqrt{377} \sin \left(6.27 \times 10^3 t-2.09 \times 10^{-5} x\right) \mathrm{N} / \mathrm{C}$

The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^2$. The value of $\alpha$ is

$$ \left(\text { Take } \sqrt{\frac{\mu_0}{\varepsilon_o}}=377 \text { in SI units }\right) $$

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4
JEE Main 2026 (Online) 23rd January Morning Shift
Numerical
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In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $ 2 : 1 $ and $1: 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ( $D_1 / D_2$ ) is $\_\_\_\_$。

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