1
JEE Main 2026 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
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Four persons measure the length of a rod as $20.00 \mathrm{~cm}, 19.75 \mathrm{~cm}, 17.01 \mathrm{~cm}$ and 18.25 cm . The relative error in the measurement of average length of the rod is :

A

0.18

B

0.24

C

0.06

D

0.08

2
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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3
JEE Main 2026 (Online) 23rd January Morning Shift
Numerical
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-1
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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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4
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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