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1
JEE Main 2026 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
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-1
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A LCR series circuit driven with $E_{r m s}=90 \mathrm{~V}$ at frequency $f_{\mathrm{d}}=30 \mathrm{~Hz}$ has resistance $R=80 \Omega$, an inductance with inductive reactance $X_L=20.0 \Omega$ and capacitance with capacitive reactance $X_C=80.0 \Omega$. The power factor of the circuit is $\_\_\_\_$ .

A

0.8

B

0.64

C

0.9

D

0.5

2
JEE Main 2026 (Online) 6th April Morning Shift
Numerical
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Refer to the circuit diagram given below. The heat generated across the $6 \Omega$ resistance in 100 second is $\frac{\alpha}{100} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer)

JEE Main 2026 (Online) 6th April Morning Shift Physics - Current Electricity Question 3 English
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3
JEE Main 2026 (Online) 6th April Morning Shift
Numerical
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An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0^{\circ}$ and intensity of light emerged from analyser is $\frac{3}{8} I_0$, the angle of rotation of the light by the solution with respect to analyser is $\_\_\_\_$ degrees.

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4
JEE Main 2026 (Online) 6th April Morning Shift
Numerical
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The energy released when $\frac{7}{17.13} \mathrm{~kg}$ of ${ }_3^7 \mathrm{Li}$ is converted into ${ }_2^4 \mathrm{He}$ by proton bombardment is $\alpha \times 10^{32} \mathrm{eV}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer) (Mass of ${ }_3^7 \mathrm{Li}=7.0183 \mathrm{u}$, mass of ${ }_2^4 \mathrm{He}=4.004 \mathrm{u}$, mass of proton $=1.008 \mathrm{u}$ and $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$ and Avogadro number $=6.0 \times 10^{23}$ )

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