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In a fluorescent lamp choke ( a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _____.
Option: 1 10
Option: 2 20
Option: 3 30
Option: 4 40

$V = L\frac{\text d i}{\text d t}\Rightarrow 100 = L\frac{0.25}{0.025} \\ \Rightarrow L = 10 \ \text{mH}$

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Posted by

avinash.dongre

Magnetic field in a plane electromagnetic wave is given by Expression for corresponding electric field will be : Where c is speed of light.
Option: 1
Option: 2
Option: 3
Option: 4

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Posted by

vishal kumar

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Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln(k2/k1) is equal to : (R=8.314 J mol⁻¹K⁻¹)
Option: 1 6
Option: 2 4
Option: 3 8
Option: 4 12

$K=Ae^{\frac{-Ea}{RT}}$

$\frac{K_{1}}{K_{2}}=e\left ( \frac{Ea_{1}-Ea_{2}}{RT} \right )$

Taking log both the sides

$ln\left ( \frac{K_{2}}{K_{1}} \right )=\frac{E_{a1}-E_{a2}}{RT}= \frac{10000}{8.314\times 300}=4$

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Posted by

vishal kumar

Four resistances of $15\Omega ,12\Omega,4\Omega\:\:and\:\:10\Omega$ respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $10\Omega$ to balance the network is ___________ $\Omega$ .
Option: 1 10
Option: 2 5
Option: 3 15
Option: 4 20

For balanced Wheatstone bridge $\frac{R_1}{R_2}=\frac{R_3}{R_4} \ \ or \ \ \frac{R_1}{R_3}=\frac{R_2}{R_4}$

$R_2=12 \Omega \ \ and \ \ R_4=4 \Omega$

As $\frac{R_2}{R_4}=\frac{12}{4}=3$

So using $R_1=15 \Omega$

We get $R_3=R_{AD}=5 \Omega$

let we connected x-ohms in parallel to 10-ohm resistance

i.e $R_3=5 \Omega=\frac{x*10}{x+10}$

we get $x=10 \Omega$

So the answer will be 10.

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Posted by

vishal kumar

JEE Main high-scoring chapters and topics

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At time $t=0$ magnetic field of $1000$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to $500\: Gauss$ , in the next $5s$ , then induced EMF (in $\mu V$ ) in the loop is :
Option: 1 56

Option: 3 48

Option: 5 36

Option: 7 28

Area of the given loop

$\\A=Area\ of\ rectangle-Area\ of \ 2\ triangles\ with\ height\ 2cm \\ A=16\times4-2\times4=56cm^2$

Magnitude of emf

$\\E=A\frac{dB}{dt}=56\times10^{-4}\times(\frac{(1000-500)\times10^{-4}}{5})\\ E=5600\times10^{-8}V=56\mu \ V$

So the correct graph is given in option 1.

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Posted by

vishal kumar

The length of a potentiometer wire is $1200cm$ and it carries a current of $60mA$ . For a cell of emf $5V$ and internal resistance of $20\Omega$ , the null point on it is found to be at $1000cm$ . The resistance of whole wire is $(in \ \Omega)$ :
Option: 1 100
Option: 2 80
Option: 3 604
Option: 7 120

In the potentiometer, a battery of known emf E is connected in the secondary circuit. A constant current I is flowing through AB from the driver circuit. The jockey is a slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let $l$ be the length at which galvanometer shows null deflection.

Since the potential of wire AB (V) is proportional to the length AB(L).

Similarly $E \ \ \alpha \ \ \ l$

So we get

$\frac{V}{E}=\frac{L}{l}$

$V=E\frac{L}{l}$

$V=5\times\frac{1200}{1000}=6V$

given current through potentiometer wire =60mA

V=iR

$\\60\times10^{-3}R=6\\\Rightarrow R=100\Omega$

So the correct option is 4.

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Posted by

vishal kumar

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

A proton with kinetic energy of $1MeV$ moves from south to north. It gets acceleration of $10^{12}m/s^2$ by an applied magnetic field (west to east). The value of magnetic field ( in mT): (Rest mass of a proton is $1.6\times 10^{-27}kg$ )
Option: 1 0.71
Option: 2 71
Option: 3 0.071
Option: 4 7.1

$\begin{array}{l}{\because \mathrm{K.E.}=1.6 \times 10^{-13}=\frac{1}{2} \times 1.6 \times 10^{-27} \mathrm{V}^{2}} \\ \\ {\mathrm{V}=\sqrt{2} \times 10^{7}} \\ \\ {\therefore \mathrm{Bqv}=\mathrm{ma}} \\ \\ {\mathrm{B}=\frac{1.6 \times 10^{-27} \times 10^{12}}{1.6 \times 10^{-19} \times \sqrt{2} \times 10^{7}}} \\ \\ {=0.71 \times 10^{-3} \mathrm{T}} \\ \\ {\text { so } 0.71 \mathrm{mT}}\end{array}$

So option (1) is correct

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Posted by

vishal kumar

A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is $\vec B=(3\hat{i}+4\hat{j})T$ .The quantity of flux through the loop ABCDEFA (in Wb) is:-
Option: 1 175
Option: 2 60
Option: 3 25
Option: 4 170

Magnetic flux - - wherein

Magnetic flux-

The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to
the magnetic flux linked with that area.

I.e for the below figure

Net magnetic flux through the surface is given by

$\phi_B = \oint \vec{B}\cdot \vec{dA}= BA\cos \Theta$