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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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vishal kumar

Two reactions R1 and R2 have identical pre-exponential factors.  Activation energy of R1 exceeds 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


\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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vishal kumar

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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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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 




given current through potentiometer wire =60mA


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

So the correct option is 4.


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vishal kumar

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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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vishal kumar

In a meter bridge experimental S is a standard resistance, R is resistance wire. It is found that balancing length is l=25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance {l}' (in cm) will now be _____.
Option: 1 40
Option: 2  50
Option: 3 20
Option: 4 30





R=\frac{P^{l}}{A}=\frac{4P^{l}}{\pi d^{2}}

{R}'=\frac{4\rho \left ( \frac{l}{2} \right )}{\pi \left ( \frac{d}{2} \right )^{2}}=2R

then, \frac{x}{{R}'}=\frac{X}{2R}=\frac{3}{2}

l=40.00\; cm


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Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t=0. If the charge on capacitors A and B at time t=CR is Q_{A} and Q_{B} respectively , then (Here e is the base of natural logarithm)
Option: 1 Q_{A}=VC,\; Q_{B}=CV      
Option: 2  Q_{A}=\frac{CV}{2},\; Q_{B}=\frac{VC}{e}      
Option: 3   Q_{A}=\frac{VC}{e},\; Q_{B}=\frac{CV}{2}     
Option: 4  Q_{A}=VC,\; Q_{B}=\frac{VC}{e}      



Maximum charge on the capacitor=5CV

(a) is reverse biased and (b) is forward biased

(a)                                                           (b)


So, q=q_{max}\left [ e^{-t/RC} \right ]


Hence the correct option is (4). 

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A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when it rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :
Option: 1 T=\sqrt{\frac{\pi m}{IB}}      
Option: 2 T=\sqrt{\frac{\pi m}{2IB}}      
Option: 3   T=\sqrt{\frac{2m}{IB}}   
Option: 4 T=\sqrt{\frac{2\pi m}{IB}}



\tau =MB\sin \theta =I\alpha (using \; M=IA)

\Rightarrow \pi R^{2}IB\theta =\frac{mR^{2}}{2}\alpha \; (Using \ \alpha =-\omega ^{2}\theta )

\Rightarrow \omega =\sqrt{\frac{2\pi IB}{m}}=\frac{2\pi }{t}

\Rightarrow T =\sqrt{\frac{2\pi m}{IB}}


Hence the correct option is (4).

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The current I1 (in A) flowing through 1\Omega resistor in the following circuit is:-                                
Option: 1 0.25
 Option: 2 0.40
Option: 3 0.2  
Option: 4 0.5



Ohms Law -

  1. Ohm’s law

In a conductor, if all external physical conditions like temperature and pressure are kept constant the Current flowing through a conductor is directly proportional to the Potential difference across two ends.

 V\propto I


R- Electric Resistance


  • The graph between V and I

The slope gives the resistance


  • The graph between V and I at different temperatures

Here T1>T2. The resistance of a conductor increases with increase in temperature









 \Delta V_{AB}= \Delta V_{CD}\\ \Rightarrow 2.5*I_2 =2*I_3...(1)

and I_2+I_3=I=\frac{10}{9 }...(2)

From equation (1) and (2)

I_3=\frac{50}{81} , I_2=\frac{40}{81} ,

and I_1= \frac{I_2}{2}=\frac{20}{81} = 0.2469

I_1= 0.25  ( when considered up to two decimal point) 


I_1= 0.2 (when considered up to one decimal point)

Answer given by the NTA is I_1= 0.2 But most appropriate answer is I_1= 0.25

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Ritika Jonwal

An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns / length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed \nu . If the electron does not hit the surface of the solenoid, maximum possible value of \nu is (all symbols have their standard meaning) :  
Option: 1 \frac{2e\mu _{0}nIR}{m}      
Option: 2 \frac{e\mu _{0}nIR}{2m}

Option: 3 \frac{e\mu _{0}nIR}{4m}

Option: 4 \frac{e\mu _{0}nIR}{m}



Malus's Law -

Malus's Law-

This law states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer.



                                                    I=I_{0} \cos ^{2} \theta


output intensity is given by I=I_0Cos^2(\theta )

Initial output intensity=10% of I_0

I.e \frac{10I_0}{100}=I_0Cos^2(\theta )\Rightarrow \theta =71.57

Final output intensity=O

means new angle is 90^0

the angle by which the analyser needs to be rotated further is 90^0-\theta =18.4^0


So Option (3) is correct.

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Ritika Jonwal

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