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Freundlich adsorption isotherm

$\frac{x}{m}=k{p}^{1/n}$

$log\frac{x}{m}=\frac{1}{n}logp+logk$

Comparing with y = mx + C

so, slope of given graph is $\frac{1}{n}$ $\frac{}{}$ whose value varies from 0 to 1.

Sphalerite is ore of zinc consists of ZnS during its concentration group 1 cyanides are used as depressants like NaCN

Major components of gun metal are Cu, Sn and Zn

There are different types of electromagnetic waves and each of them have different wavelengths. Human eye can only see the light rays which have a minimum wavelength of around 400-700nm. Rays which don't meet this threshold will eventually not be visible to the humans.

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

 =   $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$        

Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC\times AL$

$\Delta =\frac{1}{2}\times 2\sqrt[]{2hr-a^2}\times h$       

then   $x=\sqrt[]{2\times \frac{3r}{2}\times r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$      

So . $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$  

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${?}_1=\frac{3}{5}{E}_0\left(0.2\right)N{m}^2{C}^{-1},and{?}_2=\frac{4}{5}{E}_0\left(0.3\right)N{m}^2{C}^{-1}$

$\Rightarrow \frac{{?}_1}{{?}_2}=\frac{3\times 0.2\times 5}{5\times 0.3\times 4}=\frac{1}{2}$

Let v is velocity at the highest position.

$T_{max}=5T_{min}\Rightarrow mg+\frac{m\left(v^2+4g\mathcal{l}\right)}{\mathcal{l}}=5\left(\frac{m{v}^2}{\mathcal{l}}-mg\right)\Rightarrow 4.\frac{v^2}{\mathcal{l}}=10g$

$\Rightarrow v=\sqrt[]{\frac{5}{2}g\mathcal{l}}=\sqrt[]{\frac{5}{2}\times 10\times 1}=5m/s$

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It depends on the frequency of the wave. Rays such as Gamma rays, X-Rays etc. have higher frequencies which can lead to harmful impacts if exposed to the human body. These kind of rays cannot be even produced easily unlike low frequency rays.

$\Delta U+\Delta KE=0$

$\Rightarrow \frac{f}{2}nR\Delta T=\frac{1}{2}m{v}^2\Rightarrow \Delta T=\frac{m}{n}\frac{v^2}{fR}=4\times 10^{-3}\times \frac{30^2}{3R}=\frac{3.6}{3R}K.$

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux. 

For equilibrium,

$\frac{dU}{dr}=0\Rightarrow \frac{-10\alpha }{r^{11}}+\frac{5\beta }{r^6}=0\Rightarrow r={\left(\frac{2\alpha }{\beta }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$           

$={\left(Sinx\right)}_0^{\pi /2}=1$           

$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$           

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$

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$L\frac{di}{dt}=V\Rightarrow 2di=3tdt\Rightarrow 2{\displaystyle \underset{0}{\overset{i}{\int }}di=3}{\displaystyle \underset{0}{\overset{4}{\int }}tdt\Rightarrow i=12A}$

$U=\frac{1}{2}L{i}^2=\frac{1}{2}\times 2\times 12^2=144J$

For JEE Main exam, the Electric Charges and Fields is an important chapter. It is a part of Electrostatics unit. The weightage of this unit is around 6-9% of the total Physics section. You should expect around 3-5 questions in this unit including the questions from the class 12 physics chapter 1.

Reach 30 minutes prior to reporting time

Co-operate with exam centre officials

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Do not involve in mischievous activities

$L\frac{di}{dt}=V?2di=3tdt?2{\displaystyle \underset{0}{\overset{i}{?}}di=3}{\displaystyle \underset{0}{\overset{4}{?}}tdt?i=12A}$

$U=\frac{1}{2}L{i}^2=\frac{1}{2}*2*12^2=144J$

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$       

= 2 cos² θ + 2 sin² θ + 6 sin θ + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin q = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.