Class 12th

18. Option (iv) is correct since in antiferromagnetic substances the domains  are oppositely oriented and hence they cancel out each other's magnetic moments. 

17. Option (ii) Quartz glass (SiO₂) is correct since quartz glass (SiO₂)is amorphous in nature as there is no long range ordered arrangement of the constituent particles being present in it and hence it is an amorphous solid.

16. Option (ii) Isotropic nature is correct since crystalline solids exhibit anisotropic properties like refractive index, electrical resistance etc. Since these are found to have different values when measured along different directions in the same crystal and hence they are not isotropic in nature. 

The given D.E. is

$\left(1-y^2\right)\frac{dx}{dy}+yx=\propto y\left(-1<y<1\right)$

$\frac{dx}{dy}+\frac{y}{1-y^2}*x=\frac{\propto y}{1-y^2}$ which is of form

$\begin{array}{l}\frac{dx}{dy}+Px=Q\&P=\frac{y}{1-y^2}\&Q=\frac{\propto y}{1-y^2}\\ {\displaystyle \int pdx={\displaystyle \int \frac{y}{1-y^2}dx=-\frac{1}{2}{\displaystyle \int \frac{-2y}{1-y^{2}dx}}}}\end{array}$

$\begin{array}{l}=-\frac{1}{2}log\left|1-y^2\right|\\ =log{\left[1-y^2\right]}^{-\frac{1}{2}}\end{array}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log{\left[1-y^2\right]}^{-\frac{1}{2}}}\\ ={\left[1-y^2\right]}^{-\frac{1}{2}}$

$\therefore$ option (D ) is correct.

15. Option (ii) Low temperature is correct since at sufficiently low temperature, the thermal energy is low, so the intermolecular forces bring the molecules of a substance closer so that they cling to one another and occupy fixed positions. They keep on vibrating about their fixed positions. Such conditions favours the existence of the substance in solid state. 

The given D.E is

$\begin{array}{l}x\frac{dy}{dx}-y=2x^2\\ \frac{dy}{dx}-\frac{1}{x}y=2x\end{array}$

Which is of form $\frac{dy}{dx}+Py=Q$

So,  $P=-\frac{1}{x}$

$\therefore I.E=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{x}dx}}=e^{log-x}=e^{log-x^{-1}}=x^{-1}=\frac{1}{x}$

$\therefore$ Option (c) is correct

We know that slope of tangent to the curve is $\frac{dy}{dx}$

$\therefore x+y=\frac{dy}{dx}+5$

$\frac{dy}{dx}-y=x-5$ Which has form $\frac{dy}{dx}+Px=Q$

$where,P=1\&Q=x-5$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -1dx}}=e^{-x}$

Thus the solution has the form

$\begin{array}{l}y{e}^{-x}={\displaystyle \int \left(x-5\right)e^{-x}dx+c}\\ ={\displaystyle \int x{e}^{-x}dx-{\displaystyle \int 5e^{-x}dx+c}}\\ =y{e}^{-x}=I+5e^{-x}+c\\ where,I={\displaystyle \int x{e}^{-x}dx}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int e^{-x}dxdx}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx}\\ =-x{e}^{-x}-e^{-x}\end{array}$

$\begin{array}{l}\therefore y{e}^{-x}=-x{e}^{-x}-e^{-x}+5e^{-x}+c\\ =y{e}^{-x}=-x{e}^{-x}+4e^{-x}+c\\ =y=-x+4+\frac{c}{e^{-x}}\\ =y+x=4+c{e}^x\end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l}2+0=4c{e}^0\\ 2-4=c\\ c=-2\end{array}$

$\therefore$ The particular solution is

$y+x=4-2e^x$

We know the slope of tangent to curve is $\frac{dy}{dx}$ .

$\therefore$ $\frac{dy}{dx}=x+y$

$=\frac{dy}{dx}-y=x$ which has form $\frac{dy}{dx}+Py=Q$

So, $P=-1\&Q=x$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -dx}}=e^{-x}$

Thus the solution is of the form .

$\begin{array}{l}y*e^{-x}={\displaystyle \int x.e^{-x}dx+c}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int e^{-x}dxdx+c}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx+c}\\ =y{e}^{-x}=-x{e}^{-x}-e^{-x}+c\\ =y=-x-1+\frac{c}{e^{-x}}\\ =y+x+1=c{e}^x\end{array}$

Given, the curve passes through origin (0,0) i.e, $y=0,when,x=0$

$0+0+1=c{e}^0=c=1$

$\therefore$ Thus, equation of the curve is

$y+x+1=e^x$

The given D.E. is

$\frac{dy}{dx}-3ycotx=sin2x$

$\frac{dy}{dx}-3cotx.y=sin2x$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=-3cotx\&Q=sin2x$

$\therefore I.F=e^{-{\displaystyle \int Pdx}}=e^{{\displaystyle \int -3cotxdx}}=e^{-3{\displaystyle \int cotxdx}}=e^{-3log\left|cotxdx\right|}=e^{log{\left(sin\right)}^{-3}}=\frac{1}{si{n}^{3}x}$

Thus the solution is of the form.

$\begin{array}{l}y*\frac{1}{si{n}^{3}x}={\displaystyle \int sin2x.\frac{1}{si{n}^{3}x}dx+c}\\ ={\displaystyle \int \frac{2sinxcosx}{si{n}^{3}x}}dx+c\left\{?sin2x=2sinxcosx\right\}\\ ={\displaystyle \int 2cosecxcotxdx+c}\\ =-2cosecx+c\\ =\frac{y}{si{n}^{3}x}=\frac{-2}{sinx}+c\\ =-2y=-2si{n}^{2}x+csi{n}^{3}x\end{array}$

Given, $y=2,when,x=\frac{\pi }{2}$

$\begin{array}{l}2=-2si{n}^2\frac{1}{2}+csi{n}^3\frac{\pi }{2}\\ =2=-2+c\\ =e=2+2=4\end{array}$

$\therefore$ The particular solution is, $y=-2si{n}^{2}x+4si{n}^{3}x$