Class 12th

This is a short answer type question as classified in NCERT Exemplar

p =K_Hx (where p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas in solution) is Henry's law expressed mathematically.

As a result of the aforementioned equation, "the lower the solubility of the gas in the liquid, the greater the value of Henry's law constant K_H at a given pressure."

We have, f(x)= $\frac{4sinx-2x-xcosx.}{2+cosx}$

$=\frac{4sinx-x(2+cosx)}{2+cosx}$

$f(x)=\frac{4sinx}{2+cosx}-x.$

So, $f^{\prime }(x)=\frac{(2+cosx)\frac{d}{dx}(4sinx)-4sinx\frac{d}{dx}(2+cosx)}{{(2+cosx)}^2}-\frac{dx}{dx}.$

$=\frac{(2+cosx)(4cosx)-(4sinx)(-sinx)}{{(2+cosx)}^2}-1.$

$=\frac{8cosx+4co{s}^{2}x+4si{n}^{2}x}{{(2+cosx)}^2}-1$

$=\frac{8cosx+4\left(co{s}^{2}x+si{n}^{2}x\right)}{{(2+cosx)}^2}-1.$

$=\frac{8cosx+4}{{(2+cosx)}^2}-1.$

$=\frac{8\cdot cosx+4-{(2+cosx)}^2}{{(2+cosx)}^2}$

$=\frac{8cosx+4-4-4cosx-co{s}^{2}x}{{(2+cosx)}^2}$

$=\frac{4cosx-co{s}^{2}x}{{(2+cosx)}^2}=\frac{cosx(4-cosx)}{{(2+cosx)}^2}.$

Now, ${(2+cosx)}^2>0.$

And, $4-cosx>0$ as cos x lies in [1, 1].

So, (i) for increasing, f(x) ≥ 0.

cosx ≥ 0.

x lies in Ist and IVth quadrant.

i.e., f(x) is increasing for $0\le x\le \frac{x}{2}$ and $\frac{3x}{2}\le x\le 2x.$

(ii) for decreasing, f(x) ≤ 0.

cosx ≤ 0.

x lies in IInd and IIIrd quadrant.

i.e., f(x) is decreasing for $\frac{x}{2}\le x\le \frac{3x}{2}$ .

We have $x=acos\theta +a\theta sin\theta$

$\begin{array}{l}\therefore \frac{dx}{d\theta }=-asin\theta +asin\theta +a\theta cos\theta =a\theta cos\theta \\ y=asin\theta -a\theta cos\theta \\ \therefore \frac{dy}{d\theta }=acos\theta -acos\theta +a\theta sin\theta =a\theta sin\theta \\ \therefore \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\frac{a\theta sin\theta }{a\theta cos\theta }=tan\theta \end{array}$

$\therefore$ Slope of the normal at any point $\theta$ is $-\frac{1}{tan\theta }$

The equation of the normal at a given point $\left(x,y\right)$ is given by,

$\begin{array}{l}y-asin\theta +a\theta cos\theta =\frac{-1}{tan\theta }\left(x-acos\theta -a\theta sin\theta \right)\\ \Rightarrow ysin\theta -asi{n}^2\theta +a\theta sin\theta cos\theta =-xcos\theta +aco{s}^2\theta +a\theta sin\theta cos\theta \\ \Rightarrow xcos\theta +ysin\theta -a\left(si{n}^2\theta +co{s}^2\theta \right)=0\\ \Rightarrow xcos\theta +ysin\theta -a=0\end{array}$

Now, the perpendicular distance of the normal from the origin is

This is a short answer type question as classified in NCERT Exemplar

The solubility rule "like dissolves like"  is based on the intermolecular forces of that exist in solution as follows:

If the intermolecular interactions in both components are similar, a substance (solute) dissolves in a solvent (ie. solvent and solute particles or molecules). When polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents, this is a regular occurrence. 

Equation of the curve is $y^2=4x..........(i)$

$\begin{array}{l}2y\frac{dy}{dx}=4\\ \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}\\ \therefore {\frac{dy}{dx}]}_{\left(1,2\right)}=\frac{2}{2}=1\end{array}$

Now, the slope of the normal at point $\left(1,2\right)$ is $\frac{-1}{{\frac{dy}{dx}]}_{\left(1,2\right)}}=\frac{-1}{1}=1$

$\therefore$ Equation of the normal at $\left(1,2\right)$ is $y-2=-1\left(x-1\right).$

$\begin{array}{l}\Rightarrow y-2=-x+1\\ \Rightarrow x+y-3=0\end{array}$

Let 'b' and 'x' be the fixed base and equal side of isosceales triangle.

Then,  $\frac{dx}{dt}=-3$ cm/s (Ø decreasing).

This is a short answer type question as classified in NCERT Exemplar

The vapour pressure of a liquid in comparison to air pressure determines its boiling point. At a constant atmospheric pressure, the lower the vapour pressure, the higher the boiling point of a liquid, and vice versa.

Because NaCl is a nonvolatile solute, it reduces the vapour pressure of water when added to it. The boiling point of water rises as a result. Methyl alcohol, on the other hand, is more volatile than water, therefore adding it to the solution raises the overall vapour pressure, lowering the boiling point of water.

Because both components exist in the distillate and the liquid and vapour compositions are the same, this indicates that the liquids have formed an azeotropic combination that cannot be separated at this stage by fractional distillation.

We have, f(x) $\frac{logx}{x},x>0.$

f(x) = $\begin{array}{l}\underset{\_}{x\frac{d}{dx}logx-logx\frac{d}{dx}x}\\ x^2.\end{array}$

$=\frac{x*\frac{1}{x}-logx}{x^2}=\frac{1-logx}{x^2}$

f(x) = $\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{d}{dx}{x}^2}{x^4}$

= $\frac{x^2\left(-\frac{1}{x}\right)-(1-logx)(2x)}{x^4}$

= $\frac{-1-2+2logx}{x^3}$ = $\frac{2logx-3}{x^3}$

At extreme points, f(x) = 0.

$\Rightarrow \frac{1-logx}{x^2}=0.$

$\Rightarrow logx=1=loge.$

$\Rightarrow x=e.$

At x = e, f"(e) = $\frac{2loge-3}{e^3}=\frac{2-3}{e^3}=-\frac{1}{e^3}<0$

x = e is a point of maximum.