Class 12th

Let r, h, l and Ø be the radius, height, slant height and semi-vertical angle respectively of the cone. i.e., r, h, l>0.

Then, Volume V of the cone is

$=\frac{1}{3}\pi r^{2}h.$ 

$=\frac{1}{3}\pi \left(l^2-h^2\right)h$

$=\frac{1}{3}\pi \left(l^{2}h-h^3\right).$

So,  $\frac{\mathrm{dV}}{dh}=\frac{1}{3}\pi \left(l^2-3h^2\right)$

$\frac{d^{\mathrm{2V}}}{d{h}^2}=-2\pi h$

Let r and h be the radius and height of the cone.

The volume V of the cone is.

$=\frac{1}{3}\pi r^{2}h$

$r^{2}h=\frac{\mathrm{3V}}{\pi }=\mathrm{k\; (SAY)}\Rightarrow r^2=\frac{k}{h}.$

And curve surface area S is

Let r and h be the radius and height of the one in scribed in the sphere of radius R.

Then, is ΔOBC, rt angle at B (h-r)² + r² = R²

$\Rightarrow$ h² + R²- 2hR + h² = R²

$\Rightarrow$ r² = 2hR -h²

Then the volume v of the cone is, $V=\frac{1}{3}*r^{2}h$ $=\frac{1}{3}\pi \left(2hR-h^2\right)h$

$=\frac{1}{3}\pi \left(2{\mathrm{Rh}}^2-h^3\right).$

$\Rightarrow \frac{\mathrm{dV}}{dh}=\frac{\pi }{3}\left(4\mathrm{Rh}-3h^2\right).$

$\frac{d^{\mathrm{2V}}}{d{h}^2}=\frac{\pi }{3}(\mathrm{4R}-6h).$

At $\frac{\mathrm{dV}}{dh}=0$

$-\frac{\pi }{3}\left(\mathrm{4R}h-3h^2\right)=0.$

4Rh – 3h² = 0.

h(4R – 3h) = 0.

h = 0 and $h=\frac{4R}{3}$

As h> 0, $h=\frac{4}{3}.$

At $h=\frac{\mathrm{4R}}{3},\frac{d^{\mathrm{2v}}}{d{h}^2}=\frac{\pi }{3}\left[4R-6*\frac{\mathrm{4rR}}{3}\right]$

$=\frac{1}{3}[\mathrm{4R}-8R]=-\frac{4\mathrm{R\pi }}{3}<0$

$h=\frac{\mathrm{4R}}{3}$ is a point of maxima.

and $r^2=2\mathrm{hR}-h^2=2*\frac{\mathrm{4R}}{3}R-{\left(\frac{\mathrm{4R}}{3}\right)}^2$

$=\frac{3*\mathrm{8R}{}^2}{9}-\frac{1\mathrm{6R}{}^2}{9}$

$r^2=\frac{\mathrm{8R}{}^2}{9}$

Hence, Volume of Cone, $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\pi *\frac{\mathrm{8R}{}^2}{9}*\frac{\mathrm{4R}}{3}$

$=\frac{8}{27}*\frac{4}{3}\mathrm{\pi R}{}^3=\frac{8}{27}*$ Volume of sphere.

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

(i) For a binary solution having both components as volatile liquids (viz. CHCl₃ and CH₂Cl₂), the total pressure will be

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (ie. CHCl₃)

p₂= partial vapour pressure of component 2 (ie. CH₂Cl₂)

(ii) For a solution containing non-volatile solute ie. NaCl (s) and H₂O (l), the Raoult's law is applicable only to vaporisable component (1) ie. H₂O (l) and total vapour pressure is written as 

P= P₁= x₁P₁⁰+ x₂P₂⁰

= x₁P₁⁰+ (1-x₁) P₂⁰

(P₁⁰-P₂⁰)X¹ + P₂⁰

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (CHCl₃ ) 

Let x and y in 'm' be the length of side of the square the radius of the circle respectily

Then, length of wire = perimeter of square + circumference of circle

$\Rightarrow$ 28 = 4x + 2πy

$\Rightarrow$ 2x + πy = 14

$\Rightarrow y=\frac{14-2x}{x}$

The combine area A of the square and the circle is

A = x² + πy²

$=x^2+\pi {\left[\frac{14-2x}{\pi }\right]}^2$

${=x^2+\frac{[2(7-x)}{x}]}^2$

$=x^2+\frac{4}{\pi }{(7-x)}^2.$

So, $\frac{\mathrm{dA}}{dx}=2x+\frac{4\cdot 2}{\pi }(7-x)+(-1)=2x-\frac{8}{\pi }(7-x)$

$\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{8}{\pi }$

At, $\frac{\mathrm{dA}}{dx}=0$

$\Rightarrow 2x-\frac{8(7-x)}{\pi }=0$

$\Rightarrow 2x=\frac{8}{\pi }(7-x)$

$\Rightarrow 2\pi x=56-8x$

$\Rightarrow (2x+8)x=56$

$\Rightarrow x=\frac{56}{2\pi +8}=\frac{28}{x+4}.$

At, $x=\frac{28}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{\gamma }{\pi }>0$

$\therefore x=\frac{28}{\pi +4}$ isa point of minima

Hence, length of square = $4x=4\left(\frac{28}{1+4}\right)=\frac{112}{\pi +4}m$

and length of circle = 2πy

$=2\pi \left(\frac{14-2x}{\pi }\right)$

$=2\pi \left[\frac{14}{\pi }-\frac{2}{1}\left[\frac{28}{\pi +4}\right]\right]$

$=28-4\left[\frac{28}{\pi +4}\right]=28-\frac{112}{1+4}$

$=\frac{28x+112-112}{x+4}$

$=\frac{28\pi }{\pi +4}$

The volume v of a cylinder of height h and radius r is

V = πr²h = 100

$\Rightarrow h=\frac{100}{\pi r^2}.$

Let, s be the surface area then

S = 2r²hr(r + h) = $2x\left(r^2+\frac{100r}{\pi r^2}\right)$

$=2\pi \left(n^2+\frac{100}{\pi n}\right).$

$\therefore \frac{dS}{dr}=2\pi \left[2x-\frac{100}{1r^2}\right]$

$\Rightarrow \frac{d^{2}S}{d{r}^2}=2\pi \left[2+\frac{200}{\pi r^3}\right]$

At, $\frac{ds}{dr}=2\pi \left[2r-\frac{100}{\pi r^2}\right]=0$

$\Rightarrow 2r=\frac{100}{\pi r^2}$

$\Rightarrow M{}^3=\frac{50}{\pi }\Rightarrow M={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},>0.$

At, $x={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},\frac{d^{\mathrm{2S}}}{d{r}^2}=2\pi \left[2+\frac{200}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}*3\right]$

$=2\pi [2+4]=12\pi >0$

$\therefore r={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ isa point of minimum

And $h=\frac{100}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=2{\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

Let r and h be the radius and height of the cylinder

So, r, h > 0

The total surface area s is given by

S = 2πr(h + r) = content .

$\Rightarrow \frac{S}{2\pi }=r(h+r)$ = content = x (say)

$\Rightarrow h+r=\frac{u}{r}$

$\Rightarrow h=\frac{k}{r}-r=\frac{k-r^2}{r.}$

Then, the volume v of the cylinder

$V=\pi r^{2}h=\pi r^2\left(\frac{k-r^2}{r}\right)=\pi r\left(x-x^2\right)=\pi (kx-{\pi }^3)$

So, $\frac{dv}{dr}=\pi \left(x-3r^2\right)$

$\frac{d^2}{d{r}^2}=-6\pi r.$

For maximum, $\frac{dv}{dr}=0$

$\Rightarrow \pi \left[x-3x^2\right)=0$

$\Rightarrow r^2=\frac{x}{3}$

Let A.B.C.D.be the square increased in a given fixed circle with radius x

Let 'x' and 'y' be the length and breadth of the rectangle

∴x, y> 0

In ABC, right angle at B,

x² + y² = (2x)²

$\Rightarrow$ x² + y² = 4x²

Let 'x' cm be the length of side of the square to be cut off from the rectangular surface

Then, the volume v of the box is v = (45 - 2x) (24 - 2x) x

= 1080x - 138x² + 4x³

So,  $\frac{dv}{dx}=1080-276x+12x^2$

$\frac{d^2}{d{x}^2}=-276+24x$

At,  $\frac{dv}{dx}=0\to 1080-276x+12x^2=0$

$\Rightarrow$ x²- 23x + 90 = 0

$\Rightarrow$ x²- 5x - 18x + 90 = 0

$\Rightarrow$ x (x - 5) - 18 (x - 5) = 0

$\Rightarrow$ (x- 5) (x- 18) = 0

$\Rightarrow$ x = 5 and x = 18

At x = 18, breadth = 24 - 2 (18) = 24 - 36 = -12 which is not possible

At,  $x=5,\frac{d^{2}v}{d{x}^2}=-276+24(5)=-276+120=-150<0$

Hence, x = 5 is the point of maximum

So, '5' cm length of square seeds to be cut from each corner of the secthgle