Class 12th

41. Option (i) and (iii)

Explanation: Haematite is an ore of iron that can be calcined and reduced by carbon. In the metallurgical process, calamine ore is calcined ore that can be reduced by carbon. Hence, option (i) and (ii) are correct.

As $y=3$ intersect $y^2=4x$ at Athen,

$\begin{array}{l}{3}^2=4x\\ \Rightarrow x=\frac{9}{4}\end{array}$

$\therefore$ A has coordinate $\left(\frac{a}{4},3\right)$

Hence, area of curve = ${\displaystyle \underset{y=0}{\overset{y=3}{\int }}x}dy$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{3}{\int }}\frac{y^2}{4}dy={\left[\frac{y3}{4*3}\right]}_0^3}\\ =\frac{3^3}{12}-\frac{0}{12}\\ =\frac{27}{12}=\frac{9}{4}uni{t}^2\end{array}$

$\therefore$ Option (B) is correct

Given equation of the circle is

∴ option (A) is correct

20. Ce  =  4f¹ 5d¹ 6s²

Ce^{3 +}  =  4f¹ 5d⁰ 6s⁰ 

As in Ce³⁺ ,

4f has only single electron. Cerium holds the capacity to lose this electron also and attain 4f⁰ configuration which is more stable. 

Ce⁴⁺  =  4f⁰ 5d⁰ 6s⁰

Hence it shows 4⁺ oxidation states.

The given equation of the curve is

$\begin{array}{l}{y}^2=4x\\ \Rightarrow y=\pm \surd 3\end{array}$

$\begin{array}{l}\mathrm{\to \; y}=+\surd 2x\end{array}$ in Ist quadrant

So, area of curve enclosed by $y^2=4x$

And $x=3=2*\; area$ area (AOCA)

19. As an effect of lanthanoid contraction, zirconium and hafnium have similar radius of 160 pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii. 

40. Option (i) and (iii)

Calcination is the process of heating when the volatile matter escapes, leaving the metal oxide behind. It is usually done in the absence of air.

Given curve is $x^2=4y$ and the equation of line is $x=4y-2$

The point of intersection of the curve and the line can be determine as follows.

Put, $x=4y-2\Rightarrow x+2=4y\Rightarrow y\frac{x+2}{4}$

In $x^2=4y$ to determine value of x

i.e, $x^2=4*\frac{\left(x+2\right)}{4}=x+2$

$\begin{array}{l}\Rightarrow x^2-x-2=0\Rightarrow x^2+x-2x-2=0\\ \Rightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Rightarrow \left(x+1\right)\left(x-2\right)=0\end{array}$

$\Rightarrow x=2$ and $x=-1$

$x=2$ , we have ${\left(2\right)}^2=4y\Rightarrow 4-4y\Rightarrow y=1$

And at $x=-1$ we have ${\left(-1\right)}^2=4y\Rightarrow y=\frac{1}{4}$

So, the coordinates A and B are (2,1) and ( $-1,\frac{1}{4}$ )

$\therefore$ The required area before the line & the curve is area $B{D}^{\iota }AB$ = area of trapezium (BNMAB)- area under curve BDA

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{line}}dx-{\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{curve}}dx\\ ={\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x+2}{4}da-{\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x^2}{4}}}\\ =\frac{1}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}xda+\frac{2}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}dx-\frac{1}{4}{\displaystyle \underset{-1}{\overset{2}{\int }}{x}^{2}dx}}}\end{array}$

$\begin{array}{l}=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_{-1}^2+\frac{2}{4}{\left[x\right]}_{-1}^2-\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^2\\ =\frac{1}{8}\left[2^2-{\left(-1\right)}^2\right]+\frac{1}{2}\left[2-\left(-1\right)\right]-\frac{1}{12}\left[2^3-{\left(-1\right)}^3\right]\end{array}$

$\begin{array}{l}=\frac{3}{8}+\frac{3}{2}-\frac{9}{12}=\frac{3}{8}+\frac{3}{2}-\frac{3}{4}=\frac{3+4*3-2*3}{8}\\ =\frac{3+12-6}{8}=\frac{9}{8}uni{t}^2\end{array}$

18. Ce, Pr and Nd belong to the lanthanide series whereas Th, Pa and U belong to the actinides family. 

When electrons start accommodating the 4f and 5f orbitals, the 5f electrons penetrate less into the inner core. They are more effectively shielded nuclei in comparison to 4f-electrons in lanthanides. This leads to the fact that 5f-electrons experience reduced nuclear force of attraction and hence they have Lower ionisation enthalpies than lanthanoids. 

Given that equation of

curve $y=x^2$

line $y=\left|x\right|=\left\{x,ifx\ge 0\&-x,ifx\angle 0\right\}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y=x^2$

$\Rightarrow x=x^2$ for Ist quadrant and

$\Rightarrow -x=x^2$ for IInd t quadrant

So, $\Rightarrow x^2-x=0$ and $x^2+x=0$

$\Rightarrow x\left(x-1\right)=0$ and $x\left(x+1\right)=0$

$\Rightarrow x=0,1$ $x=0,-1$

$x=0,y=0$

$x=1,y=1$ i.e, A has coordinate (1,1)

$x=-1,y=1$ i.e, B has coordinate (1,1)

Now, area of AODA = area (AOM)-area (ADOM)

$={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{line}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{curve}dx}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}xdx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx={\left[\frac{x^2}{2}\right]}_0^1-}{\left[\frac{a^3}{3}\right]}_0^1$

$=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}unit{s}^2$

$\therefore$ The required area of the region bounded by curve $y=x^2$ and line $y=\left|x\right|$ is $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}unit{s}^2$