Class 12th

The given equation of the lines are

$\begin{array}{l}2x+y=4------(1)\\ 3x-2y=6-----(2)\\ x-3y+5=0----(3)\end{array}$

$\therefore$ Area of $?ABC=area(PCBQ)-area(?APC)-area(?AQB)$

$\begin{array}{l}={\displaystyle \underset{1}{\overset{4}{\int }}{y}_{(3)}dx-}{\displaystyle \underset{1}{\overset{2}{\int }}{y}_{(1)}dx-{\displaystyle \underset{2}{\overset{4}{\int }}{y}_{(2)}dx-}}\\ ={\displaystyle \underset{1}{\overset{4}{\int }}\frac{\left(x+5\right)}{3}dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(4-2x\right)dx-{\displaystyle \underset{2}{\overset{4}{\int }}\left(\frac{3x-6}{2}dx\right)}}}\end{array}$

$\therefore$ The point of intersection of the circle and the parabola is . $A\left(\frac{1}{2},\right)\&C\left(\frac{1}{2},-\right)$

Taking in first quadrant

Area of $(OABO)=area(OADO)+area(BADB)$

34. (B) 3d⁵

Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

3d⁷ =  3 unpaired electrons 

3d⁵ =  5 unpaired electrons

3d⁸ =  2 unpaired electrons 

3d² =   2 unpaired electrons 

Out of all, 3d⁵ has five unpaired electrons that is maximum and hence it has the highest magnetic moment.

52. (i) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: Van Arkel method is generally used to obtain pure forms of Zirconium (Zr) and Titanium (Ti).

The metal iodide is heated at 1800 K and is decomposed on a tungsten filament. The pure metal is dropped on the tungsten filament. This proves that ZrI4 is volatile and decomposes at 1800 K.

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is $y-0=\frac{5-0}{4-2}\left(x-2\right)$

$=y=\frac{5}{2}\left(x-2\right)$

Similarly equation of BC is

$\begin{array}{l}y-5=\frac{3-5}{6-4}\left(x-4\right)\\ =y=5+\left(-x+4\right)\\ y=9-x\end{array}$

And equation of AC is $y-0=\frac{3-0}{6-2}\left(x-2\right)$

= $y=\frac{3}{4}\left(x-2\right)$

$\therefore$ Area of $?ABC$

= $area(?ABE)+area(BCDE)-area(?ACD)$

$\begin{array}{l}={\displaystyle \underset{2}{\overset{4}{\int }}{y}_{AB}dx+}{\displaystyle \underset{4}{\overset{6}{\int }}{y}_{BC}dx-{\displaystyle \underset{2}{\overset{6}{\int }}{y}_{BC}dx}}\\ ={\displaystyle \underset{2}{\overset{4}{\int }}\frac{5}{2}\left(x-2\right)dx+{\displaystyle \underset{4}{\overset{6}{\int }}\left(9-x\right)dx-{\displaystyle \underset{2}{\overset{6}{\int }}\frac{3}{4}}}}\left(x-2\right)dx\\ =\frac{5}{2}{\left[\frac{x^2}{2}-2x\right]}_2^4+{\left[9x-\frac{x^2}{2}\right]}_4^6-\frac{3}{4}{\left[\frac{x^2}{2}-2x\right]}_2^6\end{array}$

$\begin{array}{l}=\frac{5}{2}\left[\left(\frac{4^2}{2}-2.4\right)-\left(\frac{2^2}{2}-2.2\right)\right]+\left[\left(9.6-\frac{6^2}{2}\right)-\left(9.4-\frac{4^2}{2}\right)\right]-\frac{3}{4}\left[\left(\frac{6^2}{2}-2.6\right)-\left(\frac{2^2}{2}-2.2\right)\right]\\ =\frac{5}{2}\left[\left(8-8\right)-\left(2-4\right)\right]+\left[54-18-36+8\right]-\frac{3}{4}\left[18-12-2+4\right]\\ =5+8-6=7uni{t}^2\end{array}$

51.  (i) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: In the Mond process Nickel is reacted with carbon monoxide reversibly to give Nickel carbonyl,  Ni (CO)₄. Nickel carbonyl is a volatile compound. It decomposes to nickel and carbon monoxide at 460 K.

33.  (A) Mn₂O₇

2KMnO₄ +  2H₂SO₄ (conc.) → Mn₂O₇ +    2KHSO₄ + H₂O

Manganese heptoxide is an acid anhydride of permanganic acid ( HMnO₄). It is a dangerous oxidiser, volatile and highly reactive liquid.

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

Given equation of the curve is $\left|x\right|+\left|y\right|=1$ , which can be break down into each quadrant .

For Ist quadrant,  $\left|x\right|=x,\left|y\right|=y$

i.e.,  $x+y=1$ - (1)

Similarly for IInd, IIIRd nad IVth quadrant

$-x+y=1$ - (2)

$-x-y=1$ - (3)

$x-y=1$ - (4)

We draw the above focus lines on a graph and find the area enclosed which is a square.

$\therefore$ Required area $\left(?ABCD\right)=4*area\left(?AOB\right)$ .

$\begin{array}{l}=4*{\displaystyle \underset{0}{\overset{1}{\int }}y}dx\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}\left(-x\right)}dx\\ =4{\left[x-\frac{x^2}{2}\right]}_0^1\\ =4\left[1-\frac{1}{2}\right]\\ =4*\frac{1}{2}\\ =2uni{t}^2\end{array}$

The given equation of the parabola is $x^2=y$ ---------(1)

and that the line is $y=x+2$ --------------(2)

Solving (1) and (2) for x and y

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

When $x=-1,y={\left(-1\right)}^2=1$

And $x=2,y=2^2=4$

$\therefore$ The point of intersection of the parabola and the lines $A(-1,1)\&B(2,4)$

Hence the required area enclosed region is $ea(AOBA)=area(DABCD)-area(DAOBCD)$

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{line}dx-}{\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{curve}dx}\\ ={\displaystyle \underset{-1}{\overset{2}{\int }}\left(x+2\right)dx}-{\displaystyle \underset{-1}{\overset{2}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^2}{2}2x\right]}_{-1}^2-{\left[\frac{x^3}{3}\right]}_{-1}^2\end{array}$

$\begin{array}{l}=\left[\left(\frac{2^2}{2}2.2\right)-\left(\frac{{\left(-1\right)}^2}{2}+2\left(-1\right)\right)\right]-\left[\frac{2^3}{3}-\frac{{\left(-1\right)}^3}{3}\right]\\ =\left[2+4-\frac{1}{2}+2\right]-\left[\frac{8+1}{3}\right]=\frac{15}{2}-3=\frac{9}{2}uni{t}^2\\ \end{array}$