Class 12th

The equation of the curve is $y=x^2+2\Rightarrow x^2=y-2$ - (1) and

lines are

$y=x$ - (2)

$x=0$ - (3)

$x=3$ - (4)

Equation (1)is a parabola with vertex (0,2)

Equation (2)is a straight line passing origin with shape = $tan\theta =1=\theta =45^{?}$

$\therefore$ The required area enclosed OBCDO = area (ODCAO)-area (OBAO)

24. As an effect of lanthanide contraction, the second and third rows of transition elements resemble each other. For example, zirconium and hafnium have similar radius i.e. 160pm and 159pm 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.

43. Option (i) and (ii)

In the vapor phase refining method, the metal is converted into its volatile compound and then is collected elsewhere. This involves two techniques:

1. Mond Process for refining Nickel: in this process, a volatile complex, nickel tetracarbonyl is formed when nickel is heated with a stream of carbon monoxide.

2. Van Arkel Method for refining Zirconium or Titanium: this method is basically used for removal of oxygen and nitrogen present as impurities in the Zr or Ti metal. These are heated in an evacuated metal with iodine. As iodine is more covalent than these metals, it volatilizes.

The equation of the given circle is

$=x^2+y^2=1$ - (1)

$=\; (x-1)2^{}+y^2=1$ - (1) - (2)

Equation (1) is a circle with centre 0 (0,0) and radius 1. Equation (2) is a circle with centre c (1,0) and radius 1.

Solving (1) and (2)

23. KMnO₄ act as an oxidising agent however it's activity as oxidising agent is influenced by the pH of the solution i.e. acidic, basic or neutral solution. 

K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O

(Orange)                      (yellow)

 2K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇ + K₂SO₄ + H₂O

(yellow)                                       (orange) 

22. This is due to the inter conversion of dichromate (orange) to chromate ion (yellow). 

Cr₂O₇^2-   CrO₄^2-  

(orange)     (yellow) 

42. Option (i) and (iv)

(i) Carbon monoxide is the primary reducing agent in the furnace.

(ii) This is an endothermic reaction in which heat is absorbed from the furnace. As a result, it is critical not to add too much limestone, as this will cool the furnace. Calcium oxide is a basic oxide that reacts with acidic oxides in the rock, such as silicon dioxide. Calcium silicate is formed when calcium oxide reacts with silicon dioxide.

The equation given circle is

$\begin{array}{l}4x^2+4y^2=9\\ \Rightarrow x^2+y^2=\frac{9}{4}\\ \Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2\end{array}$

i.e, centre (0,0), radius $r=\frac{3}{2}$

since $x^2=4y$ intersect the circle

we can put $x^2=4y$ in $x^2+y^2={\left(\frac{3}{2}\right)}^2$

$\begin{array}{l}\left(4y\right)+y^2=\frac{9}{4}\\ \Rightarrow y^2+4y-\frac{9}{4}=0\\ \Rightarrow 4y^2+16y-9=0\\ \Rightarrow 4y^2+18y-2y-9=0\end{array}$

$\begin{array}{l}\Rightarrow 2y\left(2y+9\right)-1\left(2y+9\right)=0\\ \Rightarrow \left(2y+9\right)\left(2y-1\right)=0\\ \Rightarrow y=\frac{-9}{2}\&y=\frac{1}{2}\\ \end{array}$

$y=\frac{-9}{2},x^2=4*\left(\frac{-9}{2}\right)=-18$ which is not possible or $x^2$ cannot be (-)ve

21. 2MnO₄^-  +  16H⁺ +  5C₂O₄^2-  → 2Mn₂ +  +  8H₂O  +  10CO₂

KMnO₄ oxidises the oxalic acid to CO₂ and reduces itself to Mn²⁺  state. Mn²⁺ is colourless that's why it seems that KMnO₄ has been disappeared. 

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.