Class 12th

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.

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.