Class 12th

38. (C) Tm

In the periodic table, actinides lie from atomic number 90 to 103. 

Thulium is not an actinide; it is a lanthanide. 

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{0}{\int }}ydx}={\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx}\\ ={\left[\frac{-x^3}{-3}\right]}_{-1}^0\\ =\frac{1}{3}\end{array}$

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.

37. (D) Mn²⁺ acts as auto catalyst

Mn²⁺  formed in the reaction acts as an autocatalyst. Auto catalyst is a compound that is one of the products that are formed in the reaction itself catalyses the reaction.

Reaction: 2MnO₄^-  +  16H⁺  +  5C₂O₄^2- → 2Mn₂ +  +  10CO₂ +  8H₂O

55.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: Hydrometallurgy is used to extract copper from low-grade ore. Hydrometallurgy entails dissolving the ore in a suitable reagent and then precipitating it. In this method, more electropositive metal is used, allowing pure metal to be displaced.

Given is $y=x^3$ and the ines $x=-2\&x=1$

For $y=x^3$

$\begin{array}{l}area\left(OAB\right)={\displaystyle \underset{0}{\overset{1}{\int }}ydx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}\\ area\left(ODC\right)={\displaystyle \underset{-2}{\overset{0}{\int }}ydx}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}{x}^{3}dx}\\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|=\left|\left[\frac{0^4}{4}-\frac{{\left(-2\right)}^4}{4}\right]\right|=4\end{array}$

$\therefore$ Total area of the bounded region $=\frac{1}{4}+4$

$=\frac{17}{4}uni{t}^2$

36.  (A)i, ii

Disproportionation/redox reaction is a reaction in which one compound is oxidised as well as reduced at the same time.

(A) Cu⁺  ?  Cu²⁺  +  Cu 

  ( + 1)       ( + 2)        (0) 

(B) 3MnO₄^- +  4H⁺  ?  2MnO₄^- +  MnO₂ +  2H₂O 

  ( + 6)                          ( + 7)           ( + 4) 

54.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: The zone refining method is very useful for producing high-purity semiconductors and other metals, such as germanium.

35. (B)+3

Out of all +3 is the most common oxidation state for all the lanthanide elements.

The given equation of curve $y^2\le 4x$ i.e,  $y^2=4x$ - (1) is a parabola and

$4x^2+4y^2\le 9\Rightarrow x^2+y^2\le \frac{9}{4}\Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2$ - (2) is a circle

With centre (0,0)and radius 

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

$\begin{array}{l}{x}^2+4x=\frac{9}{4}\\ =x^2+4x-\frac{9}{4}=0\\ =4x^2+16x-9=0\end{array}$