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49. Option (i) A (3) B (4) C (2) D (1)

Sapphire is a gemstone containing Co.

Hence, option (A) from column I is matched with option (3) from column II. The Sphalerite single is ZnS.

Hence, option (B) from column I is matched with option (4) from column II. NaCN is also used as a depressant.

Hence, option (C) from column I is matched with option (2) from column II. Al2O3 is also called corundum.

Hence, option (D) from column I is matched with option (1) from column II.

30. (A) Cu (II) is more stable .

Electronic configuration of Cuis [Ar] 3d¹⁰ 4s¹

Cu (I) - [Ar] 3d¹⁰ 4s0

Cu (II)- [Ar] 3d⁹ 4s0

Despite the fact that Cu (I)  has fully filled 3d-orbital but Cu (II) is more stable than Cu (I) due to the greater effective nuclear charge of Cu (II) as nucleus has to hold 17 electrons rather than 18 electrons like in Cu (I).

The equation of the parabola is $y^2=4a^2$ -----------(1)

and that of line is $y=mx$ ------(2)

The Point of intersection of(1)and (2) is given by

$\begin{array}{l}{\left(mx\right)}^2=4ax\\ \Rightarrow m^2{x}^2-4ax=0\\ \Rightarrow x\left(m^{2}x-4a\right)=0\\ \Rightarrow x=0\&x=\frac{4a}{m^2}\end{array}$

For, $x=0,y^2=4a\times 0\to y=0$ i.e, O(0,0)

For, $x=\frac{4a}{m^2},y^2=4a\times \frac{4a}{m^2}\to y=\frac{4a}{m}$ (in first quadrant)

i.e, $A\left(\frac{4a}{m^2},\frac{4a}{m}\right)$

Hence, the required area enclosed by the curve and the lines is

$area(DACO)=area(OCABO)-area(?OAB)$

29.  (B) 26

Positive oxidation states indicate the loss of electrons from the atom. If X is in +3 oxidation state, then three electrons have been removed from it. Find the atomic number of the parent atom, we will add three in the given electronic number. i.e.

X³⁺  (Z = 23)  =  [Ar] 3d⁵

X  =  23 + 3  = 26  

To get into Vikram Sarabhai Institute of Engineering & Technology for MBA course, candidates must meet the eligibility criteria set by the institute. Students must pass graduation with a minimum of 50% aggregate (45% aggregate for ST/SC).  VSIOET offers full-time MBA courses of two years' duration. Students can get into this course based on entrance-exam. 

The given equation of the curve is $y=sinx$

$\therefore$ The required area bounded by the curve

$\begin{array}{l}={\displaystyle \underset{0}{\overset{\pi }{\int }}y}dx+{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}y}dx\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}sinxdx+{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}sinxdx}}\\ =\left|{\left[-cosx\right]}_0^{\pi }\right|+\left|{\left[-cosx\right]}_0^{\pi }\right|\\ =|-\left[cosx-cos\theta \right]|+|-\left[cos2\pi -cos\pi \right]|\\ =|-\left[-1-1\right]|+|-\left[1+1\right]|\\ =\left|2\right|+|-2|=2+2=4uni{t}^2\end{array}$

Given equation of lines is $y=|x+3|$ -------(1)

The point (x,y)satisfying (1)are

Hence plotting the above in graph we get     

Now, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}$

We know that, $\begin{array}{l}{\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}\\ y=|x+3|=\left\{\begin{array}{l}x+3,if,x+3\ge 0\Rightarrow x\ge -3\\ -\left(x+3\right),if,x+3\angle 0\Rightarrow x\angle -3\end{array}\right\}\end{array}$

So, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}-\left(x+3\right)dx+}{\displaystyle \underset{-3}{\overset{0}{\int }}\left(x+3\right)dx}$

$\begin{array}{l}=-{\left[\frac{x^2}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0\\ =-\left\{\left[\frac{{\left(-3\right)}^2}{2}+3\left(-3\right)\right]-\left[\frac{{\left(-6\right)}^2}{2}+3\left(-6\right)\right]\right\}+\left\{\left[\frac{0^2}{2}+3\times 0\right]-\left[\frac{{\left(-3\right)}^2}{2}+3\times \left(-3\right)\right]\right\}\\ =-\left\{\frac{9}{2}-9-\frac{36}{2}+18\right\}+\left\{-\frac{9}{2}+9\right\}\\ =-\frac{9}{2}+9+18-18-\frac{9}{2}+9\\ =9uni{t}^2\end{array}$

To be eligible for the MBA programme at Mohamed Sathak College of Arts and Science, candidates must hold a three-year Bachelor's degree in any discipline from a recognised university or an equivalent professional qualification approved by the University of Madras. Selection is based on scores from entrance exams like TANCET, CET, GMAT, or CAT, followed by the university's selection process.

Given curve is $y=4x^2\Rightarrow x^2=\frac{1}{4}y$ - (1)

$x=0$  i.e, y-axis and y=4 and y=1

Hence, the required area in Ist quadrant i.e, area ABCD = ${\displaystyle \underset{y=1}{\overset{y=4}{\int }}xdy}$

No, it is not mandatory to pass the model exam to get the Karnataka 2nd PUC hall ticket. Model exams are meant for students to be familiar with the board exam. Clearing the model exams is not a criteria for obtaining the Karnataka 2nd PUC hall ticket.

Mohamed Sathak College of Arts and Science offers a two-year MBA with diverse specialisations across four semesters. Students can choose from Financial Management, Hospital Management, Human Resource Management, Infrastructure Management, and Supply Chain Management. These specialisations cater to various industry needs, helping students gain in-depth knowledge and prepare for a dynamic career in their chosen field.

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

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

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

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

Where, $x=0,y=0^2=0$

And when $x=1,y=1^2=1$

$\therefore$ The point of intersection of the parabola $y=x^2$ and the line $y=x$

Is O(0,0) and B (1,1)

Hence, area between the curve and the line is

$area\left(DCAO\right)=area\left(?OAB\right)-area\left(OABO\right)$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{line}}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{curve}}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}xdx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}}\\ ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\end{array}$

$\begin{array}{l}=\frac{1}{2}-\frac{1}{3}\\ =\frac{3-2}{6}=\frac{1}{6}uni{t}^2\end{array}$

48. Correct code (i) A (4) B (2) C (3) D (1)

The cyanide process is used in the extraction of Au.

Hence, option (A) from column I is matched with option (4) from column II. Froth floatation process is used in the dressing of ZnS.

Hence, option (B) from column I is matched with option (2) from column II. Electrolytic reduction is used in the extraction of AI.

Hence, option (C) from column I is matched with option (3) from column II. Zone refining is used to get ultrapure Ge.

Hence, option (D) from column I is matched with option (1) from column II.

(i) Given of curve is $y=x^2\Rightarrow x^2=y$ and the equation are $x=1\&x=2.$

$\therefore$ Area enclosed

$\begin{array}{l}={\displaystyle \underset{x=1}{\overset{x=2}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{2}{\int }}{x}^2}dx\\ ={\left[\frac{x^3}{3}\right]}_1^2=\frac{2^3}{3}-\frac{1^3}{3}\\ =\frac{8-1}{3}=\frac{7}{3}uni{t}^2\end{array}$

(ii) Given equation of curve is $y=x^4$ and the lines are $x=1\&x=5$

So, area enclosed

$\begin{array}{l}={\displaystyle \underset{1}{\overset{5}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{5}{\int }}{x}^4}dx\\ ={\left[\frac{x^5}{5}\right]}_1^5=\left(\frac{5^5-1^5}{5}\right)\\ =\frac{\left(3125-1\right)}{5}\\ =\frac{3124}{5}=624.8uni{t}^2\end{array}$

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Hall ticket for the Karnataka 2nd PUC board exam is issued by the Karnataka School Examination and Assessment Board (KSEAB) to which you are enrolled via your school. So your school cannot deny you the admit card even if you have failed to pass in their pre-boards exam. If they still goes on to deny you the hall ticket you can straight away write it to KSEAB asking them to issue your PUC hall ticket in duplicate as your school has failed to give you.

Mithibai College BSc cutoff 2025 stood at 45.67 across multiple specialisations for the students belonging to the Home State quota. Aspirants seeking admission must target to secure a cutoff percentage of 50 and above to improve their chances of securing a seat at Mithibai College for BSc admission. Refer to the table below to view the BSc specialisation-wise first-round cutoff percentages: