NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

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# NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

Edited By Ramraj Saini | Updated on Sep 14, 2023 10:14 PM IST | #CBSE Class 12th

## NCERT Application Of Integrals Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals are discussed here. These NCERT solutions are created by expert team at Creers360 keeping in mind of latest syllabus of CBSE 2023-24. In geometry, you must have learned some formulas to calculate areas of simple geometrical figures like triangles, rectangles, trapeziums, circles, etc. But how do you calculate areas enclosed by curves? In this article, you will get NCERT solutions for class 12 maths chapter 8 application of integrals. This article also includes application of integrals class 12 solutions. Important topics that are going to be discussed in this chapter 8 class 12 maths are the area under simple curves, the area between lines and arcs of circles, parabolas, and ellipses. Interested students can find all NCERT Solutions for Class 12 Maths in one place

In the applications of integrals class 12 ncert solutions, questions from all these topics are covered. In this composition of Class 12 Maths Chapter 8 NCERT solutions application of integrals, you will learn some important applications of integrals class 12. If you are interested to check all NCERT solutions from classes 6 to 12 in a single place, which will help you to learn CBSE maths and science. Here you will get NCERT solutions for class 12.

## Application Of Integrals Class 12 NCERT Solutions - Important Formulae

>> Area Enclosed by a Curve and Lines: The area enclosed by the curve y = f(x), the x-axis, and the lines x = a and x = b (where b > a) is given by the formula:

Area = ∫[a, b]y.dx = ∫[a, b]f(x).dx

>> Area Bounded by Curve and Horizontal Lines: The area of the region bounded by the curve x = φ(y) as its y-axis and the lines y = c and y = d is given by the formula: Area = ∫[c, d]x.dy = ∫[c, d]φ(y).dy

>> Area Between Two Curves and Vertical Lines: The area enclosed between two given curves y = f(x) and y = g(x), and the lines x = a and x = b is given by the formula:

Area = ∫[a, b][f(x) - g(x)].dx (Where f(x) ≥ g(x) in [a, b])

>> Area Between Curves with Different Intervals: If f(x) ≥ g(x) in [a, c] and f(x) ≤ g(x) in [c, b], where a < c < b, then the resultant area between the curves is given as:

Area = ∫[a, c][f(x) - g(x)].dx + ∫[c, b][g(x) - f(x)].dx

Free download Application Of Integrals Class 12 NCERT Solutions for CBSE Exam.

## NCERT Application Of Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT class 12 maths chapter 8 question answer Exercise: 8.1

Area of the region bounded by the curve $\dpi{100} y^2=x$ and the lines $\dpi{100} x=1,x=4$ and the $\dpi{100} x$ -axis in the first quadrant

Area = $\int_{1}^{4}ydy = \int_{1}^{4}\sqrt{x}dx$

$\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_1 = \frac{2}{3}\left [ (4)^\frac{3}{2}- (1)^\frac{3}{2} \right ]$

$= \frac{2}{3}\left [ 8 -1 \right ]$

= 14/3 units

Area of the region bounded by the curve $\dpi{100} y^2=9x,x=2,x=4$ and the $\dpi{100} x$ -axis in the first quadrant

Area = $\int_{2}^{4}ydy = \int_{2}^{4}\sqrt{9x}dx = 3\int_{2}^{4}\sqrt{x}dx$

$3\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_2 = 3.\frac{2}{3}\left [ (4)^\frac{3}{2}- (2)^\frac{3}{2} \right ]$

$= 2\left [ 8 -2\sqrt2 \right ]$

$= \left [ 16 -4\sqrt2 \right ]$ units

The area bounded by the curves $\dpi{100} x^2=4y,y=2,y=4$ and the $\dpi{100} y$ -axis in the first quadrant is ABCD.

$= \int^4_{2} 2\sqrt{y} dy$

$= 2\int^4_{2} \sqrt{y} dy$

$=2\left \{ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right \}^4_{2}$

$= \frac{4}{3}\left \{ (4)^{\frac{3}{2}}-(2)^{\frac{3}{2}} \right \}$

$= \frac{4}{3} \left \{ 8 -2\sqrt 2 \right \}$

$= \left \{ \frac{32-8\sqrt 2}{3} \right \}\ units.$

The area bounded by the ellipse : $\dpi{100} \frac{x^2}{16}+\frac{y^2}{9}=1.$

Area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^4_{0} y dx$

$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$

$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$

$= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}$

$= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]$

$= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]$

$= \frac{3}{4}\left [ 4\pi \right ] =3\pi$

Therefore the area bounded by the ellipse will be $= 4\times {3\pi} = 12\pi\ units.$

The area bounded by the ellipse : $\dpi{100} \small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^2_{0} y dx$

$= \int^2_{0}3\sqrt{1-\frac{x^2}{4}} dx$

$= \frac{3}{2}\int^2_{0}\sqrt{4-x^2} dx$

$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$

$= \frac{3}{2}\left [ \frac{2\pi}{2} \right ]$

$= \frac{3\pi}{2}$

Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi\ units.$

The area of the region bounded by $\dpi{100} \small x=\sqrt{3}y$ and $\dpi{100} \small x^2+y^2=4$ is ABC shown:

The point B of the intersection of the line and the circle in the first quadrant is $(\sqrt3,1)$ .

Area ABC = Area ABM + Area BMC where, M is point in x-axis perpendicular drawn from the line.

Now,area of $ABM = \frac{1}{2}\times AM\times BM = \frac{1}{2}\times \sqrt{3}\times 1 =\frac{\sqrt3}{2}$ ............(1)

and Area of $BMC = \int^2_{\sqrt{3}} ydx$

$= \int^2_{\sqrt3} \sqrt{4-x^2} dx$

$= \left [ \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{\sqrt3}$

$= \left [ 2\times\frac{\pi}{2}-\frac{\sqrt3}{2}\sqrt{4-3}-2\sin^{-1}\left ( \frac{\sqrt3}{2} \right ) \right ]$

$= \left [ \pi - \frac{\sqrt3\pi}{2}-2\frac{\pi}{3} \right ]$

$= \left [ \pi-\frac{\sqrt3}{2}-\frac{2\pi}{3} \right ]$

$= \left [ \frac{\pi}{3}-\frac{\sqrt3}{2} \right ]$ ..................................(2)

then adding the area (1) and (2), we have then

The Area under ABC $= \frac{\sqrt3}{2} +\frac{\pi}{3}-\frac{\sqrt3}{2} = \frac{\pi}{3}\ units.$

we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC
Area of ABC = $\int_{\frac{a}{\sqrt2}}^{a} ydx= \int_{\frac{a}{\sqrt2}}^{a} \sqrt{a^2-x^2}dx= \left [ \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right ]^{a}_\frac{a}{\sqrt2}\\ \\$
$=\left [ \frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{a}{a}- \frac{a}{2\sqrt2}\sqrt{a^2-(\frac{a}{\sqrt2})^2}-\frac{a^2}{2}\sin^{-1}\frac{a}{a\sqrt2}\right ]$
$=\left [ \frac{a}{2}\sqrt{0}+\frac{a^2}{2}\sin^{-1}1- \frac{a}{2\sqrt2}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt2}\right ]$
$=\left [ 0+\frac{a^2}{2}\frac{\pi}{2}- \frac{a^2}{4}-\frac{a^2}{2}\frac{\pi}{4}\right ]$
$=\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )$
Area of ABCD = 2 X Area of ABC
$=2\times\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )= \frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$
Therefore, the area of the smaller part of the circle is $\frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$

we can clearly see that given area is symmetrical about x - axis
It is given that
Area of OED = Area of EFCD
Area of OED = $\int_{0}^{a}ydx = \int_{0}^{a}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^{a}= \frac{a^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2a^{\frac{3}{2}}}{3}$
Area of EFCD = $\int_{a}^{4}ydx = \int_{a}^{4}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_a^{4}= \frac{4^{\frac{3}{2}}-a^\frac{3}{2}}{\frac{3}{2}} = \frac{2(8-a^\frac{3}{2})}{3}=\frac{2(8-a^\frac{3}{2})}{3}$
Area of OED = Area of EFCD
$\frac{2a^{\frac{3}{2}}}{3}= \frac{2(8-a^{\frac{3}{2}})}{3}\\ \\ 2a^\frac{3}{2} = 8\\ a^\frac{3}{2} = 4\\ a = (4)^\frac{2}{3}$
Therefore, the value of a is $a = (4)^\frac{2}{3}$

We can clearly see that given area is symmetrical about y-axis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of $y=x^2 \ and \ y = |x|$ is (1 , 1) and (-1 , 1)
Now,
Area od OCAO = Area OAM - Area of OCMO
Area of OAM = $\frac{1}{2}.OM.AM = \frac{1}{2}.1.1 = \frac{1}{2}$
Area of OCMO = $\int_{0}^{1}ydx= \int_{0}^{1}x^2dx= \left [ \frac{x^3}{3} \right ]_{0}^{1}= \frac{1}{3}$
Therefore,
Area od OCAO $=\frac{1}{2}- \frac{1}{3}= \frac{1}{6}$
Now,
Area of the region bounded by the parabola $\dpi{100} \small y=x^2$ and $\dpi{100} \small y=|x|$ is = 2 X Area od OCAO $=2\times \frac{1}{6} = \frac{1}{3}$ Units

Points of intersections of $y = x^2 \ and \ x = 4y-2$ is
$A\left ( -1,\frac{1}{4} \right ) \ and \ B(2,1)$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO- Area of OMBO

Area of OMBCO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{0}^{2}+\left [ \frac{x}{2} \right ]_{0}^{2}= \frac{4}{8}+\frac{2}{2}=\frac{3}{2}$

Area of OMBO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{0}^{2}= \frac{8}{12}= \frac{2}{3}$

Area of OBCO = Area of OMBCO- Area of OMBO
$= \frac{3}{2}-\frac{2}{3}= \frac{5}{6}$
Similarly,
Area of OCAO = Area of OCALO - Area of OALO

Area of OCALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{-1}^{0}+\left [ \frac{x}{2} \right ]_{-1}^{0}=- \frac{1}{8}-\frac{(-1)}{2}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$

Area of OALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{-1}^{0}= -\frac{(-1)}{12}= \frac{1}{12}$

Area of OCAO = Area of OCALO - Area of OALO
$=\frac{3}{8}- \frac{1}{12}= \frac{9-2}{24}= \frac{7}{24}$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
$=\frac{5}{6}+ \frac{7}{24}= \frac{20+7}{24}= \frac{27}{24} = \frac{9}{8}$

Therefore, area bounded by the curve $\dpi{100} \small x^2=4y$ and the line $\dpi{100} \small x=4y-2$ is $\frac{9}{8} \ units$

The combined figure of the curve $y^2=4x$ and $x=3$

The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2 $\times$ Area of OAB
$\\=2[\int_{0}^{3}ydx]\\ =2\int^3_02\sqrt{x}dx\\ =4[\frac{x^{3/2}}{3/2}]^3_0\\ =8\sqrt{3}$
therefore the required area is $8\sqrt{3}$ units.

Question: 12 Choose the correct answer in the following

The area bounded by circle C(0,0,4) and the line x=2 is

The required area = area of OAB
$\int^2_0ydx = \int^2_0\sqrt{4-x^2}dx$
$\\=[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}]^2_0\\ =2(\pi/2)\\ =\pi$

Question: 13 Choose the correct answer in the following.

The area bounded by the curve $y^2=4x$ and y =3

the required area = OAB =
$\\\int ^3_0xdy\\ =\int ^3_0\frac{y^2}{4}dy\\ =\frac{1}{4}.[\frac{y^3}{3}]^3_0\\ =\frac{9}{4}$

NCERT application of integrals class 12 solutions Exercise: 8.2

The area bounded by the circle $\dpi{100} \small 4x^2+4y^2=9$ and the parabola $\dpi{100} \small x^2=4y$ .

By solving the equation we get the intersecting point $D(-\sqrt{2},\frac{1}{2})$ and $B(\sqrt{2},\frac{1}{2})$
So, the required area (OBCDO)=2 times the area of (OBCO)
Draw a normal on the x-axis (M = $\sqrt{2},0$ )

Thus the area of OBCO = Area of OMBCO - Area of OMBO

$\\\int_{0}^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-\int_{0}^{\sqrt{2}}{\frac{x^2}{4}}dx\\ =\frac{1}{2}\int_{0}^{\sqrt{2}}\sqrt{9-4x^2}-\frac{1}{4}\int_{0}^{\sqrt{2}}x^2dx\\ =\frac{1}{4}[x\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}]_0^{\sqrt{2}}-\frac{1}{4}[\frac{x^3}{3}]_0^{\sqrt{2}}\\ =\frac{1}{4}[\sqrt{2}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}]-\frac{1}{12}(\sqrt{2})^3\\ =\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\\ =\frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})$
S0, total area =
$\\=2\times \frac{1}{2}(\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3})\\ =\frac{\sqrt{2}}{6}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$

Given curves are $\dpi{100} \small (x-1)^2+y^2=1$ and $\dpi{100} \small x^2+y^2=1$

Point of intersection of these two curves are

$A = \left ( \frac{1}{2},\frac{\sqrt3}{2} \right )$ and $B = \left ( \frac{1}{2},-\frac{\sqrt3}{2} \right )$

We can clearly see that the required area is symmetrical about the x-axis

Therefore,

Area of OBCAO = 2 $\times$ Area of OCAO

Now, join AB such that it intersects the x-axis at M and AM is perpendicular to OC

Coordinates of M = $\left ( \frac{1}{2},0 \right )$

Now,

Area OCAO = Area OMAO + Area CMAC

$=\left [ \int_{0}^{\frac{1}{2}}\sqrt{1-(x-1)^2}dx +\int_{\frac{1}{2}}^{1}\sqrt{1-x^2}dx \right ]$
$=\left [ \frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac{1}{2}\sin^{-1}(x-1) \right ]_{0}^{\frac{1}{2}}+\left [ \frac{x}{2}+\frac{1}{2}\sin^{-1}x \right ]_{\frac{1}{2}}^{1}$

$=\left [- \frac{1}{4}\sqrt{1-(-\frac{1}{2})^2}+\frac{1}{2}\sin^{-1}(\frac{1}{2}-1)-0-\frac{1}{2}\sin^{-1}(-1) \right ]+\left [ 0+\frac{1}{2}\sin^{-1}(1)- \frac{1}{4}\sqrt{1-\left ( \frac{1}{2} \right )^2}-\frac{1}{2}\sin^{-1}\left ( \frac{1}{2} \right )\right ]$
$=\left [ -\frac{\sqrt3}{8}+\frac{1}{2}\left ( -\frac{\pi}{6} \right )-\frac{1}{2}\left ( -\frac{\pi}{2} \right ) \right ]+\left [ \frac{1}{2}\left ( \frac{\pi}{2} \right ) -\frac{\sqrt3}{8}-\frac{1}{2}\left ( \frac{\pi}{6} \right )\right ]$
$= \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]+\left [ \frac{\pi}{6}-\frac{\sqrt3}{8} \right ]$
$=2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
Now,
Area of OBCAO = 2 $\times$ Area of OCAO

$=2\times 2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
$=\frac{2\pi}{3}-\frac{\sqrt3}{2}$

Therefore, the answer is $\frac{2\pi}{3}-\frac{\sqrt3}{2}$

The area of the region bounded by the curves,

$\dpi{100} \small y=x^2+2,y=x,x=0$ and $\dpi{100} \small x=3$ is represented by the shaded area OCBAO as

Then, Area OCBAO will be = Area of ODBAO - Area of ODCO

which is equal to

$\int_0^3(x^2+2)dx - \int_0^3x dx$

$= \left ( \frac{x^3}{3}+2x \right )_0^3 -\left ( \frac{x^3}{2} \right )_0^3$

$= \left [ 9+6 \right ] - \left [ \frac{9}{2} \right ] = 15-\frac{9}{2} = \frac{21}{2}units.$

So, we draw BL and CM perpendicular to x-axis.

Then it can be observed in the following figure that,

$Area(\triangle ACB) = Area (ALBA)+Area(BLMCB) - Area (AMCA)$

We have the graph as follows:

Equation of the line segment AB is:

$y-0 = \frac{3-0}{1+1}(x+1)$ or $y = \frac{3}{2}(x+1)$

Therefore we have Area of $ALBA$

$=\int_{-1}^1 \frac{3}{2}(x+1)dx =\frac{3}{2}\left [ \frac{x^2}{2}+x \right ]_{-1}^1$

$=\frac{3}{2}\left [ \frac{1}{2}+1-\frac{1}{2}+1 \right ] =3units.$

So, the equation of line segment BC is

$y-3 = \frac{2-3}{3-1}(x-1)$ or $y= \frac{1}{2}(-x+7)$

Therefore the area of BLMCB will be,

$=\int_1^3 \frac{1}{2}(-x+7)dx =\frac{1}{2}\left [ -\frac{x^2}{2}+7x \right ]_1^3$

$= \frac{1}{2}\left [ -\frac{9}{2}+21+\frac{1}{2}-7 \right ] =5units.$

Equation of the line segment AC is,

$y-0 = \frac{2-0}{3+1}(x+1)$ or $y = \frac{1}{2}(x+1)$

Therefore the area of AMCA will be,

$=\frac{1}{2}\int_{-1}^3 (x+1)dx =\frac{1}{2}\left [ \frac{x^2}{2}+x \right ]_{-1}^3$

$=\frac{1}{2}\left ( \frac{9}{2}+3-\frac{1}{2}+1 \right ) = 4units.$

Therefore, from equations (1), we get

The area of the triangle $\triangle ABC =3+5-4 =4units.$

The equations of sides of the triangle are $y=2x+1, y =3x+1,\ and\ x=4$ .

ON solving these equations, we will get the vertices of the triangle as $A(0,1),B(4,13),\ and\ C(4,9)$

Thus it can be seen that,

$Area (\triangle ACB) = Area (OLBAO) -Area (OLCAO)$

$= \int_0^4 (3x+1)dx -\int_0^4(2x+1)dx$

$= \left [ \frac{3x^2}{2}+x \right ]_0^4 - \left [ \frac{2x^2}{2}+x \right ]_0^4$

$=(24+4) - (16+4) = 28-20 =8units.$

So, the smaller area enclosed by the circle, $x^2+y^2 =4$ , and the line, $x+y =2$ , is represented by the shaded area ACBA as

Thus it can be observed that,

Area of ACBA = Area OACBO - Area of $(\triangle OAB)$

$=\int_0^2 \sqrt{4-x^2} dx -\int_0^2 (2-x)dx$

$= \left ( \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}{\frac{x}{2}} \right )_0^2 - \left ( 2x -\frac{x^2}{2} \right )_0^2$

$= \left [ 2.\frac{\pi}{2} \right ] -[4-2]$

$= (\pi -2) units.$

Thus, the correct answer is B.

The area lying between the curve, $\dpi{100} \small y^2=4x$ and $\dpi{100} \small y=2x$ is represented by the shaded area OBAO as

The points of intersection of these curves are $O(0,0)$ and $A (1,2)$ .

So, we draw AC perpendicular to x-axis such that the coordinates of C are (1,0).

Therefore the Area OBAO = $Area(\triangle OCA) -Area (OCABO)$

$=2\left [ \frac{x^2}{2} \right ]_0^1 - 2\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^1$

$=\left | 1-\frac{4}{3} \right | = \left | -\frac{1}{3} \right | = \frac{1}{3} units.$

Thus the correct answer is B.

NCERT application of integrals class 12 solutions Miscellaneous: Exercise

The area bounded by the curve $\dpi{100} \small y=x^2,x=1,x=2$ and $\dpi{100} \small x$ -axis

The area of the required region = area of ABCD
$\\=\int_{1}^{2}ydx\\ =\int_{1}^{2}x^2dx\\ =[\frac{x^3}{3}]_1^2\\ =\frac{7}{3}$
Hence the area of shaded region is 7/3 units

The area bounded by the curev $\dpi{100} \small y=x^4,x=1,x=5$ and $\dpi{100} \small x$ -axis

The area of the required region = area of ABCD
$\\=\int_{1}^{5}ydx\\ =\int_{1}^{2}x^4dx\\ =[\frac{x^5}{5}]_1^2\\ =625-\frac{1}{5}\\ =624.8$
Hence the area of the shaded region is 624.8 units

the area between the curves $\dpi{100} \small y=x$ and $\dpi{100} \small y=x^2$ .

The curves intersect at A(1,1)
Draw a normal to AC to OC(x-axis)
therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO)
$\\=\int_{0}^{1}xdx-\int_{0}^{1}x^2dx\\ =[\frac{x^2}{2}]_0^1-[\frac{x^3}{3}]_0^1\\ =1/2-1/3\\ =\frac{1}{6}$
Thus the area of shaded region is 1/6 units

the area of the region lying in the first quadrant and bounded by $\dpi{100} \small y=4x^2,x=0,y=1$ and $\dpi{100} \small y=4$ .

The required area (ABCD) =
$\\=\int_{1}^{4}xdy\\ =\int_{1}^{4}\frac{\sqrt{y}}{2}dy\\ =\frac{1}{2}.\frac{2}{3}[y^{3/2}]_1^4\\ =\frac{1}{3}[8-1]\\ =\frac{7}{3}$
The area of the shaded region is 7/3 units

y=|x+3|

the given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

Integral to be evaluated is

$\\\int_{-6}^{0}|x+3|dx\\ =\int_{-6}^{-3}(-x-3)dx+\int_{-3}^{0}(x+3)dx\\ =[-\frac{x^{2}}{2}-3x]_{-6}^{-3}+[\frac{x^{2}}{2}+3x]_{-3}^{0}\\ =(-\frac{9}{2}+9)-(-18+18)+0-(\frac{9}{2}-9)\\ =9$

The graph of y=sinx is as follows

We need to find the area of the shaded region

ar(OAB)+ar(BCD)

=2ar(OAB)

$\\=2\times \int_{0}^{\pi }sinxdx\\ =2\times [-cosx]_{0}^{\pi }\\ =2\times [-(-1)-(-1)]\\ =4$

The bounded area is 4 units.

We have to find the area of the shaded region OBA

The curves y=mx and y 2 =4ax intersect at the following points

$\left ( 0,0 \right )and\left ( \frac{4a}{m^{2}},\frac{4a}{m} \right )$

$\\y^{2}=4ax\\ \Rightarrow y=2\sqrt{ax}$

The required area is

$\\\int_{0}^{\frac{4a}{m^{2}}}(2\sqrt{ax}-mx)\\ =2\sqrt{a}[\frac{2x^{\frac{3}{2}}}{3}]_{0}^{\frac{4a}{m^{2}}}-m[\frac{x^{2}}{2}]_{0}^{\frac{4a}{m^{2}}}\\ =\frac{32a^{2}}{3m^{3}}-\frac{8a^{2}}{m^{3}}\\ =\frac{8a^{2}}{3m^{3}}units$

We have to find the area of the shaded region COB

$\\2y=3x+12\\ \Rightarrow y=\frac{3}{2}x+6\\ 4y=3x^{2}\\ \Rightarrow y=\frac{3x^{2}}{4}$

The two curves intersect at points (2,3) and (4,12)

Required area is

$\\\int_{-2}^{4}(\frac{3}{2}x+6-\frac{3x^{2}}{4})dx\\ =[\frac{3x^{2}}{4}+6x-\frac{x^{3}}{4}]{_{-2}}^{4}\\ =[12+24-16]-[3-12+2]\\ =20-(-7)\\ =27\ units$

We have to find the area of the shaded region

The given ellipse and the given line intersect at following points

$\left ( 0,2 \right )and \left ( 3,0 \right )$

$\\\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\\ y=\frac{2}{3}\sqrt{9-x^{2}}$

Since the shaded region lies above x axis we take y to be positive

$\\\frac{x}{3}+\frac{y}{2}=1\\ y=\frac{2}{3}(3-x)$

The required area is

$\\\frac{2}{3}\int_{0}^{3}\left ( \sqrt{9-x^{2}}-(3-x) \right )dx\\ =\frac{2}{3}[\frac{x}{2}(\sqrt{9-x^{2}})+\frac{9}{2}sin^{-1}\frac{x}{3}-3x+\frac{x^{2}}{2}]_{0}^{3}\\ =\frac{2}{3}\left ( \left [ \frac{9}{2}\times \frac{\pi }{2}-9+\frac{9}{2} \right ]-0 \right )\\ =\frac{2}{3}(\frac{9\pi }{4}-\frac{9}{2})\\ =\frac{3}{2}(\pi -2)units$

The area of the shaded region ACB is to be found

The given ellipse and the line intersect at following points

$\left ( 0,b \right )and\left ( a,0 \right )$

$\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ \Rightarrow y=\frac{b}{a}\sqrt{a^{2}-x^{2}}$

Y will always be positive since the shaded region lies above x axis

$\\\frac{x}{a}+\frac{y}{b}=1\\ \Rightarrow y=\frac{b}{a}(a-x)$

The required area is

$\\\frac{b}{a}\int_{0}^{a}(\sqrt{a^{2}-x^{2}}-(a-x))dx\\ =\frac{b}{a}[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}-ax+\frac{x^{2}}{2}]_{0}^{a}\\ =\frac{b}{a}[(\frac{a^{2}}{2}\times \frac{\pi }{2}-a^{2}+\frac{a^{2}}{2})]\\ =\frac{b}{a}(\frac{\pi a^{2}}{4}-\frac{a^{2}}{2})\\ =\frac{ab}{4}(\pi -2)units$

We have to find the area of the shaded region BAOB

O is(0,0)

The line and the parabola intersect in the second quadrant at (-1,1)

The line y=x+2 intersects the x axis at (-2,0)

$\\ar(BAOB)=ar(BAC)+ar(ACO)\\ =\int_{-2}^{-1}(x+2)dx+\int_{-1}^{0}(x^{2})dx\\ =[\frac{x^{2}}{2}+2x]_{-2}^{-1}+[\frac{x^{3}}{3}]_{-1}^{0}\\ =(\frac{1}{2}-2)-(2-4)+0-(-\frac{1}{3})\\ =\frac{5}{6}\ units$

The area of the region enclosed by the parabola $\dpi{100} \small x^2=y,$ the line $\dpi{100} \small y=x+2$ and the $\dpi{100} \small x$ -axis is 5/6 units.

We need to find the area of the shaded region ABCD

ar(ABCD)=4ar(AOB)

Coordinates of points A and B are (0,1) and (1,0)

Equation of line through A and B is y=1-x

$\\ar(AOB)=\int_{0}^{1}(1-x)dx\\ =[x-\frac{x^{2}}{2}]_{0}^{1}\\ =(1-\frac{1}{2})-0 \\=\frac{1}{2}\ units\\ ar(ABCD)=4ar(AOB)\\ =4\times \frac{1}{2}\\ =2\ units$

The area bounded by the curve $\dpi{100} \small |x|+|y|=1$ is 2 units.

We have to find the area of the shaded region

y=|x|=x

$\\=2\int_{0}^{1}(x-x^{2})dx\\ =2[\frac{x^{2}}{2}-\frac{x^{3}}{3}]{_{0}}^{1} \\=2(\frac{1}{2}-\frac{1}{3})-0\\ =1-\frac{2}{3}\\ =\frac{1}{3}\ units$

The area bounded by the curves is 1/3 units.

Equation of line joining A and B is

$\\\frac{y-0}{x-2}=\frac{5-0}{4-2}\\ y=\frac{5x}{2}-5$

Equation of line joining B and C is

$\\\frac{y-5}{x-4}=\frac{5-3}{4-6}\\ y=9-x$

Equation of line joining A and C is

$\\\frac{y-0}{x-2}=\frac{3-0}{6-2}\\ y=\frac{3x}{4}-\frac{3}{2}$

ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)

$\\ar(ABL)=\int_{2}^{4}(\frac{5x}{2}-5)dx\\ =[\frac{5x^{2}}{4}-5x]_{2}^{4}\\ =(20-20)-(5-10)\\ =5\ units$

$\\ar(LBCM)=\int_{4}^{6}(9-x)dx\\ =[9x-\frac{x^{2}}{2}]_{4}^{6}\\ =(54-18)-(36-8)\\ =8\ units$

$\\ar(ACM)=\int_{2}^{6}(\frac{3x}{4}-\frac{3}{2})dx\\ =[\frac{3x^{2}}{8}-\frac{3x}{2}]_{2}^{6}\\ =(\frac{27}{2}-9)-(\frac{3}{2}-3)\\ =6\ units$

ar(ABC)=8+5-6=7

Therefore the area of the triangle ABC is 7 units.

We have to find the area of the shaded region ABC

ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC)

The lines intersect at points (1,2), (4,3) and (2,0)

$\\x-3y=-5\\ y=\frac{x}{3}+\frac{5}{3}$

$\\ar(ACLM)=\int_{1}^{4}(\frac{x}{3}+\frac{5}{3})dx\\ =[\frac{x^{2}}{6}+\frac{5x}{3}]_{1}^{4}\\ =(\frac{4^{2}}{6}+\frac{5\times 4}{3})-(\frac{1}{6}+\frac{5}{3})\\ =\frac{15}{2}\ units$

$\\2x+y=4\\ y=4-2x$

$\\ar(ALB)=\int_{1}^{2}(4-2x)dx\\ =[4x-x^{2}]_{1}^{2}\\ =(8-4)-(4-1)\\ =1\ unit$

$\\3x-2y=6\\ y=\frac{3x}{2}-3$

$\\ar(BMC)=\int_{2}^{4}(\frac{3x}{2}-3)dx\\ =[\frac{3x^{2}}{4}-3x]_{2}^{4}\\ =(12-12)-(3-6)\\ =3\ units$

$\\ ar(ABC)=\frac{15}{2}-1-3\\ =\frac{7}{2}\ units$

Area of the region bounded by the lines is 3.5 units

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

$\\4x^{2}+4y^{2}=9\\ y=\sqrt{\frac{9}{4}-x^{2}}$

$\\y^{2}=4x\\ y=2\sqrt{x}$

In the first quadrant, the curves intersect at a point $\left ( \frac{1}{2},\sqrt{2} \right )$

$\\\int_{0}^{\frac{1}{2}}\left ( \sqrt{\frac{9}{4}-x^{2}} -2\sqrt{x}\right )dx\\ =[\frac{x}{2}\sqrt{\frac{9}{4}-x^{2}}+\frac{9}{8}sin^{-1}\frac{2x}{3}]{_{0}}^{\frac{1}{2}}-4[\frac{x^{3/2}}{3}]_0^{1/2}\\ =\frac{\sqrt{2}}{4}+\frac{9}{8}sin^{-1}\frac{1}{3}-\frac{\sqrt{2}}{3}$

The total area of the shaded region is-
= Area of half circle - area of the shaded region in the first quadrant

$\\\frac{\pi }{2}\times (\frac{3}{2})^{2}-2\left ( \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{3}+\frac{9}{8}sin^{-1}\frac{1}{3}\right )\\ =\frac{9 }{8}\left ( \pi-2sin^{-1}\frac{1}{3} \right )+\frac{\sqrt{2}}{6}\ units$

Hence the required area

$=\int_{-2}^1 ydx$

$=\int_{-2}^1 x^3dx = \left [ \frac{x^4}{4} \right ]_{-2}^1$

$= \left [ \frac{x^4}{4} \right ]^0_{-2} + \left [ \frac{x^4}{4} \right ]^1_{0}$

$= \left [ 0-\frac{(-2)^4}{4} \right ] + \left [ \frac{1}{4} - 0 \right ]$

$= -4+\frac{1}{4} = \frac{-15}{4}$

Therefore the correct answer is B.

The area of the shaded region is to be found.

Required area =ar(DOC)+ar(DOA)

The region to the left of the y-axis is half of the circle with radius 4 units and centre origin.

Area of the shaded region to the left of y axis is ar(1) = $\frac{\pi 4^{2}}{2}=8\pi\ units$

For the region to the right of y-axis and above x axis

$\\x^{2}+y^{2}=16\\ y=\sqrt{16-x^{2}}$

$\\y^{2}=6x\\ y=\sqrt{6x}$

The parabola and the circle in the first quadrant intersect at point

$\left ( 2,2\sqrt{3} \right )$

Remaining area is 2ar(2) is

$\\ar(2)=\int_{0}^{2}\left ( \sqrt{16-x^{2}}-\sqrt{6x} \right )dx\\ =[ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}(\frac{x}{4})-\frac{2\sqrt{6}}{3}x^{\frac{3}{2}} ]{_{0}}^{2}\\ =[\sqrt{12}+\frac{16}{2}\times \frac{\pi }{6}-\frac{4\sqrt{12}}{3}]\\ =\frac{4\pi }{3}-\frac{2\sqrt{3}}{3}$

Total area of shaded region is

$\\ar(1)+2ar(2)\\ =8\pi +\frac{8\pi}{3}-\frac{4\sqrt{3}}{3}\\ =\frac{4}{3}(8\pi -\sqrt{3})\ units$

Given : $\dpi{100} \small y=\cos x$ and $\dpi{100} \small y=\sin x$

Area of shaded region = area of BCDB + are of ADCA

$=\int_{\frac{1}{\sqrt{2}}}^{1}x dy +\int_{1}^{\frac{1}{\sqrt{2}}}x dy$

$=\int_{\frac{1}{\sqrt{2}}}^{1} cos^{-1} y .dy +\int_{1}^{\frac{1}{\sqrt{2}}} sin^{-1}x dy$

$=[y. cos^{-1}y - \sqrt{1-y^2}]_\frac{1}{\sqrt{2}}^1 + [x. sin^{-1}x + \sqrt{1-x^2}]_1^\frac{1}{\sqrt{2}}$

$= cos^{-1}(1)-\frac{1}{\sqrt{2}} cos^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}+\frac{1}{\sqrt{2}} sin^{-1}(\frac{1}{\sqrt{2}})+\sqrt{1-\frac{1}{2}}-1$

$=\frac{-\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

$=\frac{2}{\sqrt{2}} - 1$

$=\sqrt{2} - 1$

Hence, the correct answer is B.

If you are looking for exercises solutions for chapter application of integrals class 12 then these are listed below.

#### More about NCERT Solutions for Class 12 Maths chapter 8

Generally, one question (5 marks) is asked from this ch 8 maths class 12 in the 12th board final exam and you can score these 5 marks very easily with help of these NCERT solutions for class 12 maths chapter 8 application of integrals. In this chapter 8 class 12 maths, there are 2 exercises with 20 questions. In the NCERT class 12 maths ch 8 question answer, these questions are prepared and explained in a detailed manner using diagrams.

There are a total of 14 solved examples are given in the NCERT textbook to give a better understanding of the concepts related to the application of integrals class 12. Applications of integrals class 12 are indispensable for the Board exam. At the end of the chapter, 19 questions are given in a miscellaneous exercise. In the NCERT solutions for class 12 maths chapter 8 application of integrals article, you will get solutions to miscellaneous exercises too.

## Application of integrals class 12 - Topics

8.1 Introduction

8.2 Area under Simple Curves

8.2.1 The area of the region bounded by a curve and a line

8.3 Area between Two Curves

Let's understand these topics with help of examples

• How to find the area under simple curves- To find the area bounded by the curve and the ordinates and . Let assume that area under the curve as composed of large numbers of very thin vertical strips. Consider an arbitrary strip of width dx and height y ,then area of the elementary strip(dA) = ydx , where, y = f(x). This small area called the elementary area.

• How to find the area of the region bounded by a curve and a line- In this subsection, we will find the area of the region bounded by a line and a circle, a line and an ellipse, a line and a parabola. Equations of the above-said curves will be in their standard forms only.

## NCERT solutions for class 12 maths - Chapter wise

 chapter 1 NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions chapter 2 NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 NCERT solutions for class 12 maths chapter 4 Determinants chapter 5 NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals chapter 8 NCERT solutions for class 12 maths chapter 8 Application of Integrals chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra chapter 11 NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

## Key Features of NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

The application of integrals class 12 ncert solutions offers several key features, including

1. Expertly crafted by knowledgeable subject specialists.

2. Designed to enhance students' fundamental understanding of the applications of integrals.

3. Clear and organized material.

4. All questions are answered using informative diagrams.

5. Supports students in completing their assignments and preparing for competitive exams.

## NCERT Solutions Class Wise

Benefits of NCERT solutions:

• These NCERT maths chapter 8 class 12 solutions are very easy to understand as they are explained and are prepared in a detailed manner.
• Scoring good marks in the 12th board exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 8 application of integrals.
• NCERT solutions for Class 12 Maths Chapter 8 PDF download provides in-depth knowledge of the concepts as well as different ways to solve the problems.
• NCERT maths chapter 8 class 12 solutions are prepared by experts who know how to answer to score well in the board exam.
• You will also get some short tricks to check whether the answer is correct or not.

## NCERT Books and NCERT Syllabus

1. What is the weightage of the chapter Application of integrals for CBSE board exam ?

Generally, one question of 5 marks is asked from this chapter in the 12th board final exam but if you want to obtain the full 5 five marks that demands practice, good strategy, and command of concepts, and therefore NCERT textbooks, NCERT syllabus becomes very important. it is advised for students to precisely go through the basics of class 12 maths chapter 8 question answer to score good marks.

2. How does the NCERT solutions are helpful in the CBSE board exam ?

Only knowing the answer is not enough to score good marks in the exam. One should know how to write the answer to board exam in order to get good marks. NCERT solutions are provided by experts who knows how best to write the answer in the board exam in order to get good marks. Practice class 12 maths chapter 8 question answer to command the concepts.

3. What are the strategies for achieving maximum marks in class 12 math chapter 8 application of integrals ncert solutions?

The NCERT Solutions for the application of integration class 12 at Careers360 are created by a team of experts through extensive research on each concept. The solutions are presented in a thorough and comprehensive manner, enabling students to perform well on class exams and board exams. Additionally, the solutions aid students in completing their assignments efficiently and on time. Interested students can download the application of integrals class 12 pdf.

4. Where can I find the complete solutions of NCERT for class 12 maths ?

Here you will get the detailed and comprehensive NCERT Solutions for class 12 maths by clicking on the link. if  you want to study other material like NCERT Syllabus, NCERT Notes, NCERT Exercise Solutions, etc then you can browse the careers360 official website.

5. What are the important topics in chapter Application of integrals ?

Some applications of integrals like finding area under simple curves, area of the region bounded by a curve and a line, and area between two curves are the important topics covered in this chapter. practice class 12 maths chapter 8 ncert solutions to score well in the exam.

6. Which is the official website of NCERT ?

NCERT official is the official website of the "https://ncert.nic.in" where you can get NCERT textbooks and syllabus from class 1 to 12 that can be downloaded and studied offline. Careers360 created class 12 application of integrals ncert solutions. practice them to command the concepts.

7. What are the benefits of practicing class 12 application of integrals ncert solutions?

The class 12 application of integrals ncert solutions can be challenging as it incorporates numerous formulas and intricate concepts. To gain a solid understanding of each topic, students must regularly test their mathematical skills by practicing the questions provided. With consistent revision and diligent effort, students can confidently ace any exam, be it for their board exams or in a competitive setting.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
##### Carpenter

Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood.  Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.

2 Jobs Available
##### Welder

An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it.

2 Jobs Available
##### Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Dentist

Those who wish to make a dentist career in India must know that dental training opens up a universe of expert chances. Notwithstanding private practice, the present dental school graduates can pick other dental profession alternatives, remembering working in medical clinic crisis rooms, leading propelled lab examinations, teaching future dental specialists, or in any event, venturing to the far corners of the planet with International health and relief organizations.

2 Jobs Available
##### Health Inspector

Individuals following a career as health inspectors have to face resistance and lack of cooperation while working on the sites. The health inspector's job description includes taking precautionary measures while inspecting to save themself from any external injury and the need to cover their mouth to avoid toxic substances. A health inspector does the desk job as well as the fieldwork. Health inspector jobs require one to travel long hours to inspect a particular place.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
##### Fashion Blogger

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns.

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available