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NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

Edited By Ramraj Saini | Updated on Dec 04, 2023 01:24 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 8 class 12 Application of Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 8 Miscellaneous exercise deals with the area finding of the various curves and intersection of two curves. Class 12 Maths chapter 8 miscellaneous exercise is a good source to practice good questions before the exam. NCERT book Class 12 Maths chapter 8 miscellaneous exercise is relatively easier than the chapter 7 exercise as it provides questions based on the application part. Class 12 Maths chapter 8 miscellaneous exercise solutions with other two exercises are sufficient to score good marks.

This Story also Contains
  1. NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise
  2. Application of Integrals Class 12 Chapter 8 Miscellaneous: Exercise
  3. More About NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise
  4. Also Read| Application of Integrals Class 12 Notes
  5. Benefits of Ncert Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises
  6. Key Features Of NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise
  7. Also see-
  8. NCERT Solutions Subject Wise
  9. Subject Wise NCERT Exemplar Solutions

Miscellaneous exercise class 12 chapter 8 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Application of Integrals Class 12 Chapter 8 Miscellaneous: Exercise

Question:1 Find the area under the given curves and given lines:

(i) \small y=x^2,x=1,x=2 and \small x -axis

Answer:

The area bounded by the curve \small y=x^2,x=1,x=2 and \small x -axis
15947281267411594728124007
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

Question:1 Find the area under the given curves and given lines:

(ii) \small y=x^4,x=1,x=5 and \small x -axis

Answer:

The area bounded by the curev \small y=x^4,x=1,x=5 and \small x -axis

15947282868341594728284206
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

Question:2 Find the area between the curves \small y=x and \small y=x^2 .

Answer:

the area between the curves \small y=x and \small y=x^2 .
AOI graph1594728523616
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

Question:3 Find the area of the region lying in the first quadrant and bounded by \small y=4x^2,x=0,y=1 and \small y=4 .

Answer:

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

15947286788651594728676557
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

Question:4 Sketch the graph of \small y=|x+3| and evaluate \small \int_{-6}^{0}|x+3|dx.

Answer:

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)

1654760706138

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

Question:5 Find the area bounded by the curve \small y=\sin x between \small x=0 and \small x=2\pi .

Answer:

The graph of y=sinx is as follows

1654760755958

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.

Question:6 Find the area enclosed between the parabola \small y^2=4ax and the line \small y=mx .

Answer:

1654760804965

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

Question:7 Find the area enclosed by the parabola \small 4y=3x^2 and the line \small 2y=3x+12 .

Answer:

1654764435919

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

Question:8 Find the area of the smaller region bounded by the ellipse \small \frac{x^2}{9}+\frac{y^2}{4}=1 and the line \small \frac{x}{3}+\frac{y}{2}=1 .

Answer:

1654764622123

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

Question:9 Find the area of the smaller region bounded by the ellipse \small \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and the line \small \frac{x}{a}+\frac{y}{b}=1 .

Answer:


1654764663614

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


Question:10 Find the area of the region enclosed by the parabola \small x^2=y, the line \small y=x+2 and the \small x -axis.

Answer:

1654764778111

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 \small x^2=y, the line \small y=x+2 and the \small x -axis is 5/6 units.

Question:11 Using the method of integration find the area bounded by the curve \small |x|+|y|=1.

[ Hint: The required region is bounded by lines \small x+y=1,x-y=1,-x+y=1 and \small -x-y=1 ]

Answer:

1654764830021

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 \small |x|+|y|=1 is 2 units.

Question:12 Find the area bounded by curves \small \left \{ (x,y);y\geq x^2\hspace{1mm} and \hspace{1mm}y=|x| \right \} .

Answer:

1654764925341

We have to find the area of the shaded region

In the first quadrant

y=|x|=x

Area of the shaded region=2ar(OADO)

\\=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.

Question:13 Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are \small A(2,0),B(4,5)\hspace{1mm}and\hspace{1mm}C(6,3).

Answer:

1654764957923

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.

Question:14 Using the method of integration find the area of the region bounded by lines:
\small 2x+y=4,3x-2y=6\hspace{1mm}and\hspace{1mm}x-3y+5=0.

Answer:

1654764994031

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

Question:15 Find the area of the region \small \left \{ (x,y);y^2\leq 4x,4x^2+4y^2\leq 9 \right \} .

Answer:

1654765056298

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

For the fist quadrant

\\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 )

Area of the unshaded region in the first quadrant is

\\\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

Question:16 Choose the correct answer.

Area bounded by the curve \small y=x^3 , the \small x -axis and the ordinates \small x=-2 and \small x=1 is

(A) \small -9 (B) \small \frac{-15}{4} (C) \small \frac{15}{4} (D) \small \frac{17}{4}

Answer:

1654765098486

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.

Question:17 Choose the correct answer.

T he area bounded by the curve \small y=x|x| , \small x -axis and the ordinates \small x=-1 and \small x=1 is given by

(A) \small 0 (B) \small \frac{1}{3} (C) \small \frac{2}{3} (D) \small \frac{4}{3}

[ Hint : y=x^2 if x> 0 and y=-x^2 if x<0 . ]

Answer:

The required area is

\\2\int_{0}^{1}x^{2}dx\\ =2\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ =\frac{2}{3}\ units

Question:18 Choose the correct answer.

The area of the circle \small x^2+y^2=16 exterior to the parabola \small y^2=6x is

(A) \small \frac{4}{3}(4\pi -\sqrt{3} ) (B) \small \frac{4}{3}(4\pi +\sqrt{3} ) (C) \small \frac{4}{3}(8\pi -\sqrt{3} ) (D) \small \frac{4}{3}(8\pi +\sqrt{3} )

Answer:

1654765195847

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

Question:19 Choose the correct answer The area bounded by the \small y -axis, \small y=\cos x and \small y=\sin x when \small 0\leq x\leq \frac{\pi }{2} is

(A) \small 2(\sqrt{2}-1) (B) \small \sqrt{2}-1 (C) \small \sqrt{2}+1 (D) \small \sqrt{2}

Answer:

Given : \small y=\cos x and \small y=\sin x

1654765304392


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.

More About NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise

Miscellaneous exercise chapter 8 Class 12 chapter application of integrals is quite interesting and important from exam perspective. Class 12 Maths chapter 8 miscellaneous solutions mainly deals with application based questions like area finding between two curves etc. NCERT solutions for Class 12 Maths chapter 8 miscellaneous exercise will not take much time to solve provided chapter 7 integrals are done well before chapter 8.

Also Read| Application of Integrals Class 12 Notes

Benefits of Ncert Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises

  • The Class 12th Maths chapter 8 exercise has some good questions which must be done before examination.
  • NCERT syllabus Class 12 Maths chapter 8 miscellaneous exercises can be done along with chapter 7 as basic concepts used are the same.
  • These Class 12 Maths chapter 8 miscellaneous exercises are also useful in Physics and chemistry problems.
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Key Features Of NCERT Solutions For Class 12 Chapter 8 Miscellaneous Exercise

  • Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 8, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 chapter 8 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 8 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this class 12 maths ch 8 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for class 12 chapter 8 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 8 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Happy learning!!!


Frequently Asked Questions (FAQs)

1. How many questions are there in Miscellaneous exercise Chapter 8 ?

There are 19 questions total in Miscellaneous exercise Chapter 8.

2. Can we find an area without using integrals ?

Simple figures like triangle, circle etc. can be tackled without integration but not the complex ones. 

3. Are questions repeated in the examination from this Chapter ?

Yes, in the Board exam the questions are repeated every year. 

4. What is the level of questions asked from this Chapter ?

Moderate level questions are asked from this Chapter. 

5. Can one skip Miscellaneous exercise ?

No, as it has some good questions, miscellaneous exercise must be done. 

6. What is the time it will take to complete for the first time ?

It will take around 5-6 hours to complete for the first time.

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

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

 
Read More

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)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

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

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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