Q76.The area (in sq. units) of the part of the circle π₯2 + π¦2 = 36, which is outside the parabola π¦2 = 9π₯, is equal to (1) 12π+ 3β3 (2) 24π+ 3β3 (3) 24π- 3β3 (4) 12π- 3β3
What This Question Tests
This question involves calculating the area of a region that is part of a circle but outside a parabola, requiring intersection points and careful integration or geometric subtraction.
Concepts Tested
Formulas Used
Area of a sector = 1/2 * r^2 * theta
Area of a triangle = 1/2 * base * height
Area by integration: β«y dx
π NCERT Sections This Tests
9.5 β A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A
Physics Class 12 Β· Chapter 9
9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
2.2 β A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 Β· Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
2.4 β A Spherical Conductor Of Radius 12 Cm Has A Charge Of 1.6 Γ 10β7C
Physics Class 11 Β· Chapter 2
2.4 A spherical conductor of radius 12 cm has a charge of 1.6 Γ 10β7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
π Question Details
- Chapter
- Definite Integration & Area
- Topic
- Area under curves
- Year
- 2021
- Shift
- 24 Feb Shift 1
- Q Number
- Q76
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 8: Application of Integrals
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