Q76.The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is (1) 9 (2) 7 (3) 5 (4) 3
What This Question Tests
The problem involves related rates, where the surface area of a spherical balloon increases at a constant rate, and requires integrating this rate to find the radius at a later time.
Concepts Tested
Formulas Used
A = 4ฯrยฒ
dA/dt = constant
Integration of differential equations
๐ NCERT Sections This Tests
2.4 โ A Spherical Conductor Of Radius 12 Cm Has A Charge Of 1.6 ร 10โ7C
Physics Class 11 ยท Chapter 2
2.4 A spherical conductor of radius 12 cm has a charge of 1.6 ร 10โ7C 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?
1.19 โ A Point Charge Causes An Electric Flux Of โ1.0 ร 103 Nm2/C To Pass
Physics Class 11 ยท Chapter 1
1.19 A point charge causes an electric flux of โ1.0 ร 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
12.7 โ The Radius Of The Innermost Electron Orbit Of A Hydrogen Atom Is
Physics Class 12 ยท Chapter 12
12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3ร10โ11 m. What are the radii of the n = 2 and n =3 orbits?
๐ Question Details
- Chapter
- Applications of Derivatives
- Topic
- Rates of change
- Year
- 2022
- Shift
- 24 Jun Shift 1
- Q Number
- Q76
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Applications of Derivatives
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