JEE Exam > JEE Questions > The surface area of a balloon of spherical sh...
Start Learning for Free
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
- a)9
- b)10
- c)11
- d)12
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The surface area of a balloon of spherical shape being inflated increa...
Given:
- The surface area of a balloon of spherical shape increases at a constant rate.
- Initially, the radius of the balloon is 3 units.
- After 5 seconds, the radius becomes 7 units.
To find:
- The radius of the balloon after 9 seconds.
Solution:
Let's first understand the relationship between the radius and the surface area of a sphere.
The surface area of a sphere is given by the formula:
Surface Area = 4πr^2, where r is the radius of the sphere.
Since the surface area of the balloon is increasing at a constant rate, we can write:
Rate of increase of surface area = k, where k is a constant.
Step 1:
We are given that initially the radius of the balloon is 3 units, so the initial surface area (SA1) is:
SA1 = 4π(3^2) = 36π
Step 2:
After 5 seconds, the radius of the balloon becomes 7 units, so the surface area after 5 seconds (SA2) is:
SA2 = 4π(7^2) = 196π
Step 3:
Now, let's find the rate of increase of the surface area.
Rate of increase of surface area = (SA2 - SA1) / (Time2 - Time1)
Rate of increase of surface area = (196π - 36π) / (5 - 0)
Rate of increase of surface area = 160π / 5
Rate of increase of surface area = 32π
Since the rate of increase of the surface area is constant, it will remain the same for any time interval.
Step 4:
Now, let's find the surface area after 9 seconds (SA3):
Rate of increase of surface area = (SA3 - SA2) / (Time3 - Time2)
32π = (SA3 - 196π) / (9 - 5)
32π * 4 = SA3 - 196π
SA3 = 128π + 196π
SA3 = 324π
Step 5:
Finally, let's find the radius of the balloon after 9 seconds.
SA3 = 4π(r^2)
324π = 4π(r^2)
81 = r^2
r = √81
r = 9
Therefore, the radius of the balloon after 9 seconds is 9 units.
Hence, the correct answer is option 'A', 9.
- The surface area of a balloon of spherical shape increases at a constant rate.
- Initially, the radius of the balloon is 3 units.
- After 5 seconds, the radius becomes 7 units.
To find:
- The radius of the balloon after 9 seconds.
Solution:
Let's first understand the relationship between the radius and the surface area of a sphere.
The surface area of a sphere is given by the formula:
Surface Area = 4πr^2, where r is the radius of the sphere.
Since the surface area of the balloon is increasing at a constant rate, we can write:
Rate of increase of surface area = k, where k is a constant.
Step 1:
We are given that initially the radius of the balloon is 3 units, so the initial surface area (SA1) is:
SA1 = 4π(3^2) = 36π
Step 2:
After 5 seconds, the radius of the balloon becomes 7 units, so the surface area after 5 seconds (SA2) is:
SA2 = 4π(7^2) = 196π
Step 3:
Now, let's find the rate of increase of the surface area.
Rate of increase of surface area = (SA2 - SA1) / (Time2 - Time1)
Rate of increase of surface area = (196π - 36π) / (5 - 0)
Rate of increase of surface area = 160π / 5
Rate of increase of surface area = 32π
Since the rate of increase of the surface area is constant, it will remain the same for any time interval.
Step 4:
Now, let's find the surface area after 9 seconds (SA3):
Rate of increase of surface area = (SA3 - SA2) / (Time3 - Time2)
32π = (SA3 - 196π) / (9 - 5)
32π * 4 = SA3 - 196π
SA3 = 128π + 196π
SA3 = 324π
Step 5:
Finally, let's find the radius of the balloon after 9 seconds.
SA3 = 4π(r^2)
324π = 4π(r^2)
81 = r^2
r = √81
r = 9
Therefore, the radius of the balloon after 9 seconds is 9 units.
Hence, the correct answer is option 'A', 9.
Free Test
FREE
| Start Free Test |
Community Answer
The surface area of a balloon of spherical shape being inflated increa...
S = 4πr2
r dr = k dt
⇒ r2/2 = kt + c
At t = 0, r = 3
9/2 = C
At t = 5,
At t = 9,
So, r = 9
r dr = k dt
⇒ r2/2 = kt + c
At t = 0, r = 3
9/2 = C
At t = 5,
At t = 9,
So, r = 9
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer?
Question Description
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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer?.
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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer?.
Solutions for 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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 isa)9b)10c)11d)12Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup