AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

The difference between inside and outside surfaces of a cylindrical tube $14 cm$ long is $88 sq . cm .$ If the volume of the tube is 176 cubic centimeter, find the inner and outer radii of the tube.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

a

2.5

cm

and 3.5

cm

respectively

b

1.5

cm

and 2.5

cm

respectively

c

3

cm

and 2.5

cm

respectively

d

1.5

cm

and 2

cm

respectively

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

It is given the difference between inside and outside surfaces of a cylindrical tube of length 14 cm is 88sq.cm and the volume of the tube is 176 cubic cm.
Let r and R be the radii of the inner and outer surface of the cylindrical tube respectively.
Question Image

Question Image

Put the value of the volume of the tube equal to the general formula of volume of a hollow cylinder

= π R 2 − r 2 h

Since, the volume of the tube

176 c m 3

.
So,

π R 2 − r 2 h = 176

⇒ 227 × 14 × R 2 − r 2 = 176

⇒ R 2 − r 2 = 176 × 7 22 × 14

⇒ R 2 − r 2 = 4

… … (1)

Put the value of the difference of outer and inner surface of the tube equal to the general formula of the curved surface area of a hollow cylinder

= 2 π R − r h

.
Since, the difference of outer and inner surface of the tube is 88

cm

.
So,

2 π R − r h = 88

⇒ 2 × 227 × 14 × R − r = 88

⇒ R − r = 88 × 7 2 × 22 × 14

⇒ R − r = 1

… … (2)

Expand equation (1) using

a 2 − b 2 = a − b a + b

,

⇒

R 2 − r 2 = 4

⇒ R + r R − r = 4

Put R-r =1 in the above equation from (2)

⇒

R + r 1 = 4

⇒

R + r = 4

… … (3)

Add (2) and (3),

⇒

2R = 5

⇒ R = 52 ⇒ R = 2.5 cm

Put the value of R in (3)
R + r = 4

⇒ 2.5 + r = 4

⇒ r = 4 − 2.5 ⇒ r = 1.5 c m

So, outer radius of the tube = 2.5

cm

The inner radius of tube = 1.5

cm

Therefore, the inner and outer radii of the tube are 1.5

cm

and 2.5

cm

respectively.
Hence option (2) is correct.

Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

gplay

mentor

Download the App

gplay

personalised 1:1 online tutoring