Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The external and internal diameters of a hollow cylinder are measured to be (4.23±0.01) cm and (3.89 ± 0.01) cm. The thickness of the wall of the cylinder is:

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

(0.34±0.02)cm

b

(0.17±0.02)cm

c

(0.17±0.01)cm

d

(0.34±0.01)cm

answer is C.

(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

D=(4.23±0.01)cmd=(3.89±0.01)cmΔt=(D−d)2 =(4.23±0.01)−(3.89±0.01)2 =(4.23−3.89)±(0.01+0.01)2 =(0.34±0.02)2cm =(0.17±0.01)cm
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon