Book Online Demo

Check Your IQ

Courses

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

Q.

Longitudinal stress of $1 kg / {mm}^{2}$ is applied on a wire. The percentage increase in length is $(Y = 10^{11} N / m^{2})$

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

a

0.01

b

0.001

c

0.002

d

0.003

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$Longitudinal strain \frac{l}{L} = \frac{stress}{Y} = \frac{10^{6}}{10^{11}} = 10^{- 5}$

$Percentage increase in length = 10^{- 5} \times 100 = 0 .001 %$

Watch 3-min video & get full concept clarity

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 Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

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 Books for IIT JEE

Best Books for NEET

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!