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

In an experiment, brass and steal wires of length 1 m each with areas of cross section 1mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is: {Given the young’s Modulus for steel and brass are, respectively 120 × 10⁹ N/m² and 60 × 10⁹ N/m²}

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

$0.2 \times 10^{6} N / m^{2}$

$8.0 \times 10^{6} N / m^{2}$

$1.2 \times 10^{6} N / m^{2}$

$1.8 \times 10^{6} N / m^{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given that the stress will be the same in both rods,

$stress = Y_{1} \frac{∆ l_{1}}{l} = Y_{2} \frac{∆ l_{2}}{l}________(1)$

$∆ l_{1} + ∆ l_{2} = 0.2 \times 10^{- 3} m_________(2)$

from equations 1 and 2

$stress \times (\frac{l}{Y_{1}} + \frac{l}{Y_{2}}) = 0.2 \times 10^{- 3} m$

$stress \times (\frac{l}{120 \times 10^{9}} + \frac{l}{60 \times 10^{9}}) = 0.2 \times 10^{- 3} m = 8 \times 10^{6} N / m^{2}$

Hence the correct answer is $8 \times 10^{6} N / m^{2} .$

Watch 3-min video & get full concept clarity

courses

No courses found

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!