Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A copper wire \left(y=10^{11} \mathrm{NJ}^2\right) of length 8 \mathrm{~m} and a stud wire \left(Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right) of length 4 \mathrm{~m}, each of 0.5 \mathrm{~cm}^2 cross-section are fastened end to end and stretched with a tension of 500 \mathrm{~N}.

Option: 1

Elongation in copper wire is 0.8 \mathrm{~mm}.


Option: 2

Elongation in stack is \frac{1}{4}th the elongation in copper wire.


Option: 3

Total elongation is 1.0 \mathrm{~mm}.


Option: 4

All of the above


Answers (1)

best_answer

\begin{aligned} (\Delta l)_c= & \left(\frac{\mathrm{Fl}}{\mathrm{AY}}\right)_c=\frac{500 \times 8}{0.5 \times 10^{-4} \times 10^{11}} \\ (\Delta l)_c & =0.8 \times 10^{-3} \mathrm{~m}=0.8 \mathrm{~mm} . \end{aligned}


\begin{aligned} (\Delta l)_s= & \left(\frac{\mathrm{Fl}}{\mathrm{AY}}\right)_s \\ &=\frac{500 \times 4}{0.5 \times 10^{-4 \times 2 \times 10^{11}}}=0.2 \times 10^{-3} \mathrm{~m}\\&=0.2 \mathrm{~mm} \end{aligned}

\begin{aligned} &(\Delta l)_s=\frac{1}{4}(\Delta l)_c \mathrm{~mm} . \\ & \Delta l=0 . s+0.2=1.00 \mathrm{~mm} \end{aligned}

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question