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  • A brass rod of length  and cross-sectional area  <img alt="2.0\ cm^{2}" src="https://entrancecorner.oncode

A brass rod of length 2\ m and cross-sectional area  2.0\ cm^{2} is attached end to end to a steel rod of length L and cross-sectional area 1.0\ cm^{2}. The compound rod is subjected to equal and opposite pulls of magnitude 5 \times 10^4 \mathrm{~N} at its ends. If the elongations of two rods are equal, the length of the steel (L) is-

Given,

\begin{aligned} & \text Y_{\text {Brass }}=1.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { and } \\ & \left.Y_{\text {steel }}=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right. \\ & \end{aligned}

Option: 1

1.5\ m


Option: 2

1.8\ m


Option: 3

1/ m


Option: 4

2\ m


Answers (1)

best_answer

(\Delta l)_b=(\Delta l)_s

\begin{aligned} & \left(\frac{F l}{A Y}\right)_b=\left(\frac{F l}{A Y}\right)_s \quad\left(F_b=F_s\right) \\ & \left(\frac{l}{A Y}\right)_b=\left(\frac{l}{A Y}\right)_{R_b} \\ & l_s=\left(\frac{A_s Y_s}{A_b Y_b}\right) l_b \\ & l_s=\left(\frac{1.0 \times 2.0 \times 10^{11}}{2.0 \times 1.0 \times 10^{11}}\right)(2 \mathrm{~m}) \\ & l_s=2 \mathrm{~m} \ \end{aligned}

Posted by

Sanket Gandhi

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