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A brass rod of length 2 \mathrm{~m} and crosssectional area 2.0 \mathrm{~cm}^2 is attached end to end to a steel rod of length  and cross -sectional area 1.0 \mathrm{~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 rod \left ( L \right ) is -

[Given \begin{aligned} & y_{\text {Brass }}=1.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ \\ & \end{aligned}, and \begin{aligned} & \left.y_{\text {steal }}=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right] \\ & \end{aligned}

 

Option: 1

1.5 \mathrm{~m}


Option: 2

1.8 \mathrm{~m}


Option: 3

1 \mathrm{~m}


Option: 4

2 \mathrm{~m}


Answers (1)

best_answer

\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}

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rishi.raj

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