A brass rod of length 2 m and cross-sectional area 2 cm² is attached end-to-end to a steel rod of length
L and cross-sectional area 1 cm². The compound rod is subjected to equal and opposite pulls of magnitude
5 X 10⁴ N at its ends. If the elongations of the two rods are equal, the length of the steel rod (L) is
(Y_brass= 1 x 10¹¹ Nm^-2 and Y_steel= 2 x 10¹¹ Nm^-2)

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

1.8 m

2 m

1.5 m

1 m

answer is D.

(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

Since, the elongations of the two rods are equal

$(Δ L)_{b} = (Δ L)_{s}$

$\Rightarrow {(\frac{F L}{A Y})}_{b} = {(\frac{F L}{A Y})}_{s}$ $(∵ F_{b} = F_{s})$

$or {(\frac{L}{A Y})}_{b} = {(\frac{L}{A Y})}_{s}$

$∴$ The length of the steel rod,

$L_{s} = (\frac{A_{s} Y_{s}}{A_{b} Y_{b}}) L_{b} = (\frac{1.0 \times 2.0 \times 10^{11}}{2.0 \times 1.0 \times 10^{11}}) (2) = 2 m$

Watch 3-min video & get full concept clarity