Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections and respectively are stretched by the same magnitude under a given load. The ratio of the Young’s modulus of A to that of B is x:3. T

Question

Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections $3.0 \times 10^{- 5} m^{2}$ and $4.5 \times 10^{- 5} m^{2}$ respectively are stretched by the same magnitude under a given load. The ratio of the Young’s modulus of A to that of B is $x : 3$ . The value of $x$ is ______. [2026]

Smart Achievers · Accepted Answer

(2)

$T = \frac{F / A}{Δ ℓ / ℓ} \Rightarrow Y = \frac{F ℓ}{A Δ ℓ}$

$\frac{Y_{A}}{Y_{B}} = \frac{ℓ_{A}}{ℓ_{B}} (\frac{A_{B}}{A_{A}})$

$= \frac{6}{5.4} (\frac{4.5 \times 10^{- 5}}{3 \times 10^{- 5}}) = \frac{9}{5.4} = \frac{5}{3} \Rightarrow \frac{x}{3} = \frac{5}{3}$

$x = 5$

Solution

1 4

2 5

3 1

4 2

Ans.
(2)

$T = \frac{F / A}{Δ ℓ / ℓ} \Rightarrow Y = \frac{F ℓ}{A Δ ℓ}$

$\frac{Y_{A}}{Y_{B}} = \frac{ℓ_{A}}{ℓ_{B}} (\frac{A_{B}}{A_{A}})$

$= \frac{6}{5.4} (\frac{4.5 \times 10^{- 5}}{3 \times 10^{- 5}}) = \frac{9}{5.4} = \frac{5}{3} \Rightarrow \frac{x}{3} = \frac{5}{3}$

$x = 5$

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Solution

1 4

2 5

3 1

4 2

Ans. (2) T=F/AΔℓ/ℓ ⇒ Y=FℓAΔℓ YAYB=ℓAℓB(ABAA) =65.4(4.5×10-53×10-5)=95.4=53⇒x3=53 x=5

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Ans.
(2)

$T = \frac{F / A}{Δ ℓ / ℓ} \Rightarrow Y = \frac{F ℓ}{A Δ ℓ}$

$\frac{Y_{A}}{Y_{B}} = \frac{ℓ_{A}}{ℓ_{B}} (\frac{A_{B}}{A_{A}})$

$= \frac{6}{5.4} (\frac{4.5 \times 10^{- 5}}{3 \times 10^{- 5}}) = \frac{9}{5.4} = \frac{5}{3} \Rightarrow \frac{x}{3} = \frac{5}{3}$

$x = 5$