A force is applied to a steel wire ‘A’, rigidly clamped at one end. As a result, elongation in the wire is 0.2 mm. If the same force is applied to another steel wire ‘B’ of double the length and a diameter 2.4 times that of the wire ‘A’, the elongation in

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Q.
A force is applied to a steel wire ‘A’, rigidly clamped at one end. As a result, elongation in the wire is 0.2 mm. If the same force is applied to another steel wire ‘B’ of double the length and a diameter 2.4 times that of the wire ‘A’, the elongation in the wire ‘B’ will be (wires having uniform circular cross sections). [2023]

1 $6.06 \times 10^{- 2} mm$

2 $2.77 \times 10^{- 2} mm$

3 $3.0 \times 10^{- 2} mm$

4 $6.9 \times 10^{- 2} mm$

Ans.
(4)

$Y = \frac{F / A}{\frac{Δ l}{l}}$

$\Rightarrow F = \frac{Y A}{l} Δ l$

${(\frac{A Δ l}{l})}_{1} = {(\frac{A Δ l}{l})}_{2}$

$\Rightarrow \frac{Δ l_{2}}{Δ l_{1}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$

$\Rightarrow \frac{Δ l_{2}}{0.2} = \frac{1}{2.4 \times 2.4} \times \frac{2}{1}$

$\Rightarrow Δ l_{2} = 6.9 \times 10^{- 2} mm$

Solution

1 $6.06 \times 10^{- 2} mm$

2 $2.77 \times 10^{- 2} mm$

3 $3.0 \times 10^{- 2} mm$

4 $6.9 \times 10^{- 2} mm$

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Solution

1 6.06×10-2 mm

2 2.77×10-2 mm

3 3.0×10-2 mm

4 6.9×10-2 mm

Ans. (4) Y=F/AΔll ⇒F=YAl Δl (AΔll)1=(AΔll)2 ⇒Δl2Δl1=A1A2×l2l1 ⇒Δl20.2=12.4×2.4×21 ⇒Δl2=6.9×10-2 mm

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1 $6.06 \times 10^{- 2} mm$

2 $2.77 \times 10^{- 2} mm$

3 $3.0 \times 10^{- 2} mm$

4 $6.9 \times 10^{- 2} mm$