A copper wire of length 3 m is stretched by 3 mm by applying an external force

Question

A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is $600 \times 10^{- 6} m^{3}$ . The elastic potential energy stored in the wire in stretched condition would be ______ J.

(Given Young modulus of copper $= 1.1 \times 10^{11} N / m^{2}$ ) [2026]

Smart Achievers · Accepted Answer

(33)

$U = \frac{1}{2} Y {(strain)}^{2} \times volume$

$= \frac{1}{2} \times 1.1 \times 10^{11} \times {(\frac{3 \times 10^{- 3}}{3})}^{2} \times 600 \times 10^{- 6}$

$\Rightarrow U = 33 J$

Solution

Ans.
(33)

$U = \frac{1}{2} Y {(strain)}^{2} \times volume$

$= \frac{1}{2} \times 1.1 \times 10^{11} \times {(\frac{3 \times 10^{- 3}}{3})}^{2} \times 600 \times 10^{- 6}$

$\Rightarrow U = 33 J$

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Solution

Q. A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is 600×10-6 m3. The elastic potential energy stored in the wire in stretched condition would be ______ J. (Given Young modulus of copper =1.1×1011 N/m2) [2026]

Ans. (33) U=12Y(strain)2×volume =12×1.1×1011×(3×10-33)2×600×10-6 ⇒U=33 J

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Ans.
(33)

$U = \frac{1}{2} Y {(strain)}^{2} \times volume$

$= \frac{1}{2} \times 1.1 \times 10^{11} \times {(\frac{3 \times 10^{- 3}}{3})}^{2} \times 600 \times 10^{- 6}$

$\Rightarrow U = 33 J$