Question

To double the length of a iron wire having $0.5 \mathrm {~cm} ^ { 2 }$ area of cross-section, the required force will be $\left( Y = 10 ^ { 12 } \right.$ dyne $\left. / \mathrm { cm } ^ { 2 } \right)$

• A.
• B.
• C.
• D. $0.5 \times 10 ^ { 12 }$dyne

Answer 1

Answer: (D)

$0.5 \times 10 ^ { 12 }$dyne

Answer 2

If length of wire doubled then strain = 1

⇒ $F = Y \times A$ $= 10 ^ { 12 } \times 0.5$