Question

A weight of 200 kg is suspended by vertical wire of length 600.5 cm. The area of cross-section of wire is . When the load is removed, the wire contracts by 0.5 cm. The Young's modulus of the material of wire will be

• A. $2.35 \times 10 ^ { 12 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
• B. $1.35 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
• C. $13.5 \times 10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
• D. $23.5 \times 10 ^ { 9 } \mathrm {~N} / \mathrm { m } ^ { 2 }$

Answer 1

Answer: (A)

$2.35 \times 10 ^ { 12 } \mathrm {~N} / \mathrm { m } ^ { 2 }$

Answer 2

$F = 2000 \mathrm {~N} , L = 6 \mathrm {~m} , l = 0.5 \mathrm {~cm} , A = 10 ^ { - 6 } \mathrm {~m} ^ { 2 }$

$Y = \frac { F L } { A l } = \frac { 2000 \times 6 } { 10 ^ { - 6 } \times 0.5 \times 10 ^ { - 2 } } = 2.35 \times 10 ^ { 12 } \mathrm {~N} / \mathrm { m } ^ { 2 }$