Question

Young's modulus of rubber is $10 ^ { 4 } \mathrm {~N} / \mathrm { m } ^ { 2 }$ and area of cross-section is $2 \mathrm {~cm} ^ { 2 }$. If force of $2 \times 10 ^ { 5 }$dynes is applied along its length, then its initial length l becomes

• A. 3L
• B. 4L
• C. 2L
• D. None of the above

Answer 1

Answer: (C)

2L

Answer 2

$Y = 10 ^ { 4 } \mathrm {~N} / \mathrm { m } ^ { 2 } , A = 2 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 } , F = 2 \times 10 ^ { 5 }$ dyne $= 2 \mathrm {~N}$

$l = \frac { F L } { A Y } = \frac { 2 \times L } { 2 \times 10 ^ { - 4 } \times 10 ^ { 4 } } = L$

∴ Final length = initial length + increment = 2L