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Question: The Young's modulus of a rubber string 8 cm long and density \(1.5 \mathrm {~kg} / \mathrm { m } ^ {...

The Young's modulus of a rubber string 8 cm long and density 1.5 kg/m31.5 \mathrm {~kg} / \mathrm { m } ^ { 3 } is , is suspended on the ceiling in a room. The increase in length due to its own weight will be

A

9.6×10−5 m9.6 \times 10 ^ { - 5 } \mathrm {~m}

B

9.6×10−11 m9.6 \times 10 ^ { - 11 } \mathrm {~m}

C

9.6×10−3 m9.6 \times 10 ^ { - 3 } \mathrm {~m}

D

9.6 m

Answer

9.6×10−11 m9.6 \times 10 ^ { - 11 } \mathrm {~m}

Explanation

Solution

l=L2dg2Y=(8×10−2)2×1.5×9.82×5×108l = \frac { L ^ { 2 } d g } { 2 Y } = \frac { \left( 8 \times 10 ^ { - 2 } \right) ^ { 2 } \times 1.5 \times 9.8 } { 2 \times 5 \times 10 ^ { 8 } } =9.6×10−11 m= 9.6 \times 10 ^ { - 11 } \mathrm {~m}