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Q.

An area of cross-section of rubber string is 2 cm². Its length is doubled when stretched with a linear force of $2 \times 10^{5}$ dynes. The Young's modulus of the rubber in $dyne / {cm}^{2}$ will be

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a

$4 \times 10^{5}$

b

$1 \times 10^{5}$

c

$1 \times 10^{4}$

d

$2 \times 10^{5}$

answer is B.

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Detailed Solution

detailed_solution_thumbnail

If length of the wire is doubled then strain = 1

$∴ Y = Stress = \frac{Force}{Area} = \frac{2 \times 10^{5}}{2} = 10^{5} \frac{dyne}{{cm}^{2}}$

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An area of cross-section of rubber string is 2 cm2. Its length is doubled when stretched with a linear force of 2×105 dynes. The Young's modulus of the rubber in dyne/cm2 will be