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

  • A

    $1.0 \times {10^{ - 7}}N$

  • B

    $1.0 \times {10^7}N$

  • C

    $0.5 \times {10^{ - 7}}N$

  • D

    $0.5 \times {10^{12}}$dyne

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