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