- Home
- Standard 11
- Physics
8.Mechanical Properties of Solids
easy
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
Solution
(d) If length of wire doubled then strain $= 1$
$Y = {\rm{stress}}$ ==> $F = Y \times A$$ = {10^{12}} \times 0.5$$ = 0.5 \times {10^{12}}dyne$
Standard 11
Physics
Similar Questions
hard