- Home
- Standard 11
- Physics
8.Mechanical Properties of Solids
easy
The force required to stretch a steel wire of $1\,c{m^2}$ cross-section to $1.1$ times its length would be $(Y = 2 \times {10^{11}}\,N{m^{ - 2}})$
A
$2 \times {10^6}\,N$
B
$2 \times {10^3}\,N$
C
$2 \times {10^{ - 6}}N$
D
$2 \times {10^{ - 7}}\,N$
Solution
(a) $F = A \times Y \times {\rm{strain}}$=$1 \times {10^{ – 4}} \times 2 \times {10^{11}} \times 0.1 = 2 \times {10^6}N$
Standard 11
Physics
Similar Questions
normal