If a spring extends by $x$ on loading, then the energy stored by the spring is (if $T$ is tension in the spring and $k$ is spring constant)
$\frac{{{T^2}}}{{2x}}$
$\frac{{{T^2}}}{{2k}}$
$\frac{{2x}}{{{T^2}}}$
$\frac{{2{T^2}}}{k}$
A metal wire having Poisson's ratio $1 / 4$ and Young's modulus $8 \times 10^{10} \,N / m ^2$ is stretched by a force, which produces a lateral strain of $0.02 \%$ in it. The elastic potential energy stored per unit volume in wire is [in $\left.J / m ^3\right]$
A brass rod of cross-sectional area $1\,c{m^2}$ and length $0.2\, m$ is compressed lengthwise by a weight of $5\, kg$. If Young's modulus of elasticity of brass is $1 \times {10^{11}}\,N/{m^2}$ and $g = 10\,m/{\sec ^2}$, then increase in the energy of the rod will be
If the potential energy of a spring is $V$ on stretching it by $2\, cm$, then its potential energy when it is stretched by $10 \,cm$ will be
When a $4\, kg$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $2\, cms$. The work required to be done by an external agent in stretching this spring by $5\, cms$ will be ......... $joule$ $(g = 9.8\,metres/se{c^2})$
$K$ is the force constant of a spring. The work done in increasing its extension from ${l_1}$ to ${l_2}$ will be