The force required to stretch a wire of cross-section $1 \ cm^{2}$ to double its length will be ........ $\times 10^{7} \ N$. (Given Young's modulus of the wire $= 2 \times 10^{11} \ N/m^{2}$)

$A$ uniformly tapering conical wire is made from a material of Young's modulus $Y$ and has a normal,unextended length $L$. The radii at the upper and lower ends of this conical wire have values $R$ and $3R$,respectively. The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length of this wire would be:

The pressure that has to be applied to the ends of a steel wire of length $10 \ cm$ to keep its length constant when its temperature is raised by $100 \ ^\circ C$ is: (For steel,Young's modulus $Y = 2 \times 10^{11} \ N/m^2$ and coefficient of thermal expansion $\alpha = 1.1 \times 10^{-5} \ K^{-1}$)

$A$ pendulum made of a uniform wire of cross-sectional area $A$ has a time period $T$. When an additional mass $M$ is added to its bob,the time period changes to $T_M$. If the Young's modulus of the material of the wire is $Y$,then $\frac{1}{Y}$ is equal to ($g$ = gravitational acceleration).

If in case $A$,the elongation in a wire of length $L$ is $l$,then for the same wire,the elongation in case $B$ will be:

$A$ mild steel wire of length $1.0 \; m$ and cross-sectional area $0.50 \times 10^{-2} \; cm^{2}$ is stretched,well within its elastic limit,horizontally between two pillars. $A$ mass of $100 \; g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.