The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section,one of steel and another of brass,are suspended from the same roof. If we want the lower ends of the wires to be at the same level,then the weights added to the steel and brass wires must be in the ratio of

Two wires of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are stretched by forces $F_A$ and $F_B$ respectively to produce the same extension,then the ratio $\frac{F_A}{F_B}$ is:

Column $-II$ is related to Column $-I$. Join them appropriately:

Column $-I$	Column $-II$
$(a)$ When temperature is raised,Young's modulus of a body	$(i)$ Zero
$(b)$ Young's modulus for air	$(ii)$ Infinite
	$(iii)$ Decreases
	$(iv)$ Increases

In nature,the failure of structural members usually results from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling. In cases of tall cylindrical structures like trees,the torque is caused by their own weight bending the structure,such that the vertical line through the centre of gravity does not fall within the base. The elastic torque caused by this bending about the central axis of the tree is given by $\frac{Y\pi r^4}{4R}$,where $Y$ is the Young's modulus,$r$ is the radius of the trunk,and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Why are springs made of steel instead of copper?

$A$ wire of cross-sectional area $A$,modulus of elasticity $2 \times 10^{11} \text{ N m}^{-2}$,and length $2L = 2 \text{ m}$ is stretched between two vertical rigid supports. When a mass of $2 \text{ kg}$ is suspended at the middle,it sags from its original position,making an angle $\theta = \frac{1}{100} \text{ radian}$ with the horizontal at the points of support. The value of $A$ is . . . . . . $\times 10^{-4} \text{ m}^2$. (Given: $g = 10 \text{ m/s}^2$)