Two wires $A$ and $B$ are stretched by the same force. If,for $A$ and $B$,$Y_A: Y_B = 1: 2$,$r_A: r_B = 3: 1$,and $L_A: L_B = 4: 1$,then the ratio of their extension $\left(\frac{\Delta L_A}{\Delta L_B}\right)$ will be .............

$A$ wooden wheel of radius $R$ is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross-sectional area $S$ and length $L$. $L$ is slightly less than $2\pi R$. To fit the ring on the wheel,it is heated so that its temperature rises by $\Delta T$ and it just slips over the wheel. As it cools down to the surrounding temperature,it presses the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$,and its Young's modulus is $Y$,the force that one part of the wheel applies on the other part is:

Young's modulus of rubber is $10^4 \ N/m^2$ and the area of cross-section is $2 \ cm^2$. If a force of $2 \times 10^5 \ dynes$ is applied along its length,then its initial length $L$ becomes:

The cross-sectional area of each wire is $10^{-4} \, m^2$. What is the displacement of point $B$?

$A$ wire of length $L$ is hanging from a fixed support. The length changes to $L_{1}$ and $L_{2}$ when masses $1 \, kg$ and $2 \, kg$ are suspended respectively from its free end. Then the value of $L$ is equal to:

The figure represents the extension $(\Delta l)$ of a wire of length $1\text{ m}$,suspended from the ceiling of a room at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5}\text{ m}^2$,then the Young's modulus of the wire is . . . . . . $\text{N/m}^2$.