$A$ wire of length $2L$ is fixed between two walls. When a weight $W = mg$ is applied at its midpoint,it sags by a distance $x$ $(x << L)$. Then $m$ is equal to:

$A$ compressive force, $F$ is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $l$ be the length of the rod, $A$ its area of cross-section, $Y$ its Young's modulus, and $\alpha$ its coefficient of linear expansion. Then, $F$ is equal to

$A$ 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:

$A$ $3 \ m$ long wire of radius $3 \ mm$ shows an extension of $0.1 \ mm$ when loaded vertically by a mass of $50 \ kg$ in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P \times 10^{11} \ Nm^{-2}$,where the value of $P$ is: (Take $g = 3 \pi \ m/s^2$)

$A$ steel wire of length $4.7\; m$ and cross-sectional area $3.0 \times 10^{-5}\; m^{2}$ stretches by the same amount as a copper wire of length $3.5\; m$ and cross-sectional area of $4.0 \times 10^{-5}\; m^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Two wires of the same length and radius are joined end-to-end and loaded. The Young's moduli of the materials of the two wires are $Y_{1}$ and $Y_{2}$. If the combination behaves as a single wire,then its equivalent Young's modulus is: