As shown in the figure,a light uniform rod $PQ$ of length $150 \ cm$ is suspended from the ceiling horizontally using two metal wires $A$ and $B$ tied to the ends of the rod. The ratios of the radii and the Young's moduli of the materials of the two wires $A$ and $B$ are respectively $2:3$ and $3:2$. The position at which a weight should be suspended from the rod such that the elongations of the two wires become equal is

$A$ rod of length $l$ and area of cross-section $A$ is heated from $0^{\circ}C$ to $100^{\circ}C$. The rod is so placed that it is not allowed to increase in length,then the force developed is proportional to

$A$ sphere of mass $2 \,kg$ and diameter $4.5 \,cm$ is attached to the lower end of a steel wire of $2 \,m$ length and area of cross-section $0.24 \times 10^{-6} \,m^2$. The wire is suspended from a $205 \,cm$ high ceiling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest position. Find the velocity of the sphere at the lowest position. (Young's modulus of steel $= 2 \times 10^{11} \,Nm^{-2}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-1}$)

$A$ 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

To double the length of an iron wire having $0.5 \, cm^2$ area of cross-section,the required force will be $(Y = 10^{12} \, dyne/cm^2)$.

$A$ 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: