$A$ steel ring of radius '$r$' is to be fitted over a wooden disc of radius '$R$' $(R > r)$. The force required to expand the ring so that it fits over the disc is ($Y =$ Young's modulus of steel,$A =$ area of cross-section of the wire).

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 $= 2 \times 10^{11} \ N/m^{2}$)

$A$ weight of $200 \, kg$ is suspended by a vertical wire of length $600.5 \, cm$. The area of cross-section of the wire is $1 \, mm^2$. When the load is removed,the wire contracts by $0.5 \, cm$. The Young's modulus of the material of the wire will be:

$A$ 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

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

$A$ force of $10^3 \ N$ stretches the length of a hanging wire by $1 \ mm$. The force required to stretch a wire of the same material and length,but having four times the diameter,by $1 \ mm$ is: