$5 \,m$ long aluminium wire $(Y = 7 \times 10^{10} \,N/m^2)$ of diameter $3 \,mm$ supports a $40 \,kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} \,N/m^2)$ of the same length under the same weight, the diameter should be (in $mm$):

$A$ ring of a thin wire of cross-sectional area $a$ and length $l$ is dipped into a liquid of surface tension $\sigma$ and taken out so that a film of liquid is formed in the ring. If Young's modulus of the material of the wire is $Y$,then the longitudinal strain developed in the ring will be:

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:

$A$ wire elongates by $l \ mm$ when a load $W$ is hanged from it. If the wire goes over a pulley and two weights $W$ each are hung at the two ends,the elongation of the wire will be (in $mm$)

$A$ steel wire is stretched with a definite load. If the Young's modulus of the wire is $Y$,how can the value of $Y$ be decreased?

$A$ 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: