What must be the lengths of steel and copper rods at $0^o C$ for the difference in their lengths to be $10\,cm$ at any common temperature? $(\alpha_{steel}=1.2 \times {10^{-5}} \;^o C^{-1})$ and $(\alpha_{copper} = 1.8 \times 10^{-5} \;^o C^{-1})$
$30\, cm$ for steel and $20\,cm$ for copper
$20\,cm$ for steel and $30\,cm$ for copper
$40\,cm$ for steel and $30\,cm$ for copper
$30\,cm$ for steel and $40\,cm$ for copper
Read the following two statements below carefully and state, with reasons, if it is true or false.
$(a)$ The Young’s modulus of rubber is greater than that of steel;
$(b)$ The stretching of a coil is determined by its shear modulus.
The modulus of elasticity is dimensionally equivalent to
In $CGS$ system, the Young's modulus of a steel wire is $2 \times {10^{12}}$. To double the length of a wire of unit cross-section area, the force required is
A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $F$, its length increases by $5\,cm$. Another wire of the same material of length $4 L$ and radius $4\,r$ is pulled by a force $4\,F$ under same conditions. The increase in length of this wire is $....cm$.
A wooden wheel of radius $R$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross section area $S$ and length $L$. $L$ is slighly 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 steps over the wheel.As it cools down to surronding temperature, it presses the semicircular parts together. If the coefficint of linear expansion of the metal is $\alpha$, and its young's modulus is $Y$, the force that one part of wheel applies on the other part is