A rod of length $L$ at room temperature and uniform area of cross section $A$, is made of a metal having coefficient of linear expansion $\alpha {/^o}C$. It is observed that an external compressive force $F$, is applied on each of its ends, prevents any change in the length of the rod, when it temperature rises by $\Delta \,TK$. Young’s modulus, $Y$, for this metal is
A$\frac{F}{{A\alpha \,\Delta T}}$
B$\frac{F}{{A\alpha \,\left( {\Delta T - 273} \right)}}$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
The pressure that has to be applied to the ends of a steel wire of length $10\ cm$ to keep its length constant when its temperature is raised by $100^o C$ is: (For steel Young's modulus is $2 \times 10^{11}$ $Nm^{-1}$ and coefficient of thermal expansion is $1.1 \times 10^{-5}$ $K^{-1}$ )
How much force is required to produce an increase of $0.2\%$ in the length of a brass wire of diameter $0.6\, mm$ (Young’s modulus for brass = $0.9 \times {10^{11}}N/{m^2}$)
A brass rod of length $2\,m$ and cross-sectional area $2.0\,cm^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0\,cm^2$ . The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4\,N$ at its ends. If the elongations of the two rods are equal, then length of the steel rod $(L)$ is ........... $m$ $(Y_{Brass}=1.0\times 10^{11}\,N/m^2$ and $Y_{Steel} = 2.0 \times 10^{11}\,N/m^2)$
A glass slab is subjected to a pressure of $10\, atm$. The fractional change in its volume is (Bulk modulus of glass $= 37 \times 10^9\, N\, m^{-2}$, $1\, atm = 1 \times 10^5\, N\, m^{-2}$)
The work done in increasing the length of a metre long wire of cross-sectional area ........ $J.$ $1\,mm^2$ through $1\,mm$ will be $(Y = 2 \times 10^{11}\,Nm^{-2})$
The ratio of lengths of two rods $A$ and $B$ of same material is $1 : 2$ and the ratio of their radii is $2 : 1$, then the ratio of modulus of rigidity of $A$ and $B$ will be