NCERT Solutions: Class 11 Physics Chapter 8 - Mechanical Properties of Solids

Extension is $\Delta L=FL/(AY)$. For the same load and same extension, $L_s/(A_sY_s)=L_c/(A_cY_c)$. Therefore $Y_s/Y_c=L_sA_c/(A_sL_c)=\frac{4.7\times4.0\times10^{-5}}{3.0\times10^{-5}\times3.5}=1.79$.

Steel has a much larger Young's modulus than rubber; rubber stretches more for the same stress, so its Young's modulus is smaller. A coil spring stretches mainly because the wire of the coil undergoes twisting/shearing, so its extension is governed by the shear modulus of the material.

The cross-sectional area is $A=\pi(0.0025/2)^2=4.91\times10^{-6}\,\text{m}^2$. From Fig. 8.11, the steel wire carries both masses, so $F_s=(4+6)g=98\,\text{N}$, while the brass wire carries the 6 kg mass, so $F_b=58.8\,\text{N}$. Using $\Delta L=FL/(AY)$ with $Y_s=2.0\times10^{11}\,\text{N m}^{-2}$ and $Y_b=0.91\times10^{11}\,\text{N m}^{-2}$ gives $\Delta L_s=1.50\times10^{-4}\,\text{m}$ and $\Delta L_b=1.32\times10^{-4}\,\text{m}$.

For shear, $F/A=G(\Delta L/L)$. Here $F=100g=980\,\text{N}$, $A=(0.10)^2=0.01\,\text{m}^2$, $L=0.10\,\text{m}$, and $G=25\times10^9\,\text{Pa}$. Thus $\Delta L=FL/(AG)=980(0.10)/(0.01\times25\times10^9)=3.92\times10^{-7}\,\text{m}$.

Each column carries one quarter of the load: $F=50000g/4=1.225\times10^5\,\text{N}$. Cross-sectional area of each hollow column is $A=\pi(R_o^2-R_i^2)=\pi(0.60^2-0.30^2)=0.848\,\text{m}^2$. Stress $=F/A=1.44\times10^5\,\text{N m}^{-2}$. With Young's modulus for mild steel $Y\approx2.0\times10^{11}\,\text{N m}^{-2}$, strain $=\text{stress}/Y=7.2\times10^{-7}$.

Area $A=(15.2\times10^{-3})(19.1\times10^{-3})=2.90\times10^{-4}\,\text{m}^2$. Stress $=F/A=44500/(2.90\times10^{-4})=1.53\times10^8\,\text{N m}^{-2}$. With $Y_{Cu}\approx1.1\times10^{11}\,\text{N m}^{-2}$, strain $=\text{stress}/Y=1.39\times10^{-3}$.

Maximum load is $F=\sigma A=10^8\pi r^2$. With $r=1.5\,\text{cm}=0.015\,\text{m}$, $F=10^8\pi(0.015)^2=7.07\times10^4\,\text{N}$.

For the rigid bar to remain level, the extensions of the equal-length wires must be the same. If each wire has the same tension, $\Delta L=FL/(AY)$ implies $AY$ must be the same for copper and iron. Since $A\propto d^2$, $d_{Cu}^2Y_{Cu}=d_{Fe}^2Y_{Fe}$. Thus $d_{Cu}/d_{Fe}=\sqrt{Y_{Fe}/Y_{Cu}}=\sqrt{1.9\times10^{11}/1.1\times10^{11}}=1.31$.

At the lowest point, $T-mg=m\omega^2r$, so $T=m(g+\omega^2r)$. With $\omega=2(2\pi)=4\pi\,\text{rad s}^{-1}$ and $r\approx1.0\,\text{m}$, $T=14.5[9.8+(4\pi)^2(1.0)]=2.43\times10^3\,\text{N}$. Area $A=0.065\,\text{cm}^2=6.5\times10^{-6}\,\text{m}^2$. Using $\Delta L=TL/(AY)$ with $Y_s=2.0\times10^{11}\,\text{N m}^{-2}$ gives $\Delta L=1.87\times10^{-3}\,\text{m}$.

For small compression, $\Delta V/V=\Delta p/B$, so density increases as $\rho=\rho_0/(1-\Delta p/B)$. Taking $B\approx2.2\times10^9\,\text{Pa}$ for water and $\Delta p\approx80(1.013\times10^5)=8.10\times10^6\,\text{Pa}$, $\rho=1.03\times10^3/[1-(8.10\times10^6)/(2.2\times10^9)]=1.034\times10^3\,\text{kg m}^{-3}$.

For glass, take $B\approx37\times10^9\,\text{Pa}$. The pressure is $10(1.013\times10^5)=1.013\times10^6\,\text{Pa}$. Hence $|\Delta V|/V=p/B=1.013\times10^6/(37\times10^9)=2.74\times10^{-5}$.

The cube volume is $V=(0.10)^3=1.0\times10^{-3}\,\text{m}^3$. For copper, take $B\approx140\times10^9\,\text{Pa}$. Volume contraction is $\Delta V=pV/B=(7.0\times10^6)(1.0\times10^{-3})/(140\times10^9)=5.0\times10^{-8}\,\text{m}^3$.

For water, $B\approx2.2\times10^9\,\text{Pa}$. A compression of $0.10\%$ means $|\Delta V|/V=0.001$. Therefore $\Delta p=B(|\Delta V|/V)=2.2\times10^9\times0.001=2.2\times10^6\,\text{Pa}$, which is about $2.2\times10^6/(1.013\times10^5)\approx22\,\text{atm}$.