Question

The pressure required to stop the increase in volume of a copper block when it is heated from $ 50{}^\circ C $ to $ 70{}^\circ C $ . Coefficient of linear expansion of copper is $ 8\times {{10}^{-6}}/{}^\circ C $ and bulk modulus of elasticity $ =3.6\times {{10}^{11}}N/{{m}^{2}}, $ is:

• A. $ 2.8\times {{10}^{5}}N/{{m}^{2}} $
• B. $ 1.72\times {{10}^{8}}N/{{m}^{2}} $
• C. $ 6.3\times {{10}^{3}}N/{{m}^{2}} $
• D. $ 8\times {{10}^{-6}}N/{{m}^{2}} $
• E. $ 1.57\times {{10}^{4}}N/{{m}^{2}} $

Answer 1

Answer: (B)

$ B=\frac{Change\text{ }in\text{ }pressure}{Volume\text{ }strain}=\frac{p}{\gamma ({{t}_{2}}-{{t}_{1}})} $ $ \therefore $ $ p={{B}_{\gamma }}({{t}_{2}}-{{t}_{1}}) $ Given, $ B=3.6\times {{10}^{11}}N/{{m}^{2}}, $ $ \gamma =3\alpha =3\times 8\times {{10}^{-6}}=24\times {{10}^{-6}}/{}^\circ C $ $ {{t}_{2}}-{{t}_{1}}=70-50=20{}^\circ C $ $ \therefore $ $ p=(3.6\times {{10}^{11}})(24\times {{10}^{-6}})(20) $ $ =1.728\times {{10}^{8}}N/{{m}^{2}} $