A) \[\frac{{{\rho }_{1}}}{{{\rho }_{w}}({{\rho }_{2}}-{{\rho }_{1}})}[{{w}_{1}}({{\rho }_{2}}-{{\rho }_{w}})-{{w}_{2}}{{\rho }_{2}}]\]
B) \[\frac{{{\rho }_{1}}}{{{\rho }_{w}}({{\rho }_{2}}+{{\rho }_{1}})}[{{w}_{1}}({{\rho }_{2}}-{{\rho }_{w}})+{{w}_{2}}{{\rho }_{2}}]\]
C) \[\frac{{{\rho }_{1}}}{{{\rho }_{w}}({{\rho }_{2}}-{{\rho }_{1}})}[{{w}_{1}}({{\rho }_{2}}+{{\rho }_{w}})-{{w}_{2}}{{\rho }_{1}}]\]
D) \[\frac{{{\rho }_{1}}}{{{\rho }_{w}}({{\rho }_{2}}-{{\rho }_{1}})}[{{w}_{1}}({{\rho }_{2}}-{{\rho }_{w}})-{{w}_{2}}{{\rho }_{1}}]\]
Correct Answer: A
Solution :
By Archimedes Principle \[F=v{{\rho }_{w}}g\Rightarrow \,\,({{w}_{1}}-{{w}_{2}})g=v{{\rho }_{w}}g\] Let, the total volume be vand first metal weight be \[x\] \[{{w}_{1}}-{{w}_{2}}({{v}_{1}}+{{v}_{2}}){{\rho }_{w}}\] \[{{w}_{1}}-{{w}_{2}}={{v}_{1}}{{\rho }_{w}}+{{v}_{2}}{{\rho }_{w}}\] \[\left( \because \,v=\frac{m}{\rho } \right)\] \[{{w}_{1}}-{{w}_{2}}=\left( \frac{x}{{{\rho }_{1}}}{{\rho }_{w}}+\frac{{{w}_{1}}-x}{{{\rho }_{2}}}{{\rho }_{w}} \right)\] \[{{w}_{1}}-{{w}_{2}}=\frac{x{{\rho }_{2}}{{\rho }_{w}}+({{w}_{1}}-x){{\rho }_{w}}{{\rho }_{1}}}{{{\rho }_{1}}\,{{\rho }_{2}}}\] \[{{w}_{1}}{{\rho }_{1}}{{\rho }_{2}}-{{w}_{2}}{{\rho }_{1}}{{\rho }_{2}}=x{{\rho }_{2}}{{\rho }_{w}}+{{w}_{1}}{{\rho }_{w}}{{\rho }_{1}}-x{{\rho }_{w}}{{\rho }_{1}}\] \[x({{\rho }_{2}}-{{\rho }_{1}}){{\rho }_{w}}={{\rho }_{1}}[{{w}_{1}}({{\rho }_{2}}-{{\rho }_{w}})-{{w}_{2}}{{\rho }_{2}}]\] \[x=\frac{{{\rho }_{1}}}{{{\rho }_{w}}({{\rho }_{2}}-{{\rho }_{1}})}[{{w}_{1}}({{\rho }_{2}}-{{\rho }_{w}})-{{w}_{2}}{{\rho }_{2}}]\]You need to login to perform this action.
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