A) \[{{\gamma }_{g}}+\left( \frac{{{M}_{2}}-{{M}_{1}}}{M-{{M}_{2}}} \right).\frac{1}{\left( {{t}_{2}}-{{t}_{1}} \right)}\]
B) \[{{\gamma }_{g}}-\left( \frac{{{M}_{2}}-{{M}_{1}}}{M-{{M}_{2}}} \right).\frac{1}{\left( {{t}_{2}}-{{t}_{1}} \right)}\]
C) \[{{\gamma }_{g}}-\left( \frac{M-{{M}_{2}}}{{{M}_{2}}-{{M}_{1}}} \right).\frac{1}{\left( {{t}_{2}}-{{t}_{1}} \right)}\]
D) \[{{\gamma }_{g}}+\left( \frac{{{M}_{2}}-{{M}_{1}}}{{{M}_{2}}+{{M}_{1}}} \right).\frac{1}{\left( {{t}_{2}}-{{t}_{1}} \right)}\]
Correct Answer: A
Solution :
[a]\[{{M}_{1}}g=Mg-{{V}_{1}}\rho {{\ell }_{1}}g\] | |
or \[{{M}_{1}}g=Mg-{{V}_{1}}{{\rho }_{1}}g\] | ...(i) |
and \[{{M}_{2}}g=Mg-{{V}_{1}}[1+{{\gamma }_{g}}({{t}_{2}}-{{t}_{1}})]\frac{\rho }{[1+\gamma \ell ({{t}_{2}}-{{t}_{1}})]}g\] | ?(ii) |
After simplifying, we get \[{{\gamma }_{\ell }}={{\gamma }_{g}}+\left( \frac{{{M}_{2}}-{{M}_{1}}}{M-{{M}_{2}}} \right)\frac{1}{\left( {{t}_{2}}-{{t}_{1}} \right)}\] |
You need to login to perform this action.
You will be redirected in
3 sec