A) 60
B) 80
C) 120
D) 155
Correct Answer: D
Solution :
: In air, \[^{a}{{\mu }_{g}}=1.5\] In water, \[^{w}{{\mu }_{g}}=\frac{^{a}{{\mu }_{g}}}{^{a}{{\mu }_{w}}}=\frac{1.5}{1.33}=\frac{9}{8}\]approximately\[\left( 1.33=\frac{4}{3} \right)\] \[\therefore \] \[\frac{1}{{{f}_{a}}}={{(}^{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\frac{1}{{{f}_{w}}}={{(}^{w}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] Or \[\frac{{{f}_{w}}}{{{f}_{a}}}=\frac{^{a}{{\mu }_{g}}-1}{^{w}{{\mu }_{g}}-1}\]or \[\frac{{{f}_{w}}}{40}=\frac{1.5-1}{(9/8)-1}\] Or \[\frac{{{f}_{w}}}{40}=0.5\times 8\]or \[{{f}_{w}}=40\times 4\] Or \[{{f}_{w}}=160\,cm\] \[\therefore \] \[{{f}_{w}}\approx 155\,cm\]
You need to login to perform this action.
You will be redirected in
3 sec
