Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
     F = [tex]5.20 \times 10^{5}[/tex] N
     g = 9.8 m/s
     radius = [tex]\frac{diameter}{2}[/tex]
          = [tex]\frac{30 cm}{2}[/tex] = 15 cm = 0.15 m  (as 1 m = 100 cm)
Formula to calculate depth is as follows.
    F = [tex]\rho \times g \times h \times A[/tex]
or, Â Â Â h = [tex]\frac{F}{\rho \times g \times A}[/tex] Â Â Â Â
    h = [tex]\frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}[/tex]
      = 751 m
Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.
