I believe is A) Inner core
When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).
The value of this critical angle can be derived by Snell's law, and it is equal to

where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In our problem, n1=1.47 and n2=1.33, so the critical angle is
Thew energy stored in a capacitor of capacitance
and voltage between the plates
is
.
Substituting numerical value

Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas

For a gas

Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)





Hence, The pressure of the remaining gas in the tank is 6.4 atm.