Answer:
C) Pnew=Poriginal/9
Explanation:
From the given conditions:
Original area, [tex]A_o=\pi.r_o\,^2[/tex]
New area of the barrel bottom panel after increasing by a factor of 3:
[tex]r_{new}=3r_o[/tex]
On increasing the radius the area now becomes:
[tex]A_{new}=\pi.(3r_o)^2[/tex]
[tex]A_{new}=9\pi.(r_o)^2=9A_o[/tex]
As we know that there will be no increase in the down-force since the height remains constant.
Pressure is given as:
[tex]P_{original}=\frac{Force}{A_o}[/tex]
∴New pressure:
[tex]P_{new}=\frac{Force}{9A}[/tex]
[tex]\Rightarrow P_{new}=\frac{P_{original}}{9}[/tex]