Answer:
The diameter is 0.378 ft.
Step-by-step explanation:
Given that,
Mass of shot = 12 lb
Density of fresh water = 62.4 lb/ft
Specific gravity = 6.8
We need to calculate the volume of shot
![V = (4)/(3)\pi r^3\ ft^3](https://img.qammunity.org/2020/formulas/physics/college/qzzgwhbzmeyp72r66y48mucbxw9af8hhak.png)
The density of shot is
Using formula of density
![\rho = (m)/(V)](https://img.qammunity.org/2020/formulas/physics/college/th3pnkahw5mbco887xsmksuvg9g4xkojvv.png)
Put the value into the formula
![\rho =(12)/( (4)/(3)\pi r^3)](https://img.qammunity.org/2020/formulas/physics/college/4e8g44aj6kg26ejuw50mj1d3m148qkevsz.png)
We need to calculate the radius
Using formula of specific gravity
![specific\ gravity =(density\ of\ shot)/(dnsity\ of\ water)](https://img.qammunity.org/2020/formulas/physics/college/nty4mb23nhqcn6rgjhr3ne3llpqyoafjpp.png)
Put the value into the formula
![6.8=((12)/((4)/(3)\pi r^3))/(62.4)](https://img.qammunity.org/2020/formulas/physics/college/bltiy1ko33mt71nxjhbr0ner97e1xhgg3w.png)
![r^3=(12)/((4)/(3)\pi*6.8*62.4)](https://img.qammunity.org/2020/formulas/physics/college/zp1twp0pkfcqobyke6d2x006crctgwyvkr.png)
![r^3=0.0067514](https://img.qammunity.org/2020/formulas/physics/college/x7atrcedla9m156t0ql98n14vbujoq7v7x.png)
![r =(0.0067514)^{(1)/(3)}](https://img.qammunity.org/2020/formulas/physics/college/ebd3tvgzf0le6edyvqtirerfaet8skr6yj.png)
![r=0.1890\ ft](https://img.qammunity.org/2020/formulas/physics/college/cw8ykxxdf27uu67ycgmwj0w4tgkqb4oyas.png)
The diameter will be
![d = 2* r](https://img.qammunity.org/2020/formulas/physics/college/rjotvl88ceew1406gnrqdt241i90a49nib.png)
![d =2*0.1890](https://img.qammunity.org/2020/formulas/physics/college/p6x7gh1qe2vilaui6eubp8shebijjc6tq4.png)
![d =0.378\ ft](https://img.qammunity.org/2020/formulas/physics/college/avlxqpxl8kmf56mh21yayfh46z2p7fyhn6.png)
Hence, The diameter is 0.378 ft.