Answer:
378 atmospheres
Explanation:
Solution :
Here, forceo n the phonogram needle, `F=1.2N`
radius of the circular needle, `r=0.1mm=0.1xx10^-3`m
area of cross-section of the needle, `A=pir^2`
`=3.142xx(0.1xx10^-3)^2=3.142xx10^-8m^2`
pressure exerted on the record in Pa. `P=(P)/(A)`
`=((1.2N))/((3.142xx10^-8m^2))=382xx10^5(N)/(m^)`
pressure exerted on the record in atm
`(382xx10^5(N)/(m^2))/((1.013xx10^5(N)/(m^2))/(atm))`
`=378atm`
