Force on a phonograph needle is 120 gf. The needle end has a circular cross-section of radius 0.1 mm. Find the pressure (in atm) it exerts on the record. Given,
1 atm = `1.013 xx 10^(5) Pa. Use g=10ms^(-2)`

Explanation

Here, `F=120 gf = 0.120 kg f`
`0.120 xx 10 = 1.2 N`
`r=0.1 mm =10^(-4)m`
Pressure , `P=F/(pi r^(2)) = (1.2)/((22//7)xx(10^(-4))^(2)) Pa`
`=(1.2 xx 7)/(22 xx 10^(-8)xx 1.013 xx 10^(5))atm`
`=377atm`.