The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in kPa

Pressure exerted by the needle = 78.037 kPa

Explanation:

Pressure is defined as the force exerted on a surface per unit area. The SI unit of pressure is Pa (N. m⁻²).

The force exerted by the phonograph needle is equal to the gravitational force.

Thus,

F=m*g

Where,

F is the force

m is the mass of the body

g is the acceleration due to gravity

Also, g = 9.81 ms⁻²

Given, mass of the needle = 1.00 g

The conversion of g into kg is shown below:

1 g = 10⁻³ kg

Thus, mass of the needle = 1*10⁻³ kg

Thus, F = 1*10⁻³*9.81 N = 9.81*10⁻³ N

Ares of the circle = πr²

Given : Radius of tip of the needle = 0.200 mm

The conversion of mm into m is shown below:

1 mm = 10⁻³ m

Thus, mass of the needle = 0.2*10⁻³ m

Thus, Area of the tip of the needle = π * (0.2*10⁻³) ² m² = 1.2571*10⁻⁷ m²

So, Pressure = F/A

P = (9.81*10⁻³) / 1.2571*10⁻⁷ Pa = 7.8037*10⁴ Pa

The conversion of Pa into kPa is shown below:

1 Pa = 10⁻³ kPa

Thus, pressure exerted by the needle = 7.8037*10⁴*10⁻³ kPa = 78.037 kPa