Calculation of true thickness of the bed

 CSIR-NET JUNE 2018

125.If in low pressure center of the northern hemisphere, winds are from the north at 10 m.s^-1 at a distance 250 km from the low pressure center, the centrifugal force per unit mass of air is

1. -4*10^-5 m.s^-2
2. -4*10^-4 m.s^-2
3. -4*10^-3 m.s^-2
4. -4*10^-6 m.s^-2

Solution:

Centrifugal Force: When a body of mass rotates about an axis it exerts an outward radial force called centrifugal force.

Centrifugal force (F) = $m\omega^{2}r$ ----------(1)

 $\omega^{2}= \upsilon^2/r^2$ -------(2)

Substitute eq(2) value in eq(1)

F = $m\frac{ \upsilon^2}{r^2} r$ -----(3)

In given problem The centrifugal force per unit mass 

Therefore,

$\frac{F}{m}= \frac{ \upsilon^2}{r}$

 Velocity (v) = 10 $\frac{m}{s^2}$

Distance (r) = 250 km = 250000 m

Substitute these values in eq(3)

$\frac{F}{m}= \frac{ 10^2}{250000}$

$\frac{F}{m}= 0.0004  \frac{m}{s^2}$

$\frac{F}{m}= 4*10^-4  \frac{m}{s^2}$