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} $
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