The earth, radius 6400 km. makes one revolution about its own axis in 24 hours. The centripetal acceleration of a point on its equator is nearly.

340 \frac{c m}{s e c^{2}}

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3.4 \frac{c m}{s e c^{2}}

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34 \frac{c m}{s e c^{2}}

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0.34 \frac{c m}{s e c^{2}}

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Solution

The correct option is B $3.4 \frac{c m}{s e c^{2}}$
Time period of earth's rotation is , $T = 24 h = 86400$ s
Angular velocity, $ω = 2 π / T$
Radius of earth, $r = 6400 k m = 640000 m$
Now, centripetal acceleration $ω 2 r = \frac{(2 π)}{T} 2 r = 0.034 m / s^{2} = 3.4 \frac{c m}{s^{2}}$

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The earth, radius 6400 km. makes one revolution about its own axis in 24 hours. The centripetal acceleration of a point on its equator is nearly.

The correct option is B 3.4cmsec2Time period of earth's rotation is ,T=24h=86400 sAngular velocity, ω=2π/TRadius of earth, r=6400km=640000mNow, centripetal acceleration ω2r=(2π)T2r=0.034m/s2=3.4cms2

The correct option is B $3.4 \frac{c m}{s e c^{2}}$
Time period of earth's rotation is , $T = 24 h = 86400$ s
Angular velocity, $ω = 2 π / T$
Radius of earth, $r = 6400 k m = 640000 m$
Now, centripetal acceleration $ω 2 r = \frac{(2 π)}{T} 2 r = 0.034 m / s^{2} = 3.4 \frac{c m}{s^{2}}$