If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh   \frac{3}{4}    W.  Radius of the Earth is 6400 km and g=10 m/s2.

  • Option 1)

    1.1×10−3 rad/s

  • Option 2)

     0.83×10−3 rad/s

  • Option 3)

     0.63×10−3 rad/s

  • Option 4)

     0.28×10−3 rad/s

     

 

Answers (1)

As we learnt in

Variation in 'g' due to Rotation of earth -

g'=g-\omega ^{2}R\cos ^{2}\lambda

\lambda \rightarrow latitude angle

\omega \rightarrow Angular velocity of rotation of earth

g' \rightarrow New value of 'g'.

 

 

- wherein

Apparent weight of body decrease with rotation of earth so value of g also decrease.

 

 g'= g-\omega^2R\cos ^2\theta \Rightarrow \frac{3g}{4}= g-\omega^2R

At equator \theta = 0

\frac{3}{4}g - g=\: \omega^2R

\omega^2R=\frac{g}{4}

\omega^2R=\frac{g}{4R}\Rightarrow \omega=\sqrt{\frac{g}{4R}}=\sqrt{\frac{10}{4\times 6400\times 10^3}}

\omega^2= 0.6\times 10^-3 \:rad/sec


Option 1)

1.1×10−3 rad/s

Incorrect

Option 2)

 0.83×10−3 rad/s

Incorrect

Option 3)

 0.63×10−3 rad/s

Correct

Option 4)

 0.28×10−3 rad/s

 

Incorrect

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