SATHEE: Gravitation Question 43

Gravitation Question 43

Question: The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60o latitude becomes zero is (Radius of earth = 6400 km. At the poles $[g = 10 m s^{- 2})]$ [EAMCET 2000]

Options:

A) $[2.5 \times 10^{- 3} r a d / s]$

B) $[5.0 \times 10^{- 1} r a d / s]$

C) $[10 \times 10^{1} r a d / s]$

D) $[7.8 \times 10^{- 2} r a d / s]$

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Answer:

Correct Answer: A

Solution:

$[g^{'} = g - ω^{2} R \cos^{2} λ]$

Þ $[0 = g - ω^{2} R \cos^{2} 60^{o}]$

$[0 = g - \frac{ω^{2} R}{4} \Rightarrow ω = 2 \sqrt{\frac{g}{R}} = \frac{1}{400} \frac{r a d}{\sec}]$

$[= 2.5 \times 10^{- 3} \frac{r a d}{s e c}]$

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Gravitation Question 43

Question: The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60o latitude becomes zero is (Radius of earth = 6400 km. At the poles [g=10ms−2)] [EAMCET 2000]

Options:

Answer:

Solution:

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Question: The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60o latitude becomes zero is (Radius of earth = 6400 km. At the poles $[g = 10 m s^{- 2})]$ [EAMCET 2000]