The speed with the earth have to rotate on its axis so that a person on the equator would weight (3/5)th as much as present will be (Take the equilateral radius as 6400 km.)

Question

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Answer 1

Actual weight on the equator $= W = m g$
where 'm' is the mass and 'g' is the gravity $.$
According to the given condition $,$
Weight on the equator $= W^{'} = m g^{'}$

Weight on the equator $= W^{'} = 3 / 5 m g$ ........... $(1)$
We know that, $λ = 0$ at the equator.
Now $,$
$m g^{'} = m g - m R ω ² c o s λ$
$3 / 5 m g = m g - m R ω ² c o s 0 (∵ λ = 0)$
$3 / 5 m g = m g - m R ω ²$ $(1)$ $(∵ c o s 0 = 1)$
$3 / 5 m g = m g - m R ω ²$
$m R ω ² = m g - 3 / 5 m g$
$m R ω ² = (1 - 3 / 5) m g$
$m R ω ² = 2 / 5 m g$
$R ω ² = 2 / 5 g$
$ω ² = 2 g / 5 R$
$ω ² = (2 x 10) / (5 x 6.4 x 10 ⁶) (∵ R = R a d i u s o f E a r t h = 6400 k m = 6.4 x 10 ⁶ m)$
$ω = \sqrt (6.25 x 10 ⁻ ⁷)$
$ω = 7.8 x 10 ⁻ ⁴ r a d / s e c$
Hence,
option $(B)$ is correct answer.

Answer 2

Actual weight on the equator

= W = m g

where 'm' is the mass and 'g' is the gravity

.

According to the given condition

,

Weight on the equator

= W^{'} = m g^{'}

Weight on the equator

= W^{'} = 3 / 5 m g

...........

(1)

We know that,

λ = 0

at the equator.

Now

,

m g^{'} = m g - m R ω ² c o s λ

3 / 5 m g = m g - m R ω ² c o s 0 (∵ λ = 0)

3 / 5 m g = m g - m R ω ²

(1)

(∵ c o s 0 = 1)

3 / 5 m g = m g - m R ω ²

m R ω ² = m g - 3 / 5 m g

m R ω ² = (1 - 3 / 5) m g

m R ω ² = 2 / 5 m g

R ω ² = 2 / 5 g

ω ² = 2 g / 5 R

ω ² = (2 x 10) / (5 x 6.4 x 10 ⁶) (∵ R = R a d i u s o f E a r t h = 6400 k m = 6.4 x 10 ⁶ m)

ω = \sqrt (6.25 x 10 ⁻ ⁷)

ω = 7.8 x 10 ⁻ ⁴ r a d / s e c

Hence,

option

(B)

is correct answer.