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What is the percentage change in gravity at the depth of 1/32 times of radius of earth. (In %)

Option: 1

3.125


Option: 2

1.75


Option: 3

1.25


Option: 4

3.75


Answers (1)

best_answer

 As we learnt

 

Percentage decrease in value of 'g' -

 

\frac{\Delta g}{g}\times 100\: ^{0}\! \! /\! _{0}=\frac{d}{R}\times 100\: ^{0}\! \! /\! _{0}

\Delta g\rightarrow Variation in 'g'

R\rightarrow Radius

g\rightarrow 9.8\, m/s^{2}

 

 

 

 

Using the relation,

\frac{\Delta g}{g}\times 100\: ^{0}\! \! /\! _{0}=\frac{d}{R}\times 100\: ^{0}\! \! /\! _{0}= \frac{R/32}{R}*100= \frac{100}{32}=3.125

Posted by

rishi.raj

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