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Physics

Mechanics

Easy

Question

The depth d at which the value of acceleration due to gravity becomes $1 over n$ times the value at the surface, is [R= radius of the earth]

$\frac{R}{n}$
$R (\frac{n - 1}{n})$
$\frac{R}{n^{2}}$
$R (\frac{n}{n + 1})$

The correct answer is: $R open parentheses fraction numerator n minus 1 over denominator n end fraction close parentheses$

We know $table attributes columnalign left end attributes row cell g to the power of apostrophe equals g open parentheses 1 minus d over R close parentheses rightwards double arrow g over n equals g open parentheses 1 minus d over R close parentheses end cell row cell rightwards double arrow d over R equals 1 minus 1 over n rightwards double arrow d equals open parentheses fraction numerator n minus 1 over denominator n end fraction close parentheses R end cell end table$

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