Question

The depth 'd' at which the value of acceleration due to gravity becomes `"I"/"n"` times the value at the earth's surface is ______. (R = radius of earth)

Options

• d = R`("n"/("n"-1))`

• d = R`(("n"-1)/(2"n"))`

• d = R`(("n"-1)/"n")`

• d = R²`(("n"-1)/"n")`

Answer 1

The depth 'd' at which the value of acceleration due to gravity becomes `"I"/"n"` times the value at the earth's surface is `underlinebb("d" = "R"(("n"-1)/"n"))`. (R = radius of earth)

Explanation:

Acceleration due to gravity varies with depth d as,

g' = g `(1-"d"/"R")`

And according to question, g' = `"g"/"n"`

g' = `"g"/"n"="g"(1-"d"/"R")`

⇒ `1/"n"=1-"d"/"R"`

or, `"d"/"R"=1-1/"n"=("n"-1)/"n"`

∴ d = R`(("n"-1)/"n")`