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>>Variation in Value of Acceleration due to Gravity
>> $A planet of radius R = 110 \times (radius of$

Question

A planet of radius $R = \frac{1}{1 0} \times (r a d i u s o f E a r t h)$ has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $1 0^{- 3} k g m^{- 1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 1 0^{6} m$ and the acceleration due to gravity of Earth is 10 $^{- 2}$ )

A

$96 N$

B

$108 N$

C

$120 N$

D

$150 N$

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JEE Advanced

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Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is B)

$g_{i n n e r} = g_{s u r f a c e} \frac{r}{R}$
$= \frac{4}{3} G π r ρ$
Force to keep the wire at rest (F)
$=$ weight of wire
$\int_{\frac{4 R}{5}}^{R} λ \times g_{i n n e r} d r$
Putting the values and solving we get,
$F = 108 N$

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