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JEE Main 201815 April
Exam
JEE Main
Year
2018
Date
15 April
Subject
Physics
Chapter
Gravitation

JEE Main 2018 · 15 April · Physics · Gravitation

Take the mean distance of the moon and the sun from the earth to be...

Take the mean distance of the moon and the sun from the earth to be 0.4×106km0.4 \times {10}^{6}km and 150×106km150 \times {10}^{6}km, respectively. Their masses are 8×10228 \times {10}^{22} kgkg and 2×10302 \times {10}^{30} kgkg, respectively. The radius of the earth is 6400km6400km. Let ΔF1\Delta{F}_{1} be the difference in the forces exerted by the moon at the nearest and farthest point on the earth, and ΔF2\Delta{F}_{2} be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2\dfrac{\Delta{F}_{1}}{\Delta{F}_{2}} is,

Options

  1. A66
  2. B10−2{10}^{- 2}
  3. C22
  4. D0.60 . 6

Correct answer

C. 22

Step-by-step solution

F1=GMemr13,F2=GMeMsr22{F}_{1} = \dfrac{G{M}_{e}m}{{r}_{1}^{3}},{F}_{2} = \dfrac{G{M}_{e}{M}_{s}}{{r}_{2}^{2}}

ΔF1=(−)2GMemr13Δr1,ΔF2=(−)2GMeMsr22Δr2\Delta{F}_{1} = (-) \dfrac{2G{M}_{e}m}{{r}_{1}^{3}}\Delta{r}_{1},\Delta{F}_{2} =(-) \dfrac{2G{M}_{e}{M}_{s}}{{r}_{2}^{2}}\Delta{r}_{2}

ΔF1ΔF2=mΔr1r13r23MsΔr2=(mMs)(r23r13)(Δr1Δr2)\dfrac{\Delta{F}_{1}}{\Delta{F}_{2}} = \dfrac{m\Delta{r}_{1}}{{r}_{1}^{3}}\dfrac{{r}_{2}^{3}}{{M}_{s}\Delta{r}_{2}} = \left(\dfrac{m}{{M}_{s}}\right)\left(\dfrac{{r}_{2}^{3}}{{r}_{1}^{3}}\right)\left(\dfrac{\Delta{r}_{1}}{\Delta{r}_{2}}\right)

Using Δr1=Δr2=2Rearth\Delta{r}_{1} = \Delta{r}_{2} = 2{R}_{earth}

m=8×1022kgm = 8 \times {10}^{22}kg

Ms=2×1030kgMs = 2 \times {10}^{30}kg

r1=0.4×106km{r}_{1} = 0.4 \times {10}^{6}km

r2=150×106km{r}_{2} = 150 \times {10}^{6}km

We get ΔF1ΔF2=2\dfrac{\Delta{F}_{1}}{\Delta{F}_{2}} = 2.
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