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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Physics
Chapter
Gravitation

JEE Main 2018 · 15 April · Morning Shift · Online · 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 0.4×106 km0.4 \times 10^6 \mathrm{~km} and 150×106 km150 \times 10^6 \mathrm{~km} respectively. Their masses are 8×1022 kg8 \times 10^{22} \mathrm{~kg} and 2×2 \times 1030 kg10^{30} \mathrm{~kg} respectively. The radius of the earth is 6400 km6400 \mathrm{~km}. Let ΔF1\Delta F_1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2\Delta \mathrm{F}_2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2\frac{\Delta F_1}{\Delta F_2} is:

Options

  1. A22
  2. B66
  3. C10−210^{-2}
  4. D0.60.6

Correct answer

A. 22

Step-by-step solution

As we know, Gravitational force of attraction, F=GMmR2 F1=GMemr12\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^2}\mathrm{~F}_1=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^2} and F2=GMeMsr22ΔF1=2GMemr13Δr1\mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^2} \Delta \mathrm{F}_1=\frac{2 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^3} \Delta \mathrm{r}_1 and ΔF2=GMeMsr23Δr2\Delta \mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^3} \Delta \mathrm{r}_2 Using Δr1=Δr2=2Rearth ;m=8×1022 kg\Delta \mathrm{r}_1=\Delta \mathrm{r}_2=2 \mathrm{R}_{\text{earth }} ; \mathrm{m}=8 \times 10^{22} \mathrm{~kg}; Ms=2×1030 kgr1=0.4×106 km2\mathrm{M}_{\mathrm{s}}=2 \times 10^{30} \mathrm{~kg}\mathrm{r}_1=0.4 \times 10^6 \mathrm{~km}^2 and r2=150×106 kmΔF1ΔF2=(8×10222×1030)(150×1060.4×106)3×1≅2\mathrm{r}_2=150 \times 10^6 \mathrm{~km} \frac{\Delta \mathrm{F}_1}{\Delta \mathrm{F}_2}=\left(\frac{8 \times 10^{22}}{2 \times 10^{30}}\right)\left(\frac{150 \times 10^6}{0.4 \times 10^6}\right)^3 \times 1 \cong 2
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