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STD 11 - 7. gravitation — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 7. gravitation4 Marks
MCQ
A solid sphere of radius $R$ gravitationally attracts a particle placed at $3 R$ form its centre with a force $F _{1}$. Now a spherical cavity of radius $\left(\frac{ R }{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F _{2}$. The value of $F _{1}: F _{2}$ is
  • A
    $25: 36$
  • B
    $36: 25$
  • ✓
    $50: 41$
  • D
    $41: 50$
✓

Answer

Correct option: C.
$50: 41$
c
Let initial mass of sphere is $m ^{\prime} .$ Hence mass of removed portion will be $m ^{\prime} / 8$

$F _{1}= m . E .=\frac{ m \cdot Gm ^{\prime}}{9 R ^{2}}$

$F_{2}=m\left[\frac{G \cdot m^{\prime}}{(3 R)^{2}}-\frac{G \cdot m^{\prime} / 8}{(5 R / 2)^{2}}\right]$

$=\frac{G m^{\prime}}{9 R^{2}}-\frac{G m^{\prime} \times 4}{8 \times 25}$

$=\left(\frac{1}{9}-\frac{1}{50}\right) \frac{G m^{\prime}}{R^{2}}$

$F_{2}=\frac{41}{50 \times 9} \cdot \frac{G m^{\prime}}{R^{2}}$

$\frac{F_{1}}{F_{2}}=\frac{1}{9} \times \frac{50 \times 9}{41}=\frac{50}{41}$

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