Gravitation Ques 10
- The magnitudes of the gravitational field at distance $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_{1}$ and $F_{2}$, respectively. Then
$(1994,2 \mathrm{M})$
(a) $\frac{F_{1}}{F_{2}}=\frac{r_{1}}{r_{2}}$ if $r_{1}<R$ and $r_{2}<R$
(b) $\frac{F_{1}}{F_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}$ if $r_{1}>R$ and $r_{2}>R$
(c) $\frac{F_{1}}{F_{2}}=\frac{r_{1}^{3}}{r_{2}^{3}}$ if $r_{1}<R$ and $r_{2}<R$
(d) $\frac{F_{1}}{F_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}$ if $r_{1}<R$ and $r_{2}<R$
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Answer:
Correct Answer: 10.(a,b)
Solution:
Formula:
Variation Of Acceleration Due To Gravity :
- For $r \leq R, F=\frac{G M}{R^{3}} \cdot r$
or $F \propto r$
$$ \frac{F_{1}}{F_{2}}=\frac{r_{1}}{r_{2}} \quad \text { for } \quad r_{1}<R $$
and $\quad r_{2}<R$
and for $r \geq R, F=\frac{G M}{r^{2}}$
or $\quad F \propto \frac{1}{r^{2}}$
i.e. $\quad \frac{F_{1}}{F_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}$ for $r_{1}>R$
and $\quad r_{2}>R$
BETA
