Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
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: (1) \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\]if\[{{r}_{1}}<R\]and\[{{r}_{2}}<R\] (2) \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\]if\[{{r}_{1}}>R\]and\[{{r}_{2}}>R\] (3) \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\]if\[{{r}_{1}}>R\]and\[{{r}_{2}}>R\] (4) \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\]if\[{{r}_{1}}<R\]and\[{{r}_{2}}<R\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: A
Solution :
\[F=\frac{GM'}{r{{'}^{2}}};\]for\[r<R;M'=\]mass of the sphere of radius \[r'=\frac{4}{3}\pi {{r}^{3}}\delta \] But \[\delta =\frac{M}{\frac{4}{3}\pi {{R}^{3}}}\] \[\therefore \] \[M'=\frac{4}{3}\pi {{r}^{3}}\times \frac{M}{\frac{4}{3}\pi {{R}^{3}}}=\frac{M{{r}^{3}}}{{{R}^{3}}}\] \[\therefore \] \[F=GM\frac{{{r}^{3}}}{{{R}^{3}}}\times \frac{1}{{{r}^{2}}}=\frac{GM}{{{R}^{3}}}\times r\] Or \[F\propto r\] So, \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] For, \[r>R,\]we have \[F=\frac{GM}{{{r}^{2}}}\]or\[F\propto \frac{1}{{{r}^{2}}}\] \[\therefore \] \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
