Gravitation Question 106

Question: The magnitudes of the gravitational force at distances $ [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[IIT 1994]

Options:

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_{1}^{2}}{r_{2}^{2}}] $ if $ [r_{1}>R] $ and$ [r_{2}>R] $

C) $ [\frac{F_{1}}{F_{2}}=\frac{r_{1}}{r_{2}}] $ if $ [r_{1}>R] $ and$ [r_{2}>R] $

D) $ [\frac{F_{1}}{F_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}] $ if $ [r_{1}<R] $ and$ [r_{2}<R] $

Show Answer

Answer:

Correct Answer: A

Solution:

$ [g=\frac{4}{3}\pi \rho Gr] $ \ $ [g\propto r] $ if $ [r<R] $ $ [g=\frac{GM}{r^{2}}] $ \ $ [g\propto \frac{1}{r^{2}}] $ if $ [r>R] $ If $ [r_{1}<R] $ and $ [r_{2}<R] $ then$ [\frac{F_{1}}{F_{2}}=\frac{g_{1}}{g_{2}}=\frac{r_{1}}{r_{2}}] $ If $ [r_{1}>R] $ and $ [r_{2}>R] $ then$ [\frac{F_{1}}{F_{2}}=\frac{g_{1}}{g_{2}}={{\left( \frac{r_{2}}{r_{1}} \right)}^{2}}] $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक