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        The gravitational force on a small particle lying on earths surface is F. If a concentric spherical cavity of radius R/2 is made inside earth, then force on the particle is a F/2  b. F/8  c. 7F/8  d. 17F/8
5 months ago

Ashutosh Kumar Singh
28 Points
							$F=mg=GM_{e}m/R^{2}$$g=\frac{GM_{e}}{R^2}$Mass = volume * densityafter removal of the cavity, the effective mass will be:$M'=M_{e}-\frac{4}{3}\Pi R^3\rho _{e}$$M'=\frac{7}{8}M_{e}$New gravitational acceleration will be:$g'=\frac{GM'}{R^2}=\frac{7}{8}g$New force will be:$F'=mg'=m*\frac{7}{8}g =\frac{7}{8}F$Note - while calculating the gravitational acceleration the effective distance is still R because the position of the center of mass is not changing only effective mass is changing.

4 months ago
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### Course Features

• 57 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions