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InterviewSolution
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1.3 A spherical hollow cavity is made in a lead sphere of radius R, suchthat its surface touches the outside surface of the lead sphere andpasses through its centre. The mass of the sphere before hollowingwas M. With what gravitational force will the hollowed-out lead sphereattract a small sphere of mass which lies at a distance d from thecentre of the lead sphere on the straight line connecting the centres ofthe spheres and that of the hollow, if d 2R7GMmD)7GMm18R27GMm36R27GMm(C) 9RA)(B)6the Y |
| Answer» <p> Concept:</p><p>The force of gravity on the small sphere of mass due to the lead sphere after it is made hollow is equal to the force of gravity from a solid lead sphere minus the force which would have been contributed by the smaller lead sphere which would have filled the hole.</p><p>Solution:</p><p>From the Newton’s law of universal gravitation, the force of attraction between the lead sphere and the small sphere is</p><p>F1=GMm/d2</p><p>Here, gravitational constant isG, total mass of the lead sphere isM, mass of the small sphere ismand distance between the centers of the spheres isd.</p><p>So, we need to know about the size and mass of the lead which was removed to make the hole.</p><p>The density of the lead is given by</p><p>ρ=M/[(4/3)(πR3)]</p><p>Here, radius of the lead sphere isR.</p><p>The spherical sphere has a radius ofR/2.</p><p>If the density is constant, then the mass of the hole will be the density times the volume of the hollow sphere. Thus, the mass of the hollow sphere is</p> <p>in this equation , substitute the value of d=2R and solve </p> | |