InterviewSolution
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InterviewSolution
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Statement-1 : A body released from a height equal to the radius `(R )` of the earth. The velocity of the body when it strikes the surface of the earth will be `sqrt(2 g R)`. Statement -2 : As `upsilon^(2) = u^(2) + 2` as.A. Statement -1 is true , Statement -2 is true , Statement-2 is a correct explanation of Statement -1.B. Statement -1 is true , Statement-2 is true , Statement-2 is a correct explanation of Statement-1.C. Statement-1 is true , Statement-2 is false.D. Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - D Here, statement-1 is wrong and statement-2 is correct. At height `R`, velocity of the body, `upsilon = 0`. `K.E.` of body `= 0`. `P.E.` of body `= - (Gm)/((R + R)) = - (GMm)/(2R)` Total energy `= K.E. + P.E. = - (GMm)/(2R)` On the surface of earth, let `upsilon` be the velocity of the body `K.E.` of body `= (1)/(2) m upsilon^(2)`, `P.E.` of body `= - (GMm)/(R )` Total energy `= (1)/(2) m upsilon^(2) - (GMm)/(R )`. According to law of conservation of energy `(1)/(2) m upsilon^(2) - (GMm)/(R ) = - (GMm)/(2R)` or `(1)/(2) upsilon^(2) = (GM)/(R ) = - (GM)/(2R) = (gR^(2))/(2R)` or `upsilon = sqrt(gR)` |
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