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The ratio of the masses of two planets is 1 : 10 and the ratio of their diameters is 1 : 2. If the length of a seconds pendulum on the first planet is 0.4 m, then the length of the seconds pendulum on the second planet is _______.A. 10 cmB. 0.5 mC. 10 mD. 1.0 m |
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Answer» Correct Answer - D `m_(1) : m_(2) = 1 : 10` `d_(1) : d_(2) = 1 : 2 rArr r_(1) : r_(2) = 1 : 2` `l_(1) = 0.4 m` `l_(2) = ?` `g_(1) = (GM_(1))/(R_(1)^(2)), g_(2) = (GM_(2))/(R_(2)^(2)` `(g_(1))/(g_(2))=(M_(1))/(R_(1)^(2)) xx (R_(2)^(2))/(M_(2)) = (1)/(10)xx(2^(2))/(1^(2))=(4)/(10)` `(T_(1))/(T_(2))=sqrt(l_(1)/(g_(1))xx(g_(2))/(l_(2)))=sqrt((0.4)/(l_(2))xx(10)/(4))` `(2)/(2) = sqrt((1)/(l_(2))),(1)/(l_(2)) = (1)/(1) rArr l_(2) = 1m` |
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