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The length of a second's pendulum is 1 m on the earth. If the mass and diameter of a planet than that of earth, then the length of the second's pendulum on the planet will be |
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Answer» 4m `THEREFORE 2 = 2pi sqrt((l)/(g))` on the surface of the earth `""` …….(1) On the planet `2=2pi sqrt((l')/(g'))""` …….(2) `therefore (l)/(g)=(l')/(g') "" therefore ""l'=l((g')/(g))""`......(3) But `g=(GM)/(R^(2)) "" therefore ""(g')/(g)=(GM')/(R'^(2))xx(R^(2))/(GM)""` `= 2xx((1)/(2))^(2)=(1)/(2)""` .....(4) `therefore l'=l xx (1)/(2)=(1)/(2)m = 0.5 m` |
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