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The length of the seconds pendulums is f...

The length of the seconds pendulums is first increased by 10 cm and then decreased by 5 cm. If the time period is determined in each case, find their ratio (Take `g = 10 ms^(-2), pi^(2) ~= 10`)

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` (i) T_1 = 2 pi sqrt((l+l^1)/(g )) , T_2 = 2pi sqrt((l^1 - l^(11))/( g))`
` (ii) 42 : 41`
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