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    				| 1. | 
                                    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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Answer» (i) `T_(1) = 2pi sqrt((l + l^(1))/(g)), T_(2) = 2pi (l^(1) - l^(11))/(g)` (ii) `42 : 41`  | 
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