The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )

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Vikas TU · Answer

Time Period for second’s Pendulum = 60s 60 = 2pi*root(l/g) Solving we get, l = 900 Now after increasing and decrasing respectively we get, T1/T2 = root((l+10)/(l-5)) or => root(910/895) => 30.16 : 29.91

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The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )

The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )

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The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )

The lengths of a seconds pendulum is first increased by 10 cm then decreased by 5cm .If the time period is determined in each case , find their ratio.( g=10m/s2,π^2=10 )

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