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Grade: 11
        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 )
2 years ago

Answers : (1)

Vikas TU
8729 Points
							
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
2 years ago
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