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Standard XII

Physics

Physical Pendulum

The length of...

Question

The length of the seconds pendulum is first increased by $10 c m$ and then decreased by $5 c m$ . If the time period is determined in each case, find the ratio.
(Take $g = 10 m s^{- 2}, π^{2} ≃ 10$ )

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Solution

Time Period for second’s Pendulum = $60 s$
$60 = 2 π \times \sqrt{I / G}$
Solving we get,
$l = 900$
Now after increasing and decrasing respectively we get,
$T 1 / T 2 = r o o t ((l + 10) / (l - 5))$
or
=> $r o o t (910 / 895)$
=> $30.16 : 29.91$

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The length of the seconds pendulum is first increased by 10cm and then decreased by 5cm. If the time period is determined in each case, find the ratio.(Take g=10ms−2,π2≃10)

Time Period for second’s Pendulum =60s60=2π×√I/GSolving we get,l=900Now after increasing and decrasing respectively we get,T1/T2=root((l+10)/(l−5))or=> root(910/895)=> 30.16:29.91

The length of the seconds pendulum is first increased by $10 c m$ and then decreased by $5 c m$ . If the time period is determined in each case, find the ratio.
(Take $g = 10 m s^{- 2}, π^{2} ≃ 10$ )

Time Period for second’s Pendulum = $60 s$
$60 = 2 π \times \sqrt{I / G}$
Solving we get,
$l = 900$
Now after increasing and decrasing respectively we get,
$T 1 / T 2 = r o o t ((l + 10) / (l - 5))$
or
=> $r o o t (910 / 895)$
=> $30.16 : 29.91$