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Q.

The period of oscillation of a simple pendulum is given by T = 2πlg where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is

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a

0.1%

b

1%

c

0.2%

d

0.8%

answer is C.

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Detailed Solution

T = 2πlg      ⇒T2 = 4π2lg ⇒ g = 4π2lT2Here % error in l = 1 mm100 cm ×100 = 0.1100×100 = 0.1%and % error in T = 0.12×100×100 = 0.05%∴ % error in g = % error in l + 2(% error in T)                            =0.1+2(0.05) %                            =0.2%
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