Dear student, The period of oscillation (T) of a simple pendulum of length â„“ is given by T = 2π√(â„“/g) Therefore, g = 4π2 â„“/T2 so that the fractional error in g is given by âˆ†g/g = (âˆ†â„“/â„“) + 2(âˆ†T/T) Here âˆ†â„“ = 0 1 cm and âˆ†T = 0 1 s The percentage error is 100 times the fractional error so that EI = âˆ†g/g = [(0 1/64) + 2(0 1/128)]×100 = 5/16 %, EII = âˆ†g/g = [(0 1/64) + 2(0 1/64)]×100 = 15/32 % and EIII = âˆ†g/g = [(0 1/20) + 2(0 1/36)]×100 = 19/18 % Thus EI is minimum so that the correct option is (b) Regards

Answer 10th plz

Answer 10th plz 10. (c) (d) IOHNm2 Students I, II and Ill perform experiment measuring Ecelntion due to gravity (g) a simple paidlduun, They use different lengths of the pendulum and [or reccrd time number ofoscil lations. The observatims are shown in the table. Least count fcy length = O. I cm Least count for time= O. I s StLKlent Length of the No. of Total time pendulum oscillations for (n) (2008) Time period (s) 160 160 91) 14. 15. (crn) 64 f) 64 (n) 4 .1 oscillations 128 0 64 0 36 0 If E), Evand Emare the percentage errors In g, te., Ag — for students l, II and Ill. respectively. then The diÄÄ o with Äle Verru with me ofthe is (a) 5.112mi (c) 5.136m There are two vided into 10 equ scale of me of th to 9 the (Xher caliper ( to I I main scale are shown in the calipers C, and 2 c, ¯ 11 (b) Ell.snuntrntun (d) Ell l.srnaxnntzn A '.ertuer caltpers has I mm markson It hæs2() on the Vernier 16 in:un

Dear student,

The period of oscillation (T) of a simple pendulum of length ℓ is given by
T = 2π√(ℓ/g)
Therefore, g = 4π² ℓ/T² so that the fractional error in g is given by
∆g/g = (∆ℓ/ℓ) + 2(∆T/T)
.
Here ∆ℓ = 0.1 cm and ∆T = 0.1 s
The percentage error is 100 times the fractional error so that
E_I = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,
E_II = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and
E_III = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %
Thus E_I isminimum so that the correct option is (b).
Regards

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