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# 1.If the error in the measurement of the momentum of a particle is (+100%). Then the error in the measurement of kinetic energy is:(a) 400%  (b) 300%  (c) 100%  (d) 200%2.The lenght of a simple pendulum is about 100 cm known to have an accuracy of 1 mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g ?(a) 0.2 %  (b) 0.5 %  (c) 0.1 %  (d) 2 %

Ramesh V
70 Points
11 years ago

(1) KE = 1/2 mv^2, where mv = p = momentum

KE = p2 / 2m

d(KE)/KE = 2d(p)/p

error in mesrmnt of KE = 2*100 % = 200%

option D

(2) time period is T = 2*pi*(l/g)1/2

g = 4*Pi*Pi*L/T2

log g = log C + log(L) - 2 log (T)

d(g)/g = d(L)/L  + 2 d(T)/T

=  0.1/100 + 2*0.1/(2*100)

= 2/1000 = 0.2%

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regards

Ramesh