If the length of a simple pendulum is de...

If the length of a simple pendulum is decreased by 3%, then the percentage error in its period T is

2.5

1.8

1.5

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If the length of a simple pendulum is decreased by 2%, find the percentage error in its period T.

Assertion (A): If the length (l) of the simple pendulum is decreased by 4 % then decrease in its time period (T) is 2 % Reason (R) : If y=x^(n) , ratio of percentage errors in y and x is 1:n The correct answer is

If the period of oscillation of a simple pendulum is increased by a% then the percentage increase in its length is

If there is an error of 2% in measuring its length, of a simple pendulum then the percentage error in the period will be

(A) : The percentage change in time period is 1.5%, if the length of simple pendulum increases by 3%. (R) Time period is directly proportional to square root of length of pendulum.

The error in the measurement of the length of the simple pendulum is 0.2% and the error in time period 4%. The maximum possible error in measurement of L//T^(2) is

DIPTI PUBLICATION ( AP EAMET)-APPLICATIONS OF DIFFERENTIATION-EXERCISE 1A (APPROXIMATIONS AND ERRORS)

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