L1 to L3 of Unit Measurement and Errors

Units & Measurements - L2

To estimate g (from g = 4 pi^(2)(L)/(T^(2)) ), error in measurement of L is +- 2% and error in measurement of Tis +- 3% The error in estimated g will be

In a simple pendulum experiment the percentage errors in the measurement of g and l are alpha% and beta% respectively then the maximum error in measuring T wil be "________"

The resistance R of a wire is given by the relation R= (rhol)/( pir^(2)) . Percentage error in the measurement of rho, l and r is 1 % , 2% and 3% respectively. Then the percentage error in the measurement of R is

What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of l ands g are 2% and 4% respectively?

A physical quantity is given by X=[M^(a)L^(b)T^(c)] . The percentage error in measurement of M,L and T are alpha, beta, gamma respectively. Then the maximum % error in the quantity X is

A physical quantity is given by X=M^(a)L^(b)T^(c) . The percentage error in measurement of M,L and T are alpha, beta and gamma respectively. Then maximum percentage error in the quantity X is

The fundamental frequecy of a stationary wave formed in a stretched wire is n= (1)/(2l) sqrt((1)/(m)) where l is length of the vibrating wire 'T' is the tension in the wire and 'm' is its mas per unit length . If the percentage error in measurement of l T and m are a% ,b% and c% respectively then find the maximum error in measuring n.

While measuring the acceleration due to gravity by a simple pendulum, a student makes an error of 1% in the measurement of length and an error of 2% in the measurement of time. If he uses the formula for g as g=4pi^(2)((L)/(T^(2))) , then the percentage error in the measurement of g will be

In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 3% and 2% respectively. The maximum percentage error in the value of L//T^(2) is