Repeated observations in an experiment gave the values `1.29 , 1.33 , 1.34 , 1.35 , 1.32 , 1.36, 1.30, and 1.33`. Calculate the mean value, absolute eror , relative error, and percentage error.
Text Solution
AI Generated Solution
To solve the problem step by step, we will calculate the mean value, absolute error, relative error, and percentage error based on the given observations.
### Step 1: Calculate the Mean Value
The mean value is calculated by adding all the observations and dividing by the number of observations.
Given values:
1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33
...
Topper's Solved these Questions
DIMENSIONS & MEASUREMENT
CENGAGE PHYSICS|Exercise Exercise 1.1|16 Videos
DIMENSIONS & MEASUREMENT
CENGAGE PHYSICS|Exercise Exercise 1.2|6 Videos
CENTRE OF MASS
CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
FLUID MECHANICS
CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
Similar Questions
Explore conceptually related problems
The refractive index of water as measured by the relation mu = ("Real depth")/("Apparent depth") was found to have the values 1.29 , 1.33 , 1.34 , 1.35 , 1.32 , 1.36 , 1.30 , 1.33 . Calculate (i) mean value of mu (ii) mean value of absolute error (iii) relative error (iv) percentage error .
The refractive index of water as measured by the relation p = (Real depth)/ (Apparent depth) was found to have the values 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33. Calculate mean value of p.
In an experiment the refractive index of glass was observed to be 1.45 , 1.56 , 1.54 , 1.44 , 1.54 , and 1.53 . Calculate (a). Mean value of refractive index (b). Mean absolute error ( c ) Fractional error (d) Percentage error (e) Express the result in terms of absolute error and percentage error
In an experiment refractive index of glass was observed to be 1.45, 1.56, 1.54, 1.44, 1.54 and 1.53. The mean absolute error in the experiement is
The refractive index of water measured by the relation m = (real depth)/ (apparent depth) is found to have values of 1.34, 1.38, 1.32 and 1.36, the mean value of refractive index with percentage error is
The diameter of a wire as measured by a Screw gauge was found to be 1.328, 1.330, 1.325, 1.326, 1.334 and 1.336 cm. Calculate (i) mean value of diameter (ii) absolute error in each measurement (iii) mean absolute error (iv) fractional error (v) percentage error. Also, express the result in term of absolute error and percentage error.
The refractive index (n) of glass is found to have the values 1.49,1.50,1.52,1.54 and 1.48. Calculate (i) the mean value of refractive index, (ii) absolute error in each measurement, (iii) mean absolute error, (iv) fractional error and (v) percentage error
Using screw gauge the observation of the diameter of a wire are 1.324,1.326,1.334,1.336cm respectively Find the average diameter the mean error the relative error and % error .
