Home
Class 12
PHYSICS
The length of a rod as measured in an ex...

The length of a rod as measured in an experiment is found to be 2.48 m, 2.46 m , 2.49 m , 2.49 m and 2.46 m. Find the average length , the absolute error in each observation and the percentage error.

Text Solution

Verified by Experts

Average length = Arithmetic mean of the measured values
`x_("mean")=(2.48+2.46+2.49+2.49+2.46)/5`
= `(12.38)/5m=2.476m`
`therefore` True value, `x_("mean")=2.48m`
Absolute errors in various measurements.
`|Deltax_(1)|=|x_(1)-x_("mean")|=2.48-2.48=0.00m`
`|Deltax_(2)|=|2.46-2.48|=0.02m`
`|Deltax_(3)|=|2.49-2.48|=0.01m`
`|Deltax_(4)|=|2.49-2.48|=0.01m`
`Deltax_(5)=|2.46-2.48|=+0.02m`
Mean absolute error = `(|Deltax_(1)|+|Deltax_(2)|+|Deltax_(3)|+....+|Deltax_(5)|)/5`
= `(0.00+0.02+0.01+0.01+0.02)/5=0.06/5`
`Deltax_("mean")=0.01m`
Thus, `x=2.48pm0.01m`
Percentage error `deltax=(Deltax_("mean"))/x xx100=0.01/2.48xx100=0.40%`
Promotional Banner

Similar Questions

Explore conceptually related problems

The length of a rod as measured in an experiment was found to be 2.48m, 2.46 m, 2.49 m, 2.50 m and 2.48m. Find the average length, absolute error in each observation and the percentage error.

The readings of a length come out to be 2.63 m , 2.56 m , 2.42 m , 2.71 m and 2.80 m . Calculate the absolute errors and relative errors or percentage errors.

The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate (a) mean value of diameter (b) absolute error in each measurement (c) mean absolute error (d) fractional error (e) percentage error (f) Express the result in terms of percentage error

The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate (a) mean value of diameter (b) absolute error in each measurement (c) mean absolute error (d) fractional error (e) percentage error (f) Express the result in terms of percentage error

Following observations were taken with a vernier callipers while measuring the length of a cylinder : 3.29 cm,3.28 cm,3.31 cm,3.28 cm,3.27 cm,3.29 cm,3.20 cm . Then find : (a) Most accurate length of the cylinder (b) Absolute error in each observation (c) Mean absolute error (d) Relative error (e) Percentage error Express the result in terms 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

The length of a pendulum is measuredas 1,01 m and time for 30 oscillation is measured as one minute 3 s. Error length is 0.01 m and error in the 3 s. The percentage error in the measurement of acceleration due to gravity id,

In an experiment the valno of refractive index of glass was found to be 1.54, 1.53, 1.44,1.54, 1.56 and 1.45 m in successive measurement. Calculate (i) the mean vlaue of refractive index (ii) absolute error of each measurement (iii) mean absolute error (iv) relative error and (v) percentage error.

The length and breadth of a field are measured as : l = ( 120 +- 2) m and b = ( 100 +- 5) m , respectively. What is the area of the field?

The length and breadth of a park are in the ratio 2:1 and its perimeter is 240 m. a path of 2 m wide runs inside it, along boundary. Find the cost of paving the path at Rs.3 per m^2 .