Time intervals measured by a clock give the f | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

The arithmetic mean of given values is taken as true value.

$t _{	ext {mean }}=\frac{t_{1}+ t _{2}+ t _{3}+ t _{4}+ t _{5}}{5}$

$t_{	ext {mean

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

The arithmetic mean of given values is taken as true value.

$t _{	ext {mean }}=\frac{t_{1}+ t _{2}+ t _{3}+ t _{4}+ t _{5}}{5}$

$t_{	ext {mean }}=\frac{1.25+1.24+1.27+1.21+1.28}{5}$

$t_{	ext {mean }}=1.25 s$

$\Delta t_{	ext {mean }}=\frac{\left|\Delta t_{1}ight|+\left|\Delta t_{2}ight|+\left|\Delta t_{3}ight|+\left|\Delta t_{4}ight|+\left|\Delta t_{5}ight|}{5}$

$\Delta t_{	ext {mean }}=\frac{0+0.01+0.02+0.04+0.03}{5}=0.02 s$

Percentage error $=\frac{\Delta t_{\operatorname{mean}}}{t_{	ext {mean }}} 	imes 100=\frac{0.02}{1.25} 	imes 100$

$=1.6 \%$

Time intervals measured by a clock give the following readings :

$1.25 \;s , 1.24\; s , 1.27\; s , 1.21 \;s$ and $1.28\; s$

What is the percentage relative error of the observations?

Similar Questions

If error in measuring diameter of a circle is $4\%$, the error in radius of the circle would be

If $P = \frac{{{A^3}}}{{{B^{5/2}}}}$ and $\Delta A$ is absolute error in $A$ and $\Delta B$ is absolute error in $B$ then absolute error $\Delta P$ in $P$ is

$Assertion$: In the measurement of physical quantities direct and indirect methods are used.

$Reason$ : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.

A physical quantity is $A = P^2/Q^3.$ The percentage error in measurement of $P$ and $Q$ is $x$ and $y$ respectively. Maximum error in measurement of $A$ is

Time intervals measured by a clock give the following readings : $1.25 \;s , 1.24\; s , 1.27\; s , 1.21 \;s$ and $1.28\; s$ What is the percentage relative error of the observations?