Question

A certain body weight \[22.42g\]and has a measured volume of\[4.7cc\]. The possible error in the measurement of mass and volume is \[0.01gm\]&\[0.1cc\]. The maximum error in the density will
A.\[22%\]
B.\[2%\]
C.\[0.2%\]
D.\[0.02%\]

Answer 1

Hint: Whenever we have to find maximum error in any physical quantity then we have to add all errors that occur in that quantity. Errors are always added. The difference between True Value and Experimental Value is called Absolute error. When the relative/fractional error is multiplied by \[100\] then it is expressed in percentage we will call it as percentage error.

Complete answer:
Since there is a difference between mass and weight. Mass is measured in Kilogram (Kg) which is MKS system and Weight is measured in Newton (N) which is in SI system
Here it is given in the question that,
Mass of given Body is \[(M)\]= \[22.42g\]
Volume of given Body is \[(V)=\]\[4.7cc\]
Let us assume the density of the given body is\[D\].
According to Question,
The possible error in measurement of mass is \[(\Delta m)=0.01gm\]
The possible error in measurement of volume is \[(\Delta V)=0.1cc\]
The possible error in any quantity also represents the least count of instruments through which that quantity is measured.
Since,\[Density(D)=\dfrac{Mass(M)}{Volume(V)}\]
When the unit of mass in CGS System is gm and unit of volume in CGS system is cc then the unit of density in CGS System is\[gmc{{m}^{-3}}\].
\[\dfrac{\Delta D}{D}=\dfrac{\Delta M}{M}+\dfrac{\Delta V}{V}\]
This expression represents maximum fractional error in density. When we multiply each term of this quantity with \[100\] then this expression is expressed as percentage error.
So, Maximum error in percentage of given relation can be expressed as
\[(\dfrac{\Delta D}{D}\times 100)=(\dfrac{\Delta M}{M}\times 100)+(\dfrac{\Delta V}{V}\times 100)\]
Put the values of Mass\[(M)\]; Volume\[(V)\]; Values of error in Mass and Volume, we get
\[%error(D)=(\dfrac{0.01}{22.42}\times 100)+(\dfrac{0.1}{4.7}\times 100)\]
Simplifying above equation,
\[%error(D)=0.044+2.127\]
\[%error(D)=2.168%\approx 2%\]
So maximum percentage error in density is \[2%\] (approximate).

So ,option B is the correct answer.

As this error gets reduced then results of observation get more accurate, Accuracy refers to the closeness of the measurement to the true value of the physical quantity. As we reduce the errors measurement becomes more accurate.
The measuring process is essentially a process of comparison. In Spite of our best efforts, the measured value of a quantity is always somewhat different from its actual value, or true value. This difference in the true value of a quantity is called error of measurement.

Note:
Either relation of given quantity in which we have to find error is in addition form; Subtraction form; Multiply form or Divide form we have to add errors in each quantity to find error in that particular quantity. Errors are always treated as maximum absolute/percentage errors.