Question

The density of a cube is measured by measuring its mass and length of its side. If the maximum errors in the measurement of its mass and length are 3% and 2% respectively, the maximum error in the measurement of density would be:
A. 12%
B. 14%
C. 7%
D. 9%

Answer 1

Hint: If you know the operations of errors and know how to find relative error in a measured quantity then you will be able to solve this given problem. The meaning of 2% error in length means it is the percentage relative error in length.

Complete step by step answer:
If a physical quantity is denoted by x then the error in x is denoted by $\Delta x$, where $\Delta x$ is always positive. The final measurement of that physical quantity is written as $x\pm \Delta x$. This means that the value of x is not accurate but it will be between $x-\Delta x$ and $x+\Delta x$. Relative error of x is $\dfrac{\Delta x}{x}$.
Suppose a quantity $a={{x}^{n}}$. The error in x is $\Delta x$.
Let the error in a be $\Delta a$. The relative error in a will be $\dfrac{\Delta a}{a}=\left| n.\dfrac{\Delta x}{x} \right|$.
Here ‘| |’ sign is called mode of a number. Suppose x a number. |x| means the absolute or positive value of x. If x=(-3) then |x|=|-3|=3.
Density of a body is the mass contained in one unit volume of the body. It is the ratio of mass of the given body to the volume of the body.
Here, the body is a cube.
Let the mass of the cube be m. Let the error in m be $\Delta m$.
Volume of a cube is equal to the cube of the side.
Let the length of each side of the cube a. Therefore, the volume (V) of the cube is ${{a}^{3}}$. Let the error in a be $\Delta a$
Density (D) of the cube will be $\dfrac{mass}{volume}=\dfrac{m}{V}=\dfrac{m}{{{a}^{3}}}=m{{a}^{-3}}$.
The relative error in D will be $\dfrac{\Delta D}{D}=\left| \dfrac{\Delta m}{m} \right|+\left| -3\dfrac{\Delta a}{a} \right|$.
We know that $\Delta m$ is always positive. Therefore, $\left| \dfrac{\Delta m}{m} \right|=\dfrac{\Delta m}{m}$.
The value of $-3\dfrac{\Delta a}{a}$ will be negative. Therefore, $\left| -3\dfrac{\Delta a}{a} \right|=3\dfrac{\Delta a}{a}$.
Therefore, $\dfrac{\Delta D}{D}=\dfrac{\Delta m}{m}+3\dfrac{\Delta a}{a}$ ……….(i).
It is given that relative error in mass and length is 3% and 2%, which is equal to 0.03 and 0.02 respectively.
Therefore, $\dfrac{\Delta a}{a}=0.02$ and $\dfrac{\Delta m}{m}=0.03$. Substitute these values in equation (i).
 $\Rightarrow \dfrac{\Delta D}{D}=0.03+3\left( 0.02 \right)=0.03+0.06=0.09$
Therefore, relative error in density is 0.09, which is equal to 9%, which is the maximum error.
Hence, the correct option is D.

Note: Some points important to remember.
(i) Absolute errors, relative errors and percentage errors in a measured quantity are always positive.
(ii) It is illogical to subtract errors. If we subtract errors that means we are ignoring some of the errors. Therefore, errors are always added and never subtracted.