The maximum error in the measurement of mass and length is $4 \%$ and $3 \%$ respectively. The error in the…
- $9 \%$
- $15 \%$
- $13 \%$
- $6 \%$
Solution
Let, the mass of the cube \(=m\)
length of each side of the cube is \(=a\)
Density of the cube \(=\rho\)
Given, The maximum error in the measurements of mass \(\frac{(\Delta m)}{m}=0.04\) and maximum error in the measurements of Length. \(\frac{(\Delta a)}{a}=0.03\)
Step 2. Formula used:
Volume of the cube \(v=a^3\)
Density \(=\) mass/volume \((\rho)=m / v\)
Step 2. Calculations:
Now, \(\rho=(m / v)=\left(m / a^3\right)\)
Since error has to be maximum
So, \((\Delta \rho / \rho)=(\Delta m / m)+3(\Delta a / a)\)
Now putting the given values, we get
\(\begin{aligned}
& (\Delta \rho / \rho)=0.04+3(0.03)=0.13 \\
& (\Delta \rho / \rho) \times 100=13 \%
\end{aligned}\)
Asked in: MHT CET 2020 (14 Oct Shift 2)