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Chapter 1: Problem 94

In an experiment to determine the density of a cube the percentage error in the measurement of mass is \(0.25 \%\) and the percentage error in the measurement of length is \(0.50 \%\) what will be the percentage error in the determination of its density? (a) \(2.75 \%\) (b) \(1.75 \%\) (c) \(0.75 \%\) (d) \(1.25 \%\)

Short Answer

Expert verified
The percentage error in the determination of the cube's density is \(1.75\%\).

Step by step solution

01

Recall the formula for density

The density of an object is given by: \[\rho = \frac{m}{V}\] where \(\rho\) represents density, \(m\) represents mass, and \(V\) represents volume. Since the object is a cube, its volume can be expressed as: \[V = a^3\] where \(a\) represents the side length of the cube.
02

Calculate the percentage error of the volume of the cube

We are given that the percentage error in the measurement of length is \(0.50\%\). To find the percentage error in the volume, we can use the formula: \[\% \, Error \, in \, Volume = 3 \times \% \, Error \, in \, Length\] \[ \% \, Error \, in \, Volume = 3 \times 0.50\%\] \[ \% \, Error \, in \, Volume = 1.5\%\]
03

Calculate the percentage error in density

Now, we know the percentage error in mass and volume, so we can find the percentage error in density using the formula: \[\% \, Error \, in \, Density = \% \, Error \, in \, Mass + \% \, Error \, in \, Volume\] \[ \% \, Error \, in \, Density = 0.25\% + 1.5\%\] \[ \% \, Error \, in \, Density = 1.75\%\]
04

Identify the correct option

We calculated the percentage error in density to be \(1.75\%\). Looking at the given options, we find that option (b) has the same value. Therefore, the correct answer is: (b) \(1.75\%\)

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Most popular questions from this chapter

If \(\mathrm{A}=3.331 \mathrm{~cm} \mathrm{~B}=3.3 \mathrm{~cm}\) then with regard to significant figure \(\mathrm{A}+\mathrm{B}=\ldots \ldots\) (a) \(6.6 \mathrm{~cm}\) (b) \(6.31 \mathrm{~cm}\) (c) \(6.631 \mathrm{~cm}\) (d) \(6 \mathrm{~cm}\)

Dimensional formula of \(\mathrm{CV}\) ? where C - capacitance and \(\mathrm{V}\) - potential difference (a) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{-1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\)

Match column - I with column - II $$ \begin{array}{|l|l|} \hline \multicolumn{1}{|c|} {\text { Column - I }} & \multicolumn{1}{|c|} {\text { Column - II }} \\ \hline \text { (1) capacitance } & \text { (a) } \mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1} \\ \hline \text { (2) Electricfield } & \text { (b) } \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \\ \hline \text { (3) planck's constant } & \text { (c) } \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2} \\ \hline \text { (4) Angular momentum } & \text { (d) } \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \\ \hline \end{array} $$ (a) \(a, c, b, d\) (b) \(c, a, d, b\) (c) \(c, a, b, d\) (d) \(a, b, d, c\)

Equation of physical quantity \(\mathrm{v}=\mathrm{at}+\mathrm{bt}^{2}\) where \(\mathrm{v}=\) velocity \(\mathrm{t}=\) time so write the dimensional formula of a in this equation (a) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{0}\)

Which unit of physical quantity remains same for all unit system? (a) meter (b) second (c) ampere (d) kilogram
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