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A wire has a mass 0.3+-0.003g, radius 0....

A wire has a mass `0.3+-0.003g`, radius `0.5+-0.005mm` and length `6+-0.06cm`. The maximum percentage error in the measurement of its density is

A

1

B

2

C

3

D

4

Text Solution

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The correct Answer is:
To find the maximum percentage error in the measurement of the density of the wire, we can follow these steps: ### Step 1: Understand the formula for density The density (ρ) of the wire is given by the formula: \[ \rho = \frac{m}{V} \] where \(m\) is the mass and \(V\) is the volume of the wire. ### Step 2: Determine the volume of the wire The volume \(V\) of the wire can be calculated using the formula for the volume of a cylinder: \[ V = \pi r^2 L \] where \(r\) is the radius and \(L\) is the length of the wire. ### Step 3: Calculate the maximum percentage error in density The maximum percentage error in density can be calculated using the formula: \[ \frac{\Delta \rho}{\rho} \times 100 \] where \(\Delta \rho\) is the maximum error in density. ### Step 4: Determine the errors in mass, radius, and length Given: - Mass \(m = 0.3 \pm 0.003 \, \text{g}\) - Radius \(r = 0.5 \pm 0.005 \, \text{mm} = 0.5 \times 10^{-3} \pm 0.005 \times 10^{-3} \, \text{m}\) - Length \(L = 6 \pm 0.06 \, \text{cm} = 0.06 \pm 0.0006 \, \text{m}\) ### Step 5: Calculate the percentage errors 1. **Mass error**: \[ \text{Percentage error in mass} = \frac{\Delta m}{m} \times 100 = \frac{0.003}{0.3} \times 100 = 1\% \] 2. **Radius error**: \[ \text{Percentage error in radius} = \frac{\Delta r}{r} \times 100 = \frac{0.005 \times 10^{-3}}{0.5 \times 10^{-3}} \times 100 = 1\% \] Since the volume depends on \(r^2\), we multiply this error by 2: \[ \text{Percentage error in volume due to radius} = 2 \times 1\% = 2\% \] 3. **Length error**: \[ \text{Percentage error in length} = \frac{\Delta L}{L} \times 100 = \frac{0.06}{6} \times 100 = 1\% \] ### Step 6: Combine the errors The total percentage error in density is the sum of the individual percentage errors: \[ \text{Total percentage error in density} = \text{Percentage error in mass} + \text{Percentage error in volume due to radius} + \text{Percentage error in length} \] \[ = 1\% + 2\% + 1\% = 4\% \] ### Conclusion The maximum percentage error in the measurement of the density of the wire is **4%**. ---

To find the maximum percentage error in the measurement of the density of the wire, we can follow these steps: ### Step 1: Understand the formula for density The density (ρ) of the wire is given by the formula: \[ \rho = \frac{m}{V} \] where \(m\) is the mass and \(V\) is the volume of the wire. ...
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Knowledge Check

  • A wire has a mass (0.2 +- 0.002) g, radius (0.7+-0.007) mm and length (3 +- 0.09) cm. The maximum percentage error in the measurement of its density is

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