Question:
For a particular wire of \( mass = (0.6 \pm 0.003) \) gm, \( radius = (0.50 \pm 0.01) \) cm, and \( length = (10.00 \pm 0.05) \) cm, the maximum percentage error in the measurement of its density is:
For a particular wire of \( mass = (0.6 \pm 0.003) \) gm, \( radius = (0.50 \pm 0.01) \) cm, and \( length = (10.00 \pm 0.05) \) cm, the maximum percentage error in the measurement of its density is:
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Always multiply the percentage error of a base measurement by its exponent in the physical formula.
Updated On: Jun 9, 2026
- \( 5\% \)
- \( 7\% \)
- \( 8\% \)
- \( 4\% \)
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The Correct Option is A
Solution and Explanation
Concept:
The density \(\rho\) of a wire is given by the formula \(\rho = \frac{m}{V}\). For a cylindrical wire, the volume is \(V = \pi r^2 l\), so \(\rho = \frac{m}{\pi r^2 l}\). To find the maximum percentage error, we use the rule of propagation of errors for products and quotients. If a physical quantity \(X = \frac{m^a r^b}{l^c}\), the maximum relative error is \(\frac{\Delta X}{X} = a\frac{\Delta m}{m} + b\frac{\Delta r}{r} + c\frac{\Delta l}{l}\).
Step 1: Define the error propagation formula for density.
The formula for relative error in density is: $$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} $$
Step 2: Substitute the given experimental values and errors.
We have \( m = 0.6 \), \(\Delta m = 0.003\), \( r = 0.50 \), \(\Delta r = 0.01\), \( l = 10.00 \), and \(\Delta l = 0.05\)[cite: 1118]. Calculating the individual relative errors: $$ \frac{\Delta m}{m} = \frac{0.003}{0.6} = 0.005 $$ $$ 2 \frac{\Delta r}{r} = 2 \times \frac{0.01}{0.50} = 2 \times 0.02 = 0.04 $$ $$ \frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 $$
Step 3: Sum the relative errors and convert to percentage.
$$ \frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05 $$ $$ \text{Percentage Error} = 0.05 \times 100\% = 5\% $$ $$\boxed{5\%}$$
Step 1: Define the error propagation formula for density.
The formula for relative error in density is: $$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} $$
Step 2: Substitute the given experimental values and errors.
We have \( m = 0.6 \), \(\Delta m = 0.003\), \( r = 0.50 \), \(\Delta r = 0.01\), \( l = 10.00 \), and \(\Delta l = 0.05\)[cite: 1118]. Calculating the individual relative errors: $$ \frac{\Delta m}{m} = \frac{0.003}{0.6} = 0.005 $$ $$ 2 \frac{\Delta r}{r} = 2 \times \frac{0.01}{0.50} = 2 \times 0.02 = 0.04 $$ $$ \frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 $$
Step 3: Sum the relative errors and convert to percentage.
$$ \frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05 $$ $$ \text{Percentage Error} = 0.05 \times 100\% = 5\% $$ $$\boxed{5\%}$$
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