Question:
A cuboid has volume \(V = l\times l\times \sqrt{l}\), where \(l\) is the length of one side. When the length \(l\) is measured by a meter scale of least count \(1\,\text{mm}\), the relative percentage error in the measurement of its volume is \(6.25\%\). Then the relative percentage error in measurement of its length and the value of the length \(l\) is
A cuboid has volume \(V = l\times l\times \sqrt{l}\), where \(l\) is the length of one side. When the length \(l\) is measured by a meter scale of least count \(1\,\text{mm}\), the relative percentage error in the measurement of its volume is \(6.25\%\). Then the relative percentage error in measurement of its length and the value of the length \(l\) is
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Powers multiply the relative error: V goes as l to the power 2.5.
Updated On: Oct 1, 2026
- \(2\%\) and \(2\) cm
- \(2.5\%\) and \(4\) cm
- \(1.8\%\) and \(3.6\) cm
- \(1\%\) and \(3\) cm
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The Correct Option is B
Solution and Explanation
Step 1: Understand the concept
For a quantity \(V = l^{a}\), the relative error in \(V\) is \(a\) times the relative error in \(l\). Here \(V = l \times l \times \sqrt{l} = l^{5/2}\).
Step 2: Apply the rule
\[ \frac{\Delta V}{V} = \frac{5}{2}\,\frac{\Delta l}{l} \]
Given \(\frac{\Delta V}{V} \times 100 = 6.25\%\), we get
\[ \frac{\Delta l}{l}\times100 = \frac{6.25}{2.5} = 2.5\% \]
Step 3: Find the length
The least count of the scale is 1 mm, so \(\Delta l = 1\ \text{mm} = 0.1\) cm. Then
\[ l = \frac{\Delta l}{0.025} = \frac{0.1}{0.025} = 4\ \text{cm} \]
Step 4: Check the options
Only option (B) gives both 2.5% and 4 cm. Check option (A): 2% of 2 cm is 0.04 cm, not the 0.1 cm least count. Option (D): 1% of 3 cm is 0.03 cm, again not 0.1 cm.
Final Answer:
The error is 2.5% and the length is 4 cm. This is option (B). \[ \boxed{\text{(B) }2.5\%\ \text{and}\ 4\ \text{cm}} \]
For a quantity \(V = l^{a}\), the relative error in \(V\) is \(a\) times the relative error in \(l\). Here \(V = l \times l \times \sqrt{l} = l^{5/2}\).
Step 2: Apply the rule
\[ \frac{\Delta V}{V} = \frac{5}{2}\,\frac{\Delta l}{l} \]
Given \(\frac{\Delta V}{V} \times 100 = 6.25\%\), we get
\[ \frac{\Delta l}{l}\times100 = \frac{6.25}{2.5} = 2.5\% \]
Step 3: Find the length
The least count of the scale is 1 mm, so \(\Delta l = 1\ \text{mm} = 0.1\) cm. Then
\[ l = \frac{\Delta l}{0.025} = \frac{0.1}{0.025} = 4\ \text{cm} \]
Step 4: Check the options
Only option (B) gives both 2.5% and 4 cm. Check option (A): 2% of 2 cm is 0.04 cm, not the 0.1 cm least count. Option (D): 1% of 3 cm is 0.03 cm, again not 0.1 cm.
Final Answer:
The error is 2.5% and the length is 4 cm. This is option (B). \[ \boxed{\text{(B) }2.5\%\ \text{and}\ 4\ \text{cm}} \]
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