Question

The length of a cylinder is measured with a meter rod having least count \(0.1\) cm. Its diameter is measured with vernier calipers having least count \(0.01\) cm. Length of cylinder is \(8.0\) cm and radius is \(4.0\) cm. Find the percentage error in the calculated value of the volume.

• A. \(1.25\%\)
• B. \(1.75\%\)
• C. \(1.5\%\)
• D. \(2.75\%\)

Hint:

For \(V=\pi r^2 l\), \(\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The volume of a cylinder is \(V = \pi r^2 l\). The maximum fractional error adds up the fractional errors, with the power multiplying the radius error.

Step 2: Key Formula:
\[ \frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \]

Step 3: Errors in l and r:
Length: \(\frac{\Delta l}{l} = \frac{0.1}{8.0} = 0.0125 = 1.25\%\).
Diameter: the least count is \(0.01\) cm, so \(\Delta d = 0.01\). The radius is half the diameter, so \(\Delta r = 0.005\) cm, and \(\frac{\Delta r}{r} = \frac{0.005}{4.0} = 0.125\%\).

Step 4: Combine:
\(\frac{\Delta V}{V}\times100 = 2(0.125) + 1.25 = 0.25 + 1.25 = 1.5\%\).

Final Answer:
The percentage error in the volume is \(1.5\%\), option (C). \[ \boxed{1.5\%} \]