Question

A rod has mass \((4\pm 0.01)\) kg, radius \((2\pm 0.02)\) cm and length \((50\pm 0.05)\) cm. The maximum percentage error in the measurement of density is

• A. \(1.35\ \%\)
• B. \(2.35\ \%\)
• C. \(3.35\ \%\)
• D. \(4.35\ \%\)

Hint:

Density is mass divided by volume of a cylinder, so the radius error counts twice. Add the percentage errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Density of the rod is \(\rho=\dfrac{m}{\pi r^{2}l}\).
In a product or quotient the maximum relative errors add, each multiplied by its power.

Step 2: Formula for error
\(\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\)

Step 3: Calculation
\(\dfrac{0.01}{4}\times 100=0.25\%\)
\(2\times\dfrac{0.02}{2}\times 100=2\%\)
\(\dfrac{0.05}{50}\times 100=0.1\%\)

Step 4: Total
\(0.25+2+0.1=2.35\%\), option (B).

Final Answer:
The maximum percentage error in density is \(2.35\%\). \[ \boxed{\text{(B) }2.35\ \%} \]