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

Correct answer: B. 2.35\ \% Density of the rod is given by = (M)/(V) = (M)/(π r^2 L) The maximum percentage error in density is (Δ )/( ) × 100 = ( (Δ M)/(M) + 2 (Δ r)/(r) + (Δ L)/(L) ) × 100 Substituting the given values: (Δ )/( ) × 100 = ( (0.01)/(4) + 2 × (0.02)/(2) + (0.05)/(50) ) × 100 (Δ )/( ) × 100 = (0.0025 + 0.02 + 0.001) × 100 (Δ )/( ) × 100 = 0.0235 × 100 = 2.35\ \% Answer: 2.35\ \%

MHT CET 202623 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 23 April
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

MHT CET 2026 · 23 April · Evening Shift · Physics · Mathematics in Physics

A rod has mass (4 ± 0.01) kg, radius (2 ± 0.02) cm and length (50 ±...

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

Options

A $1.35\ \%$
B $2.35\ \%$
C $3.35\ \%$
D $4.35\ \%$

Correct answer

B. $2.35\ \%$

Step-by-step solution

Density of the rod is given by

\rho = \dfrac{M}{V} = \dfrac{M}{\pi r^2 L}

The maximum percentage error in density is

\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta M}{M} + 2 \dfrac{\Delta r}{r} + \dfrac{\Delta L}{L} \right) \times 100

Substituting the given values:

\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{0.01}{4} + 2 \times \dfrac{0.02}{2} + \dfrac{0.05}{50} \right) \times 100

\dfrac{\Delta \rho}{\rho} \times 100 = (0.0025 + 0.02 + 0.001) \times 100

\dfrac{\Delta \rho}{\rho} \times 100 = 0.0235 \times 100 = 2.35\ \%

Answer:

2.35\ \%

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