Mathematics in Physics

JEE Main level, single correct, medium

54 of 121

Question 54 of 121

The density ρ\rho of a uniform cylinder is determined by measuring its mass mm, length ll and diameter dd. The measured values of mm, ll and dd are 97.42±0.0297.42 \pm 0.02 g, 8.35±0.058.35 \pm 0.05 mm and 20.20±0.0220.20 \pm 0.02 mm, respectively. Calculated percentage fractional error in ρ\rho is _______.

Show answer and solution

Answer: (B) 0.82%0.82\%

The density of a uniform cylinder is given by ρ=mV=4mπd2l\rho = \dfrac{m}{V} = \dfrac{4m}{\pi d^2 l}

The percentage fractional error in density is given by:

Δρρ×100=(Δmm+2Δdd+Δll)×100\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta m}{m} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l} \right) \times 100

Substituting the given values:

Δρρ×100=(0.0297.42+2×0.0220.20+0.058.35)×100\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{0.02}{97.42} + 2 \times \dfrac{0.02}{20.20} + \dfrac{0.05}{8.35} \right) \times 100

Δρρ×100=(0.000205+0.00198+0.005988)×100\dfrac{\Delta \rho}{\rho} \times 100 = \left( 0.000205 + 0.00198 + 0.005988 \right) \times 100

Δρρ×100=0.008173×100=0.8173%\dfrac{\Delta \rho}{\rho} \times 100 = 0.008173 \times 100 = 0.8173\%

Rounding off to two decimal places, the percentage error is 0.82%0.82\%.

Answer: 0.82%0.82\%

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Mathematics in Physics (Errors of Measurement), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.