JEE PYQ: Units & Measurements - Question ID 24243f0d904d (JEE Main 2018)

ID: 24243f0d904dJEE Main 2018Single Correct MCQ
The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:

Select Option

Step-by-step Explanation

Core Formula & Concept:

Density (ρ\rho) of a material is defined as the mass (mm) per unit volume (VV). For a cube, the volume is calculated as the cube of its side length (LL). Thus, the formula for density is:

ρ=mV=mL3\rho = \frac{m}{V} = \frac{m}{L^3}

When dealing with errors in measurements, the relative error (or percentage error) in a derived quantity like density depends on the relative errors in the measured quantities (mass and length). The key concept here is the propagation of errors for multiplication/division and exponentiation:

  • For a product or quotient of quantities, the relative errors add up.
  • If a quantity is raised to a power (nn), its relative error is multiplied by n|n|.

Mathematically, if Z=XYnZ = \frac{X}{Y^n}, then the relative error in ZZ is:

ΔZZ=ΔXX+nΔYY\frac{\Delta Z}{Z} = \frac{\Delta X}{X} + n \cdot \frac{\Delta Y}{Y} Step-by-Step Derivation:

Given:

  • Relative error in mass (Δmm\frac{\Delta m}{m}) = 1.5% = 0.015
  • Relative error in length (ΔLL\frac{\Delta L}{L}) = 1% = 0.01

The density is given by:

ρ=mL3\rho = \frac{m}{L^3}

To find the relative error in density (Δρρ\frac{\Delta \rho}{\rho}), we apply the error propagation rules:

  1. The relative error in mm is Δmm\frac{\Delta m}{m}. Since mass is in the numerator, its relative error contributes directly.
  2. The relative error in L3L^3 is 3ΔLL3 \cdot \frac{\Delta L}{L} because LL is raised to the power of 3. Since L3L^3 is in the denominator, its relative error also contributes directly (but with a positive sign due to division).

Thus, the total relative error in density is:

Δρρ=Δmm+3ΔLL\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \cdot \frac{\Delta L}{L}

Substitute the given values:

Δρρ=0.015+30.01=0.015+0.03=0.045\frac{\Delta \rho}{\rho} = 0.015 + 3 \cdot 0.01 = 0.015 + 0.03 = 0.045

Convert the relative error to a percentage:

Δρρ×100=0.045×100=4.5%\frac{\Delta \rho}{\rho} \times 100 = 0.045 \times 100 = 4.5\%

Therefore, the maximum error in determining the density is 4.5%.

Common Traps & Exam Tip:

Students often make the following mistakes in this question:

  1. Ignoring the exponent in volume: Some students forget that the side length is cubed in the volume formula and only multiply the length error by 1 instead of 3. This leads to an incorrect answer of 2.5% (Option B).
  2. Sign errors in error propagation: While the relative errors add up for division, some students mistakenly subtract the errors, leading to incorrect results.
  3. Confusing absolute and relative errors: The question provides relative errors, but some students treat them as absolute errors, leading to confusion in calculations.
  4. Rounding off errors prematurely: Students might round off intermediate steps (e.g., 0.015 + 0.03 = 0.045 as 0.05), which can lead to selecting the wrong option (e.g., 5%, which is not listed).

Exam Tip: Always write down the formula for the derived quantity first, then apply the error propagation rules systematically. For quantities raised to a power, remember to multiply the relative error by the absolute value of the exponent. Double-check your calculations to avoid arithmetic mistakes.

Related Questions from Units & Measurements

ID: 32f827e3012fJEE Main 2026

In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and 7th 7^{\text {th }} Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has

View Solution →
ID: 4c5472dca7e2JEE Main 2026

Dimensions of universal gravitational constant (GG) in terms of Planck's constant (hh), distance (LL), mass (MM) and time (TT) are _______.

View Solution →
ID: 605eaee8ed80JEE Main 2026

The time period of a simple harmonic oscillator is T=2πkmT = 2\pi \sqrt{\frac{k}{m}}. Measured value of mass (m)(m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k)(k) is ________%.

View Solution →
ID: 5561befb0f22JEE Main 2026

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 4th 4{ }^{\text {th }} mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and 5th 5^{\text {th }} division of vernier scale coincides with a main scale division. Measured length of cylinder is ____\_\_\_\_ mm.

(Least count of Vernier calliper =0.1 mm=0.1 \mathrm{~mm} )

View Solution →