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JEE Main
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Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

The density of a solid sphere is determined by measuring its mass...

The density of a solid sphere is determined by measuring its mass and radius. The mass is measured using a balance with a least count of 1 g1\ \text{g} and is found to be 100 g100\ \text{g}. The radius is measured using vernier calipers and is found to be 2.00 cm2.00\ \text{cm}. If the maximum percentage error in the calculated density is 2.5%2.5\%, the least count of the vernier calipers used to measure the radius is:

Options

  1. A0.01 cm0.01\ \text{cm}
  2. B0.03 cm0.03\ \text{cm}
  3. C0.015 cm0.015\ \text{cm}
  4. D0.09 cm0.09\ \text{cm}

Correct answer

A. 0.01 cm0.01\ \text{cm}

Step-by-step solution

The density of a solid sphere is given by the formula:

ρ=MV=M43πR3\rho = \dfrac{M}{V} = \dfrac{M}{\dfrac{4}{3}\pi R^3}

The maximum percentage error in the density is given by:

Δρρ×100=ΔMM×100+3ΔRR×100\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta M}{M} \times 100 + 3\dfrac{\Delta R}{R} \times 100

Given values:
Δρρ×100=2.5%\dfrac{\Delta \rho}{\rho} \times 100 = 2.5\%
M=100 gM = 100\ \text{g}
ΔM=1 g\Delta M = 1\ \text{g} (least count of the balance)
R=2.00 cmR = 2.00\ \text{cm}
ΔR\Delta R is the least count of the vernier calipers.

Substituting the given values into the error equation:

2.5=(1100)×100+3×(ΔR2.00)×1002.5 = \left(\dfrac{1}{100}\right) \times 100 + 3 \times \left(\dfrac{\Delta R}{2.00}\right) \times 100

2.5=1+150ΔR2.5 = 1 + 150 \Delta R

1.5=150ΔR1.5 = 150 \Delta R

ΔR=1.5150=0.01 cm\Delta R = \dfrac{1.5}{150} = 0.01\ \text{cm}

Answer: 0.01 cm0.01\ \text{cm}
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