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Units and Measurements question

2018 · 16 Apr · Shift 1 · Q64
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Units and Measurements question

2018 · 16 Apr · Shift 1 · Q64

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The relative uncertainly in the period of a satellite orbiting around the earth is 10-2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :
  1. A
    10 −-− 2
  2. B
    2 ×\times× 10 −-− 2
  3. C
    3 ×\times× 10 −-− 2
  4. D
    6 ×\times× 10 −-− 2
View written solutionFree

Correct answer: B

  1. Use the time period formula for a satellite

For a satellite orbiting Earth in a circular orbit,

T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}T=2πGMr3​​

where:

  • TTT = time period
  • rrr = orbital radius
  • GGG = gravitational constant
  • MMM = mass of Earth
  1. Rewrite to isolate MMM

Squaring both sides,

T2=4π2r3GMT^2 = 4\pi^2 \frac{r^3}{GM}T2=4π2GMr3​

So,

M=4π2r3GT2M = \frac{4\pi^2 r^3}{G T^2}M=GT24π2r3​

Thus,

M∝r3T−2M \propto r^3 T^{-2}M∝r3T−2

  1. Apply error/uncertainty relation

For powers, relative uncertainty adds with the magnitude of powers:

ΔMM=3Δrr+2ΔTT\frac{\Delta M}{M} = 3\frac{\Delta r}{r} + 2\frac{\Delta T}{T}MΔM​=3rΔr​+2TΔT​

Given that the relative uncertainty in rrr is negligible,

Δrr≈0\frac{\Delta r}{r} \approx 0rΔr​≈0

Hence,

ΔMM=2ΔTT\frac{\Delta M}{M} = 2\frac{\Delta T}{T}MΔM​=2TΔT​

  1. Substitute the given value

Given,

ΔTT=10−2\frac{\Delta T}{T} = 10^{-2}TΔT​=10−2

Therefore,

ΔMM=2×10−2\frac{\Delta M}{M} = 2 \times 10^{-2}MΔM​=2×10−2

  1. Match with options

This corresponds to Option B.

Final Answer:

2×10−22 \times 10^{-2}2×10−2

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