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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Experimental Physics

NTA Abhyas 2020 · Physics · Experimental Physics

A student measures the distance covered and time taken by a body...

A student measures the distance covered and time taken by a body under free fall which was initially at rest. He uses this data for estimating gg, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e1{e}_{1} and e2{e}_{2} respectively, then the maximum possible percentage error in the estimation of gg& is

Options

  1. Ae2−e1{e}_{2} - {e}_{1}
  2. Be1+2e2{e}_{1} + 2{e}_{2}
  3. Ce1+e2{e}_{1} + {e}_{2}
  4. De1−2e2{e}_{1} - 2{e}_{2}

Correct answer

B. e1+2e2{e}_{1} + 2{e}_{2}

Step-by-step solution

From the relation

h=ut+12gt2\text{h} = \text{ut} + \dfrac{1}{2} {\text{gt}}^{2}

h=12gt2⇒g=2ht2(∵body initially at rest)\text{h} = \dfrac{1}{2} {\text{gt}}^{2} ⇒ \text{g} = \dfrac{2 \text{h}}{{\text{t}}^{2}} \left(∵ \text{body initially at rest}\right)

Taking natural logarithms on both sides, we get

ln⁡(g)=ln⁡(h)−2ln⁡(t)+ln⁡(2)\ln \left(g\right) = \ln \left(h\right) - 2\ln \left(t\right) + \ln \left(2\right)

Differentiating,Δgg=Δhh−2Δtt\text{Differentiating,} \dfrac{\Delta \text{g}}{\text{g}} = \dfrac{\Delta \text{h}}{\text{h}} - 2 \dfrac{\Delta \text{t}}{\text{t}}

For maximum permissible errors, we add both the errors. So

(Δgg×100)max=(Δhh×100)+2×(Δtt×100){\left(\dfrac{\Delta \text{g}}{\text{g}} \times 100\right)}_{\text{max}} = \left(\dfrac{\Delta \text{h}}{\text{h}} \times \text{100}\right) + 2 \times \left(\dfrac{\Delta \text{t}}{\text{t}} \times \text{100}\right)

Δhh×100=e1andΔtt×100=e2\dfrac{\Delta \text{h}}{\text{h}} \times \text{100} = {\text{e}}_{1} \text{and} \dfrac{\Delta \text{t}}{\text{t}} \times \text{100} = {\text{e}}_{2}

Therefore,(Δgg×100)max=e1+2e2\text{Therefore,} {\left(\dfrac{\Delta \text{g}}{\text{g}} \times \text{100}\right)}_{\text{max}} = {\text{e}}_{1} + 2 {\text{e}}_{2}
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