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

NTA Abhyas 2020 · Physics · Experimental Physics

Let the end error on the LHS and RHS be equal to 1 cm. For the...

Let the end error on the LHS and RHS be equal to 1 cm. For the balance point at O, find out the % error in the value of X? (If the end error is 1 cm from both sides then it means the correct reading will become 10 cm + 1 cm from LHS and 90 cm + 1 cm from the RHS)

Let the end error on the LHS and RHS be equal to 1 cm. For the balance point at O, find out the % error in the value of

Options

  1. A4.2 %
  2. B8.1%
  3. C9.2%
  4. DNone

Correct answer

B. 8.1%

Step-by-step solution

x90=l1l2=1191\dfrac{\text{x}}{9 0} = \dfrac{{l }_{1}}{{l }_{2}} = \dfrac{1 1}{9 1}

x=1191(90)\text{x} = \dfrac{1 1}{9 1} \left(9 0\right)

\dfrac{{\text{x}}^{′}}{9 0} = \dfrac{1 0}{9 0}

\to {\text{x}}^{′} = 1 0

\to Δ \text{x} = \left(\dfrac{1 1}{9 1} \left(9 0\right) - 1 0\right)

=(990−91091)= \left(\dfrac{9 9 0 - 9 1 0}{9 1}\right)

=8091= \dfrac{8 0}{9 1}

Δ \text{x} \text{%} = \dfrac{Δ \text{x}}{\text{x}} × 1 0 0 = \dfrac{\dfrac{8 0}{9 1}}{\dfrac{9 9 0}{9 1}} × 1 0 0

= \dfrac{8 0 0}{9 9} \text{%}

= \text{8.1} \text{%}
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