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Units and Measurements — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements4 Marks
Question
Read the passage given below and answer the following questions from (i) to (v).
Maximum absolute error in the sum or difference of two quantities is equal to sum of the absolute error in the individual quantities, i.e.
Z = A + B, then, $\pm\triangle\text{Z}=\pm\triangle\text{A}\pm\text{B}$
Maximum fractional error in a product or division of quantities is equal to the sum of fractional errors in the individual quantities i.e.
AB or $\frac{\text{A}}{\text{B}},$ then, $\frac{\triangle\text{Z}}{\text{Z}}=\pm\frac{\triangle\text{A}}{\text{A}}+\frac{\triangle\text{B}}{\text{B}}$ 
Two resistors of resistances $\text{R}_1=100\pm3\Omega$ are connected (a) in series and (b) in parallel.
  1. The percentage error in the value of R1 is:
  1. 3%
  2. 4%
  3. 6%
  4. 0.3%
  1. The fractional error in the value of R2 is:
  1. $\frac{1}{40}$
  2. $\frac{1}{50}$
  3. $\frac{1}{100}$
  4. $\frac{1}{200}$
  1. Find the equivalent resistance of the series combination.
  1. $(250\pm7)\Omega$
  2. $(320\pm6)\Omega$
  3. $(300\pm7)\Omega$
  4. $(300\pm1)\Omega$
  1. The percentage error in equivalent resistance in series combination is:
  1. 2%
  2. 2.3%
  3. 2.5
  4. 3%
  1. Find the equivalent resistance of the parallel combination having error of $1.8\Omega.$
  1. $(66\pm1)\Omega$
  2. $(66.7\pm1.18)\Omega$
  3. $(66.3\pm2)\Omega$
  4. $(67\pm3)\Omega$
✓

Answer

  1. (a) 3%

Explanation:

Given, $\text{R}_1=100\pm3\Omega$

$\therefore\frac{\triangle\text{R}_1}{\text{R}_2}\times100=\frac{3}{100}\times100=3\text{%}$

  1. $\frac{1}{50}$

Explanation:

Given, $\text{R}_2=(200\pm4)\Omega$

$\therefore\frac{\triangle\text{R}_2}{\text{R}_2}=\frac{4}{200}=\frac{1}{50}$

  1. (c) $\frac{1}{100}$

Explanation:

The equivalent resistance of series combination, i.e.

$\text{R}_{\text{s}}=\text{R}_1+\text{R}_2=(100\pm3)\Omega+(200\pm4)\Omega=(300\pm7)\Omega$

  1. (b) 2.3%

Explanation:

As, $\therefore\frac{\triangle\text{R}_{\text{s}}}{\text{R}_{\text{s}}}\times100=\frac{7}{300}\times100=2.3\text{%}$

  1. (b) $(66.7\pm1.18)\Omega$​​​​​​​

Explanation:

The equivalent resistance of parallel
combination,

$\text{R}'=\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}$

$=\frac{200}{3}=66.7\Omega$

Given, $\triangle\text{R}'=1.18\Omega$

$\therefore\text{R}'=(66.7\pm1.18)\Omega$

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