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Units and Measurements physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for $\sqrt[]{A B}$ as:

(a) 1.4 m ± 0.4 m

(b) 1.41m ± 0.15 m

(c) 1.4m ± 0.3 m

(d) 1.4m ± 0.2 m

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

This is a multiple choice answer as classified in NCERT Exemplar

(d) According to question

A= 1.0m $\pm$ 0.2m, B= 2 $\pm 0.2$ m

Y= $\sqrt[]{A B} = \sqrt[]{1.0 (2.0)} = 1.414 m$

After rounding off 1.4m

= $\frac{? Y}{Y} = \frac{1}{2} [\frac{? A}{A} + \frac{? B}{B}] = \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$

= $\frac{0.6 Y}{2 (2.0)} = 0.212$

After rounding off it becomes 0.2m

Thus the correct value of $\sqrt[]{A B}$ =r+dr

so its value become 1.4 $\pm$ 0.2m

This is a multiple choice answer as classified in NCERT Exemplar (d) According to questionA= 1.0m <math> <mo>±</mo> </math> 0.2m, B= 2 <math> <mo>±</mo> <mn>0.2</mn> </math> mY= <math> <mroot> <mrow> <mrow> <mi>A</mi> <mi>B</mi> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <mn>1.0</mn> <mfenced separators="|"> <mrow> <mrow> <mn>2.0</mn> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>1.414</mn> <mi>m</mi> </math> After rounding off 1.4m= <math> <mfrac> <mrow> <mrow> <mo>? </mo> <mi>Y</mi> </mrow> </mrow> <mrow> <mrow> <mi>Y</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced open=" [" close="]" separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>? </mo> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <mo>? </mo> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced open=" [" close="]" separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>0.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.0</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <mn>0.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2.0</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>0.6</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <mn>2.0</mn> </mrow> </mrow> </mfrac> </math>  = <math><mfrac><mrow><mrow><mn>0.6</mn><mi>Y</mi></mrow></mrow><mrow><mrow><mn>2</mn><mfenced separators="|"><mrow><mrow><mn>2.0</mn></mrow></mrow></mfenced></mrow></mrow></mfrac><mo>=</mo><mn>0.212</mn></math> After rounding off it becomes 0.2mThus the correct value of <math> <mroot> <mrow> <mrow> <mi>A</mi> <mi>B</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> =r+dr so its value become 1.4 <math> <mo>±</mo> </math> 0.2m

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You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for $\sqrt[]{A B}$ as:

(a) 1.4 m ± 0.4 m

(b) 1.41m ± 0.15 m

(c) 1.4m ± 0.3 m

(d) 1.4m ± 0.2 m

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You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for A B as: (a) 1.4 m ± 0.4 m (b) 1.41m ± 0.15 m (c) 1.4m ± 0.3 m (d) 1.4m ± 0.2 m

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Your Question

You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for $\sqrt[]{A B}$ as:

(a) 1.4 m ± 0.4 m

(b) 1.41m ± 0.15 m

(c) 1.4m ± 0.3 m

(d) 1.4m ± 0.2 m