Ask & Answer Question

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

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 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

0 Upvotes 0 Downvotes

Similar Questions for you

A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is:

Vishal Baghel

According to question, we can write

Total moles of gas = n = n_Oxygen+ n_Oxygen= $\frac{16}{2} + \frac{128}{32} = 12 m o l e s$

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 10⁵ cm³ $\approx 27 \times 1 0^{4} c m^{3}$

Dimensional formula for thermal conductivity is (here K denotes the temperature):

Vishal Baghel

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

It momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:

Vishal Baghel

Viscosity = Pascal . Second

$P^{X} A^{Y} T^{Z} = [M^{1} L^{- 1} T^{- 1}]$

= ${[M^{1} L^{+ 1} T^{- 1}]}^{x} [L^{2}] y {[T^{1}]}^{z} = M^{- 1} L^{- 1} T^{- 1}$

= x = 1, x + 2y = -1, -x + z = 1

y = -1, Z = 0

Viscosity = [P¹A^-¹T⁰]

The one division of main scale of Vernier calipers reads 1mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale, When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2cm and 6^th Vernier division coincides with a main scale division. The diameter of the bob will be ________× 10^-2 cm.

Payal Gupta

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-² cm

Identify the pair of physical quantities which have different dimensions

Vishal Baghel

$Δ Q = m s Δ T \Rightarrow s = \frac{Δ Q}{m Δ T} \Rightarrow [s] = \frac{M L^{2} T^{- 2}}{M Q} = L^{2} T^{- 2} Q^{- 1}$

$Δ Q = m L \Rightarrow [L] = \frac{M L^{2} T^{- 2}}{M} = L^{2} T^{- 2}$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

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

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

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 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

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

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