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Units and Measurements physics ncert solutions class 11th Physics Class 11th

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

Option 1 -

$[P A^{- 1} T^{0}]$

Option 2 -

$[P A T^{- 1}]$

Option 3 -

$[P A^{- 1} T]$

Option 4 -

$[P A^{- 1} T^{- 1}]$

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

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]

Viscosity = Pascal . Second <math> <mrow> <msup> <mrow> <mi>P</mi> </mrow> <mrow> <mi>X</mi> </mrow> </msup> <msup> <mrow> <mi>A</mi> </mrow> <mrow> <mi>Y</mi> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mi>Z</mi> </mrow> </msup> <mo>=</mo> <mrow> <mo> [</mo> <mrow> <msup> <mrow> <mi>M</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> </math>      <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815305751"> </o:OLEObject></xml><! [endif]-> = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815305752"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <msup> <mrow> <mrow> <mo> [</mo> <mrow> <msup> <mrow> <mi>M</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mi>x</mi> </mrow> </msup> <mrow> <mo> [</mo> <mrow> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> <mi>y</mi> <msup> <mrow> <mrow> <mo> [</mo> <mrow> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mi>z</mi> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mi>M</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> </mrow> </math> = x = 1,     x + 2y = -1,      -x + z = 1y = -1,              Z = 0Viscosity = [P1A-1T0]

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

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This is a multiple choice answer as classified in NCERT Exemplar

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

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