Ask & Answer Question

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

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

4 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

(i) dimension of h = [ML²T^-1]

And c= [LT^-1] and dimension of G= [M^-1L³T^-2]

Let m= kc^xh^yG^z

[ML⁰T⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=1

x+2y+3z=0

-x-y-2z=0

On solving these equation we got x= ½ y= ½ and z= -½

So formula will coming out from this is m=k $\sqrt[]{\frac{c h}{G}}$

(ii) L=kc^xh^yG^z

ao [M⁰LT⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=0

x+2y+3z=1

-x-y-2z=0

After solving we get x= -3/2 y=1/2 and z=1/2

We got the formula is L=k $\sqrt[]{\frac{h G}{c^{3}}}$

(iii)T= kc^xh^yG^z

[M⁰L⁰T]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

On comparing powers we go

...more

This is a long answer type question as classified in NCERT Exemplar

(i) dimension of h = [ML²T^-1]

And c= [LT^-1] and dimension of G= [M^-1L³T^-2]

Let m= kc^xh^yG^z

[ML⁰T⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=1

x+2y+3z=0

-x-y-2z=0

On solving these equation we got x= ½ y= ½ and z= -½

So formula will coming out from this is m=k $\sqrt[]{\frac{c h}{G}}$

(ii) L=kc^xh^yG^z

ao [M⁰LT⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=0

x+2y+3z=1

-x-y-2z=0

After solving we get x= -3/2 y=1/2 and z=1/2

We got the formula is L=k $\sqrt[]{\frac{h G}{c^{3}}}$

(iii)T= kc^xh^yG^z

[M⁰L⁰T]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

On comparing powers we got x= -5/2 y=1/2 and z= ½

So formula for this would be T= k $\sqrt[]{\frac{h G}{c^{5}}}$

less

This is a long answer type question as classified in NCERT Exemplar (i) dimension of h = [ML2T-1]And c= [LT-1] and dimension of G= [M-1L3T-2]Let m= kcxhyGz [ML0T0]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]y-z=1x+2y+3z=0-x-y-2z=0On solving these equation we got x= ½ y= ½ and z= -½So formula will coming out from this is m=k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>c</mi> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mi>G</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> (ii) L=kcxhyGzao [M0LT0]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]y-z=0x+2y+3z=1-x-y-2z=0After solving we get x= -3/2 y=1/2  and z=1/2We got the formula is L=k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>h</mi> <mi>G</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> (iii)T= kcxhyGz [M0L0T]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]On comparing powers we got x= -5/2 y=1/2 and z= ½So formula for this would be T= k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>h</mi> <mi>G</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math>

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
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

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