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

Identify the pair of physical quantities which have different dimensions

Option 1 -

Wave number and Rydberg's constant

Option 2 -

Stress and Coefficient of elasticity

Option 3 -

Coercivity and Magnetisation

Option 4 -

Specific heat capacity and Latent heat

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

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

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

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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

Which of the following are not a unit of time?

(a) Second

(b) Parsec

(c) Year

(d) Light year

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) We know that one light year = 9.46 $\times 10^{11} m$ = distance that light travels in 1 year with a speed 3 $\times 10^{8}$ m/s

1 par sec =3.08 $\times 10^{16} m$ = distance at which average radius of earth’s orbit subtends an angle of par second.

Here second and year represent time.

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