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

Express unified atomic mass unit in kg.

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Payal Gupta | Contributor-Level 10

4 months ago

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

One atomic mass unit is 1/12 of the mass of carbon atom

Mass of one carbon atom = 12g

Number of atoms in one mole= Avogadro’s number

= 6.023 $\times$ 10²³

Mass of one ₆C¹² atom = $\frac{12}{6.023 \times 10^{23}} g$

1 amu= 1/12 of mass of carbon atom

1 amu= ( $\frac{1}{12} \times \frac{12}{6.023 \times 10^{23}}$ )g= 1.67 $\times 10^{- 27} k g$

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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):

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$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

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

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

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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

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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.

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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

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

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

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