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Q.2.1 Fill in the blanks

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alok kumar singh | Contributor-Level 10

4 months ago

Ans.2.1

(a) The volume of a cube of side 1 ${c m}^{3}$ is equal to $10^{- 6} m^{3}$

The volume of the cube = 1 cm * 1 cm * 1 cm = 1 cm³

1 cm3 = 1 * 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 * 10⁴ (mm)²

The surface area of a solid cylinder = 2? r (r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 * 22/7 * 20 * (100 +20) ${m m}^{2}$ = 1.51 * $10^{4}$ ${m m}^{2}$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 * 10³.kg m^–3.

Relative density or

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Ans.2.1

(a) The volume of a cube of side 1 ${c m}^{3}$ is equal to $10^{- 6} m^{3}$

The volume of the cube = 1 cm * 1 cm * 1 cm = 1 cm³

1 cm3 = 1 * 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 * 10⁴ (mm)²

The surface area of a solid cylinder = 2? r (r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 * 22/7 * 20 * (100 +20) ${m m}^{2}$ = 1.51 * $10^{4}$ ${m m}^{2}$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 * 10³.kg m^–3.

Relative density or Specific gravity = Density of the subject / density of water

Density of water = 1 gm/cc= 10^-3 / 10^-6

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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:

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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

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

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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

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

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

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