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# NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements

Question 2.1: Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ________ m3.

Solution:

The side of the cube is 1 cm.

Volume $$V$$ of a cube = side3

$V = (1 \, \text{cm})^3 = 1 \, \text{cm}^3$

Converting $$\text{cm}^3$$ to $$\text{m}^3$$:

$1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3$

Therefore, the volume of a cube of side 1 cm is $$1 \times 10^{-6} \, \text{m}^3$$.

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ________ (mm)2.

Solution:

Radius $$r$$ = 2.0 cm

Height $$h$$ = 10.0 cm

Surface area $$A$$ of a cylinder = $$2\pi r(h + r)$$

$A = 2\pi \times 2.0 \, \text{cm} \times (10.0 \, \text{cm} + 2.0 \, \text{cm}) = 2\pi \times 2.0 \, \text{cm} \times 12.0 \, \text{cm}$

$A = 48\pi \, \text{cm}^2$

Converting $$\text{cm}^2$$ to $$\text{mm}^2$$:

$1 \, \text{cm}^2 = 100 \, \text{mm}^2 \quad \Rightarrow \quad 48\pi \, \text{cm}^2 = 4800\pi \, \text{mm}^2$

Therefore, the surface area of a solid cylinder is $$4800\pi \, \text{mm}^2 \approx 15079.64 \, \text{mm}^2$$.

(c) A vehicle moving with a speed of 18 km h-1 covers ________ m in 1 s.

Solution:

Speed $$v$$ = 18 km/h

Converting speed to m/s:

$18 \, \text{km/h} = 18 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 5 \, \text{m/s}$

Distance covered in 1 s = speed × time

$\text{Distance} = 5 \, \text{m/s} \times 1 \, \text{s} = 5 \, \text{m}$

Therefore, a vehicle moving with a speed of 18 km h-1 covers $$5 \, \text{m}$$ in 1 s.

(d) The relative density of lead is 11.3. Its density is ________ g cm-3 or ________ kg m-3.

Solution:

Relative density (R.D.) = 11.3

Density of water = 1 g/cm3

$\text{Density of lead} = 11.3 \times 1 \, \text{g/cm}^3 = 11.3 \, \text{g/cm}^3$

Converting $$\text{g/cm}^3$$ to $$\text{kg/m}^3$$:

$1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \quad \Rightarrow \quad 11.3 \, \text{g/cm}^3 = 11300 \, \text{kg/m}^3$

Therefore, the density of lead is $$11.3 \, \text{g/cm}^3$$ or $$11300 \, \text{kg/m}^3$$.

Question 2.2: Fill in the blanks by suitable conversion of units

(a) 1 kg m2 s-2 = ________ g cm2 s-2

Solution:

We need to convert kg to g and m to cm:

$1 \, \text{kg} = 1000 \, \text{g}$

$1 \, \text{m} = 100 \, \text{cm}$

Therefore,

$1 \, \text{kg} \, \text{m}^2 \, \text{s}^{-2} = 1000 \, \text{g} \times (100 \, \text{cm})^2 \, \text{s}^{-2}$

$= 1000 \, \text{g} \times 10000 \, \text{cm}^2 \, \text{s}^{-2} = 10^7 \, \text{g} \, \text{cm}^2 \, \text{s}^{-2}$

Therefore, 1 kg m2 s-2 = $$10^7$$ g cm2 s-2.

(b) 1 m = ________ ly

Solution:

1 light year (ly) is the distance light travels in one year:

$1 \, \text{ly} = 9.461 \times 10^{15} \, \text{m}$

Therefore,

$1 \, \text{m} = \frac{1}{9.461 \times 10^{15}} \, \text{ly} \approx 1.057 \times 10^{-16} \, \text{ly}$

Therefore, 1 m = $$1.057 \times 10^{-16}$$ ly.

(c) 3.0 m s-2 = ________ km h-2

Solution:

We need to convert m/s2 to km/h2:

1 m = 0.001 km

$1 \, \text{s}^2 = \left( \frac{1}{3600} \, \text{h} \right)^2 = \frac{1}{12960000} \, \text{h}^2$

Therefore,

$3.0 \, \text{m/s}^2 = 3.0 \times 0.001 \, \text{km} \times 12960000 \, \text{h}^{-2} = 38880 \, \text{km/h}^2$

Therefore, 3.0 m s-2 = 38880 km h-2.

(d) G = 6.67 x 10-11 N m2 (kg)-2 = ________ (cm)3 s-2 g-1.

Solution:

We need to convert N to dynes, m to cm, and kg to g:

1 N = 105 dynes

$1 \, \text{m} = 100 \, \text{cm}$

$1 \, \text{kg} = 1000 \, \text{g}$

Therefore,

$G = 6.67 \times 10^{-11} \, \text{N} \, \text{m}^2 \, \text{(kg)}^{-2} \times 10^5 \, \text{dynes} \times (100 \, \text{cm})^2 \times (1000 \, \text{g})^{-2}$

$= 6.67 \times 10^{-11} \times 10^5 \times 10^4 \times 10^{-6} \, \text{(cm)}^3 \, \text{s}^{-2} \, \text{g}^{-1}$

$= 6.67 \times 10^{-8} \, \text{(cm)}^3 \, \text{s}^{-2} \, \text{g}^{-1}$

Therefore, G = $$6.67 \times 10^{-8}$$ (cm)3 s-2 g-1.

Question 2.3: Conversion of units for a calorie

A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units.

Solution:

Given: $$1 \, \text{cal} = 4.2 \, \text{J}$$

$1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$

In the new system of units:

1 unit of mass = $$\alpha \, \text{kg}$$

1 unit of length = $$\beta \, \text{m}$$

1 unit of time = $$\gamma \, \text{s}$$

We need to express 1 J in terms of the new units:

$1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} = \left( \frac{1 \, \text{kg}}{\alpha \, \text{kg}} \right) \cdot \left( \frac{1 \, \text{m}}{\beta \, \text{m}} \right)^2 \cdot \left( \frac{1 \, \text{s}}{\gamma \, \text{s}} \right)^{-2}$

$= \alpha^{-1} \cdot \beta^{-2} \cdot \gamma^{2}$

Therefore,

$4.2 \, \text{J} = 4.2 \cdot \alpha^{-1} \cdot \beta^{-2} \cdot \gamma^{2}$

Thus, a calorie has a magnitude of $$4.2 \alpha^{-1} \beta^{-2} \gamma^{2}$$ in terms of the new units.