Class 11 NCERT PHYSICS • Chapter 1: Units and Measurement • Question 1

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We need to calculate volume, surface area, distance, and density using unit conversion factors and standard geometric formulas.

Concept: Unit Conversion

Unit Conversion: The process of changing the units of a measurement without changing its magnitude, using conversion factors like $1\,\text{cm} = 10^{-2}\,\text{m}$ and $1\,\text{km/h} = \frac{5}{18}\,\text{m/s}$ .

Concept: Volume of a Cube

Volume of a Cube: Defined as $V = a^3$ , where $a$ is the side length.

Concept: Surface Area of a Cylinder

Surface Area of a Cylinder: For a solid cylinder, the total surface area is $A = 2\pi r(r + h)$ .

Concept: Relative Density

Relative Density: The ratio of the density of a substance to the density of water ( $1\,\text{g}\cdot\text{cm}^{-3}$ or $1000\,\text{kg}\cdot\text{m}^{-3}$ ).

Given:

(a) Side of cube, $a = 1\,\text{cm}$
(b) Radius, $r = 2.0\,\text{cm}$ ; Height, $h = 10.0\,\text{cm}$
(c) Speed, $v = 18\,\text{km/h}$ ; Time, $t = 1\,\text{s}$
(d) Relative density of lead, $\text{RD} = 11.3$

Solving:

Volume of the cube in $\text{m}^3$
- Convert side length to meters: $1\,\text{cm} = 10^{-2}\,\text{m}$ .

\begin{aligned} V &= a^3 \\ &= (10^{-2}\,\text{m})^3 \\ &= 10^{-6}\,\text{m}^3 \end{aligned}

Surface area of the cylinder in $\text{mm}^2$
- Convert dimensions to millimeters: $r = 20\,\text{mm}$ , $h = 100\,\text{mm}$ .

\begin{aligned} A &= 2\pi r(r + h) \\ &= 2 \cdot \pi \cdot 20 \cdot (20 + 100) \\ &= 40\pi \cdot 120 \\ &= 4800\pi \\ &\approx 1.5 \times 10^4\,\text{mm}^2 \end{aligned}

Distance covered by the vehicle in 1 s
- Convert speed to $\text{m/s}$ : $18\,\text{km/h} = 18 \cdot \frac{5}{18}\,\text{m/s} = 5\,\text{m/s}$ .

\begin{aligned} d &= v \cdot t \\ &= 5\,\text{m/s} \cdot 1\,\text{s} \\ &= 5\,\text{m} \end{aligned}

Density of lead
- Density in $\text{g}\cdot\text{cm}^{-3}$ :

\begin{aligned} \rho &= \text{RD} \cdot \rho_{\text{water}} \\ &= 11.3 \cdot 1\,\text{g}\cdot\text{cm}^{-3} \\ &= 11.3\,\text{g}\cdot\text{cm}^{-3} \end{aligned}

Density in $\text{kg}\cdot\text{m}^{-3}$ :

\begin{aligned} \rho &= 11.3 \cdot 1000\,\text{kg}\cdot\text{m}^{-3} \\ &= 1.13 \times 10^4\,\text{kg}\cdot\text{m}^{-3} \end{aligned}

Answer:

(a)

10^{-6}

(b)

1.5 \times 10^4

(c)

5

(d)

11.3

1.13 \times 10^4

NCERT Solutions

Units and Measurement

Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ..... $m^3$

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

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

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

SOLUTION

We need to calculate volume, surface area, distance, and density using unit conversion factors and standard geometric formulas.

Concept: Unit Conversion

Concept: Volume of a Cube

Concept: Surface Area of a Cylinder

Concept: Relative Density

Given:

Solving:

Volume of the cube in $\text{m}^3$

Surface area of the cylinder in $\text{mm}^2$

Distance covered by the vehicle in 1 s

Density of lead

Answer:

Units and Measurement

Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ..... m3m^3m3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... mm2{mm}^2mm2

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

(d) The relative density of lead is 11.3. Its density is .... gcm–3g {cm}^{–3}gcm–3 or .... kgm–3kg m^{–3}kgm–3.

SOLUTION

We need to calculate volume, surface area, distance, and density using unit conversion factors and standard geometric formulas.

Concept: Unit Conversion

Concept: Volume of a Cube

Concept: Surface Area of a Cylinder

Concept: Relative Density

Given:

Solving:

Volume of the cube in m3\text{m}^3m3

Surface area of the cylinder in mm2\text{mm}^2mm2

Distance covered by the vehicle in 1 s

Density of lead

Answer:

(a) The volume of a cube of side 1 cm is equal to ..... $m^3$

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

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

Volume of the cube in $\text{m}^3$

Surface area of the cylinder in $\text{mm}^2$