CBSE Class 11 Physics Chapter 2: Units and Measurement NCERT Solutions

Solution:

1. Volume of a cube:

We are given a cube with a side length of 1 cm. We need to find its volume in cubic meters (m³).

First, convert the side length from centimeters to meters: 1 cm = $\frac{1}{100}$ m = $10^{-2}$ m.

The volume of a cube is given by the formula: Volume = (side)³.

So, Volume = $(1 \text{ cm})^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3$.

Therefore, the volume of a cube of side 1 cm is equal to $10^{-6}$ m³.

2. Surface area of a cylinder:

The total surface area (S) of a solid cylinder with radius (r) and height (h) is given by the formula: $S = 2\pi r (r + h)$.

Given: radius $r = 2.0$ cm and height $h = 10.0$ cm.

We need to express the answer in (mm)². First, convert the radius and height to millimeters (mm):

$r = 2.0 \text{ cm} = 2.0 \times 10 \text{ mm} = 20 \text{ mm}$

$h = 10.0 \text{ cm} = 10.0 \times 10 \text{ mm} = 100 \text{ mm}$

Now, substitute these values into the surface area formula:

$S = 2 \times \pi \times r \times (r + h)$

$S = 2 \times \pi \times (20 \text{ mm}) \times (20 \text{ mm} + 100 \text{ mm})$

$S = 2 \times \pi \times 20 \text{ mm} \times (120 \text{ mm})$

$S = 4800 \pi \text{ mm}^2$

Using $\pi \approx 3.14$,

$S \approx 4800 \times 3.14 \text{ mm}^2 \approx 15072 \text{ mm}^2$.

So, the surface area is approximately $15072$ mm².

3. Distance covered by a vehicle:

The speed of the vehicle is given as 18 km h⁻¹.

We need to find the distance covered in 1 second (s).

First, convert the speed from km/h to m/s using the conversion factor: 1 km/h = $\frac{5}{18}$ m/s.

Speed = $18 \text{ km/h} = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s}$.

Distance is calculated using the formula: Distance = Speed × Time.

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

Therefore, the vehicle covers 5 m in 1 s.

4. Density of lead:

The relative density of lead is given as 11.3.

Relative density is defined as the ratio of the density of a substance to the density of water.

Relative density = $\frac{\text{Density of substance}}{\text{Density of water}}$

The density of water is approximately 1 g/cm³ or 1000 kg/m³.

Using the density of water as 1 g/cm³:

Density of lead = Relative density of lead × Density of water

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

Now, convert this density to kg/m³.

We know that 1 g = $10^{-3}$ kg and 1 cm³ = $(10^{-2} \text{ m})^3 = 10^{-6}$ m³.

So, 1 g/cm³ = $\frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 10^3 \text{ kg/m}^3$.

Therefore, the density of lead in kg/m³ is:

Density of lead = $11.3 \text{ g/cm}^3 = 11.3 \times 10^3 \text{ kg/m}^3 = 11300 \text{ kg/m}^3$.

Hence, the density of lead is 11.3 g/cm³ or 11300 kg/m³.