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

1. Volume of a cube:

Given the side of the cube is 1 cm.

We know that 1 cm = $10^\{-2\}$ m.

The volume of a cube is given by the formula $V\; =\; side^3$.

So, the volume of the cube is $1\; \backslash text\{\; cm\}^3$.

To convert this to cubic meters, we substitute the value of cm in meters:

$$V\; =\; (1\; \backslash text\{\; cm\})^3\; =\; (10^\{-2\}\; \backslash text\{\; m\})^3\; =\; 10^\{-6\}\; \backslash text\{\; m\}^3$$

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

2. Surface area of a cylinder:

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

The total surface area of a solid cylinder is given by the formula $S\; =\; 2\backslash pi\; r\; (r\; +\; h)$.

First, let's convert the dimensions to millimeters (mm), as the answer is required in (mm)². We know that 1 cm = 10 mm.

Radius $r\; =\; 2.0\; \backslash text\{\; cm\}\; =\; 2.0\; \backslash times\; 10\; \backslash text\{\; mm\}\; =\; 20\; \backslash text\{\; mm\}$.

Height $h\; =\; 10.0\; \backslash text\{\; cm\}\; =\; 10.0\; \backslash times\; 10\; \backslash text\{\; mm\}\; =\; 100\; \backslash text\{\; mm\}$.

Now, substitute these values into the surface area formula:

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

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

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

Using $\backslash pi\; \backslash approx\; 3.14$:

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

So, the surface area is approximately $1.5\; \backslash times\; 10^4$ mm².

3. Distance covered by a vehicle:

Given speed = 18 km h^-1 and time = 1 s.

We need to convert the speed from km/h to m/s. The conversion factor is $1\; \backslash text\{\; km/h\}\; =\; \backslash frac\{5\}\{18\}\; \backslash text\{\; m/s\}$.

Speed in m/s = $18\; \backslash times\; \backslash frac\{5\}\{18\}\; \backslash text\{\; m/s\}\; =\; 5\; \backslash text\{\; m/s\}$.

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

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

Therefore, the vehicle covers 5 m in 1 s.

4. Density of lead:

Given relative density of lead = 11.3.

The relative density of a substance is defined as the ratio of its density to the density of water at a standard temperature (usually 4°C).

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

The density of water is approximately $1\; \backslash text\{\; g/cm\}^3$.

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

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

Now, we convert this density to kg/m³.

We know that $1\; \backslash text\{\; g\}\; =\; 10^\{-3\}\; \backslash text\{\; kg\}$ and $1\; \backslash text\{\; cm\}\; =\; 10^\{-2\}\; \backslash text\{\; m\}$, so $1\; \backslash text\{\; cm\}^3\; =\; (10^\{-2\}\; \backslash text\{\; m\})^3\; =\; 10^\{-6\}\; \backslash text\{\; m\}^3$.

Therefore, $1\; \backslash text\{\; g/cm\}^3\; =\; \backslash frac\{10^\{-3\}\; \backslash text\{\; kg\}\}\{10^\{-6\}\; \backslash text\{\; m\}^3\}\; =\; 10^3\; \backslash text\{\; kg/m\}^3$.

So, the density of lead in kg/m³ is:

$$11.3\; \backslash text\{\; g/cm\}^3\; =\; 11.3\; \backslash times\; 10^3\; \backslash text\{\; kg/m\}^3\; =\; 11300\; \backslash text\{\; kg/m\}^3$$

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