Question:

Fill in the blanks 
  1. The volume of a cube of side 1 cm is equal to________\(m^3\).
  2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to__________\(\text(mm)^2\).
  3. A vehicle moving with a speed of 18 km \(h^{-1}\) covers ________m in 1 s.
  4. The relative density of lead is 11.3. Its density is ....g \(\text {cm}^{-3}\) or ________kg \(\text m^{-3}\).

Updated On: Nov 1, 2023
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Solution and Explanation

a. The volume of a cube of side 1 cm is equal to \(\underline{10^{-6}\;\text m^3}.\)
explanation: 
1 cm =\(\frac{1}{100}\)m
Volume of the cube = 1 \(\text{cm}^3\)
But, 1 \(\text{cm}^3\) = 1 cm × 1 cm × 1 cm = \(\bigg(\frac{1}{100}\bigg)\)m × \(\bigg(\frac{1}{100}\bigg)\)m × \(\bigg(\frac{1}{100}\bigg)\)m
\(\therefore\) 1 \(\text{cm}^3\) = \(10^{-6}\;\text m^3\)
Hence, the volume of a cube of side 1 cm is equal to \(10^{-6}\;\text m^3\) .


b. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to \(\underline{1.5 \times10^4 \;\text {mm}^2}.\)
explanation: 
The total surface area of a cylinder of radius r and height h is S = 2\(\pi\)r (r + h).
Given that, r = 2 cm = 2 × 1 cm = 2 × 10 mm = 20 mm
h = 10 cm = 10 × 10 mm = 100 mm 
\(\therefore\) S = 2× 3.14 × 20 × (20 + 100) = 15072 = 1.5 × \(10^4 \;\text {mm}^2\)


c. A vehicle moving with a speed of 18 km \(h^{-1}\) covers \(\underline{5\;\text m\;\text in \;1 \text s.}\)
explanation: 
Using the conversion, 1 km/h =\(\frac{5}{18}\) m / s
18 km / h = 18 ×\(\frac{5}{18}\)= 5 m / s
Therefore, distance can be obtained using the relation: 
Distance = Speed × Time = 5 × 1 = 5 m
Hence, the vehicle covers 5 m in 1 s.


d. The relative density of lead is 11.3. Its density is \(\underline{11.3\;\text{g}/\text{cm}^3}\) or \(\underline{11.3\times10^3\; \text{kg}/\text m^3}.\)
explanation: 
Relative density of a substance is given by the relation, 
Relative density =\(\frac{ \text{Density of substance}}{ \text{Density of water}}\)
Density of water = 1 \(\text{g}/\text{cm}^3\)
Density of lead = Relative density of lead × Density of water = 11.3 × 1 = 11.3 \(\text{g}/\text{cm}^3\)

Again , 1g =\(\frac{1}{1000}\) kg

\(\text{cm}^3\) = \(10^{-6}\;\text m^3\)
\(\text{g}/\text{cm}^3\) = \(\frac{10^{-3}}{10^{-6}}\) kg/ \(\text m^3\) = \(10^3\; \text{kg}/\text m^3\)
\(\therefore\) 11.3 \(\text{g}/\text{cm}^3\) = 11.3 ×\(10^3\; \text{kg}/\text m^3\)

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Concepts Used:

Units and Measurement

Unit:

A unit of a physical quantity is an arbitrarily chosen standard that is broadly acknowledged by the society and in terms of which other quantities of similar nature may be measured.

Measurement:

The process of measurement is basically a comparison process. To measure a physical quantity, we have to find out how many times a standard amount of that physical quantity is present in the quantity being measured. The number thus obtained is known as the magnitude and the standard chosen is called the unit of the physical quantity.

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System of Units:

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  3. MKS system
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Types of Units:

Fundamental Units -

The units defined for the fundamental quantities are called fundamental units.

Derived Units -

The units of all other physical quantities which are derived from the fundamental units are called the derived units.