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 $\text m^{-3}$.

Answer 1

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 $10^{-6}\;\text m^3$ .

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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 × $10^4 \;\text {mm}^2$

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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.

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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

1 $\text{cm}^3$ = $10^{-6}\;\text m^3$
1 $\text{g}/\text{cm}^3$ = $\frac{10^{-3}}{10^{-6}}$ kg/ $\text m^3$ = $10^3\; \text{kg}/\text m^3$
$\therefore$ 11.3 $\text{g}/\text{cm}^3$ = 11.3 ×$10^3\; \text{kg}/\text m^3$