Fill in the blanks: The volume of a cube of side 1cm is equal to .....$m^3$.
Answer
The volume of a cube of side 1cm is equal to $10^{-6} m^3$. Length of edge $ = 1\text{cm} = \frac{1}{100\text{m}} $ Volume of the cube = $side^3$ Putting the value of side, we get Volume of the cube $=\Big(\frac{1}{100\text{m}}\Big)^3$ $\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\Big(\frac{1}{100}\Big)\text{m}^3=\frac{1}{6^6}\text{m}^3=10^{-6}\text{m}^3$
Fill in the blanks: The relative density of lead is $11.3$. Its density is ....g $cm^{-3}$ or ....kg $m^{-3}.$
Answer
The relative density of lead is 11.3. Its density is $11.3 g cm^{-3}$ or $1.13 \times 10^3kg m^{–3}.$
Density of lead = Relative density of lead × Density of water Density of water $= 1g/ cm^3$
Putting the values, we get Density of lead
$= 11.3 \times 1g/ cm^3$
$= 11.3g cm^{-3} 1cm$
$= (1/100m) =10^{–2}m^3 1g$
$= 1/1000kg = 10^{-3}kg$
Density of lead $= 11.3g cm^{-3} = 11.3$
Putting the value of 1cm and 1 gram $11.3g/ cm^3$
$= 11.3 \times 10^{-3}kg (10^{-2}m)^{-3}$
$= 11.3 \times 10^{–3} \times 10^6kg m^{-3}$
$=1.13 \times 10^3kg m^{–3}$
Fill in the blanks: The surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to ...$(mm)^2$
Answer
The surface area of a solid cylinder of radius $2.0\ cm$ and height $10.0\ cm$ is equal to $1.5 \times 10^4mm^2$
Given, Radius, $r = 2.0\ cm = 20\ mm$ (convert cm to mm)
Height, h = 10.0cm =100mm
The formula of total surface area of a cylinder $\text{S}=2\pi\text{r}(\text{r}+\text{h})$
Putting the values in this formula, we get Surface area of a cylinder
$\text{S} = 2\pi\text{r} (\text{r} + \text{h}) = 2\times3.14 \times 20 (20+100)$
$= 15072 = 1.5\times104\text{mm}^2$
Fill in the blanks: A vehicle moving with a speed of $18 \mathrm{~km} \mathrm{~h}^{-1}$ covers .... m in 1 s
Answer
A vehicle moving with a speed of $18 \mathrm{~km} \mathrm{~h}^{-1}$ covers $\mathbf{5 m}$ in 1 s .
Using the conversion, Given, Time, $\mathrm{t}=1 \mathrm{sec}$ speed $=$ $18 \mathrm{~km} \mathrm{~h}^{-1}=18 \mathrm{~km} /$ hour $1 \mathrm{~km}=1000 \mathrm{~m}$ and 1 hour $=3600 \mathrm{sec}$ Speed $=18 \times 1000 / 3600 \mathrm{sec}=5 \mathrm{~m} / \mathrm{sec}$ Use formula
Speed $=$ distance/time Cross multiply it, we get Distance $=$ Speed $\times$ Time $=5 \times 1=5 \mathrm{~m}$
If m, v and c respectively denote mass, speed and the velocity of light, then in the equation $\text{m} = \text{m}_0\Big(1-\frac{\text{v}^2}{\text{c}^2}\Big)^{\frac{-1}{2}}$ $m_0$ has the dimensions of ______.
Answer
If m, v and c respectively denote mass, speed and the velocity of light, then in the equation $\text{m} = \text{m}_0\Big(1-\frac{\text{v}^2}{\text{c}^2}\Big)^{\frac{-1}{2}}$ $m_0$ has the dimensions of mass.
Fill in the blanks by suitable conversion of units: $1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}=\ldots . . \mathrm{g} \mathrm{cm}^2 \mathrm{~s}^{-2}$