Fill in the blanks
(a) The volume of a cube of side 1 cm isk equal to….. `m^3`
(b) the surface area fo a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …. `(mm)^2`
(c ) A vehical moving with a speed of `18km h^(-1)` covers ....m in 1s.
(d) The relative density of lead is 11.3. its density is ......g `cm^(-3) or .....kg m^(-3)`

Explanation

`1 cm =1/100 m`
volume of the cube `1 cm^(3)`
but , `1cm^(3) =1cm xx1 cm xx1cm =(1/100) m xx(1/100) m xx(1/100) m `
`:. 1 cm^(3)=10^(-6) m^(3)`
Hence , the volume of a cube of side 1 cm is equal to `10^(-6) m^(3)`
(b) the total surface area of a cylinder of radius r and height h is
`S=2pir(r+h)`
given that,
`r=2 cm =2xx1 cm =2xx10 mm =20 mm`
`h=10 cm =10xx10 mm=100 mm`
`:. s=2xx3.14 xx20xx(2+100) =15072=1.5xx10^(4) mm^(2)`
(c) Using the conservation,
`1km//h=5/18 m//s`
`18km//h =18xx5/18 =5 m//s`
Therefore, distance can be obtaied using the relation
Distance =Speed x time =5 x 1 =5 m
Hence, the vehicle covers 5 m in 1 s.
(d) Relative density of a substance is given by the releation,
Relative =density =`("Density of substance")/("Density of water")`
Density of water `=1g//cm^(3)`
Density of lead=realative density of lead x density of water
`=11.3xx1=11.3 g//cm^(3)`
Again, `1g =1/1000 kg `
`1 cm^(3) =10^(-6) m^(3)`
`1g//cm^(3) =(10^(-3))/(10^(-6)) kg /m^(3) =10^(3) kg //m^(3)`
`:. 11.3 g//cm^(3) =11.3 xx10^(3) kg //m^(3)`