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Question

Question: Fill in the blanks (a) The volume of a cube of side 1cm is equal to ____ \({m^3}\). (b) The surf...

Fill in the blanks
(a) The volume of a cube of side 1cm is equal to ____ m3{m^3}.
(b) The surface of a solid cylinder of radius 2.0cm and height 10.0cm is equal to ____ (mm)2{\left( {mm} \right)^2}
(c) A vehicle moving with a speed of 18 km/h covers ____ metres in one second.
(d) The relative density of lead is 11.3. Its density is ____ g/cm3g/c{m^3} or ____ kg/m3kg/{m^3}.

Explanation

Solution

Unit conversion is known as the conversion between different units of measurement for equivalent quantities, typically through multiplicative conversion factors. In order to find the solution to these kinds of questions, write down all the given physical quantities and then by using the units of measurement chart, we can convert from one unit to another.

Formula Used:
Volume of cube =a3 = {a^3}
Surface Area of cylinder =2πr(r+h) = 2\pi r\left( {r + h} \right)
Relative density =DensityDensity of water=ρρw = \dfrac{{Density}}{{{\text{Density of water}}}} = \dfrac{\rho }{{{\rho _w}}}

Complete answer:
(a) We know that,
1m = 100cm1m{\text{ }} = {\text{ }}100cm
1cm = 0.01m1cm{\text{ }} = {\text{ }}0.01m
Volume of the cube having side 1cm is given by,
Volume of cube =a3 = {a^3}
\eqalign{ & = {\left( {1cm} \right)^3} \cr & = {\left( {0.01m} \right)^3} \cr & = {10^{ - 6}}{m^3} \cr}
(b) Surface area of a solid cylinder is given by,
Surface Area of cylinder = Curved surface area + area of top + area of the base
\eqalign{ & = 2\pi rh + \pi {r^2} + \pi {r^2} \cr & = 2\pi r\left( {r + h} \right) \cr}
According to the question,
\eqalign{ & r{\text{ }} = {\text{ }}2cm{\text{ }} = {\text{ }}20mm \cr & h{\text{ }} = {\text{ }}10cm{\text{ }} = {\text{ }}100mm \cr}
Surface area of solid cylinder =2πr(r+h)=2×3.14×20(20+100)mm2 = 2\pi r\left( {r + h} \right) = 2 \times 3.14 \times 20\left( {20 + 100} \right)m{m^2}
=15072mm2= 15072m{m^2}
(C) It is given that the speed of a moving vehicle is 18km/hr
To find the distance covered by the moving vehicle in metres in one second, we will convert the speed of the moving vehicle from km/hr to m/s.
i.e. 1km = 1000m1km{\text{ }} = {\text{ }}1000m
1hr=(60×60)s=3600s1hr = \left( {60 \times 60} \right)s = 3600s
1km/hr =1000m3600s=518m/s = \dfrac{{1000m}}{{3600s}} = \dfrac{5}{{18}}m/s
So, 18km/h = 5m/s18km/h{\text{ }} = {\text{ }}5m/s.
Thus, distance travelled by a moving vehicle in one second is 5 metre.
(d) Mathematically relative density is given as,
Relative density =DensityDensity of water=ρρw = \dfrac{{Density}}{{{\text{Density of water}}}} = \dfrac{\rho }{{{\rho _w}}}
ρw=1000kg/m3{\rho _w} = 1000kg/{m^3}
It is given that the relative density of lead is 11.3.
So, ρ=11.3×1000kg/m3\rho = 11.3 \times 1000kg/{m^3}
Now, we know that
1kg = 1000g1kg{\text{ }} = {\text{ }}1000g
1m = 100cm1m{\text{ }} = {\text{ }}100cm
Substituting these in ‘ρ\rho ’ we get,
\eqalign{ & \rho = 1.13 \times {10^4}kg/{m^3} \cr & = 1.13 \times {10^4} \times \dfrac{{1000g}}{{{{100}^3}c{m^3}}} \cr & = 11.3g/c{m^3} \cr}
Thus, the density is 11.3g/cm311.3g/c{m^3} or 1.13×104kg/m31.13 \times {10^4}kg/{m^3}.
Hence, (a) The volume of a cube of side 1cm is equal to 106m3{10^{ - 6}}{m^3} .
(b) The surface of a solid cylinder of radius 2.0cm and height 10.0cm is equal to 15072mm215072m{m^2} .
(c) A vehicle moving with a speed of 18 km/h covers 5 metres in one second.
(d) The relative density of lead is 11.3. Its density is 11.3g/cm311.3g/c{m^3}or 1.13×104kg/m31.13 \times {10^4}kg/{m^3}.

Note:
Unit conversion is necessary because there are times when the units used in some measurement do not match the measurement preference then in such cases those units are converted to such extent that it can be understood directly and can be applied easily.