Q) 2.1 Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to .....m^3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...\small (mm)^2

(c) A vehicle moving with a speed of \small 18\ km h^{-1} covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is .... \small g cm^{-3} or .... \small kg m^{-3}.

Answers (1)

(a) We know, \small 1 cm = .01 m    (Tip: Divide by 100 to convert cm to m)

                The volume of a cube of side a  = \small a^3\ m^3

                Volume of cube of side 1 cm (i.e, .01 m ) = \small (.01)^3 = 10^{-6}\ m^3

(b) We know, 1 cm = 10 mm   (Tip: Multiply by 10 to convert cm to mm)

                The surface area of a solid cylinder of radius r and height h = 2(\pi r^2) + 2 \pi rh = 2\pi r(r + h)

                Required area = 2\pi (20)(\ 20 + 100) mm^3 = 40 \pi (120) mm^3 = 4800 \pi\ mm^3 = 1.5 \times 10^{4 }\ mm^3

(c)  (Tip: multiply by 5/18 to convert  km h^{-1}\ to\ m s^{-1} )

        Ditance covered = Speed \times time = (18 \times 5/18)ms^{-1} \times (1s) = 5 m

(d)  Density = Relative Density \times Density of water

        (Density of water = 1 \ gcm^{-3} = 1000\ kgm^{-3})

  Density of lead= 

         (11.3)\times1\ gcm^{-3} = 11.3\ gcm^{-3} \\ (11.3)\times1000\ kgm^{-3} = 11300\ kgm^{-3} = 1.13 \times 10^4\ kgm^{-3}

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