Q

# Fill in the blanks: (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...

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 .... $\small kg m^{-3}$.

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(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 = $1 \ gcm^{-3} = 1000\ kgm^{-3}$)

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