Question

Fill in the blanks
(i) The volume of a cube of side 1 cm is equal to ______$m^3$.

(ii)Fill in the blanks
The surface area of a solid cylinder of radius $2.0cm$ and height 10.0 cm is equal to _______ $(mm{)}^2$.

(iii) Fill in the blanks
A vehicle moving with a speed of $18km{h}^{-1}$ covers $m$ in $1s$

(iv) Fill in the blanks
The relative density of lead is $11.3$. Its density is $gc{m}^{-3}$ or $kg{m}^{-3}$

• A. $1.13\times {10}^{4}kg{m}^{-3}$
• B. $5m$
• C. $11.3gc{m}^{-3}$

Answer 1

Answer: (A), (B), (C)

(i) Volume of a cube of side $a,V=a^3$
Given : Side length $a=1cm$

We know that $1cm={10}^{-2}m$
Hence, volume of a cube, $V=({10}^{-2}m{)}^3={10}^{-6}{m}^3$

(ii) Total surface area(S) = Surface area of two caps + Curved surface area
Given : Radius- $r=2\text{cm}$ , Height- $h=10cm$

Surface area of two caps $=2\pi r^2$
Surface area of curved surface $=2\pi rh$
Total surface area $\left(s\right)=2\pi r^2+2\pi rh$

$S=2\pi r^2+2\pi rh\Rightarrow 2\pi (r+h)$
$S=2\times \frac{22}{7}\times 20(20+100)m{m}^2$
$\Rightarrow S=15086m{m}^2=1.5086\times {10}^{4}m{m}^2$

(iii) Given : speed- $v=18km{h}^{-1}$, Time- $t=1s$
$v=18km{h}^{-1}=\frac{18\times 1000m}{3600s}$
Formula : $x=vt$
$=5\times 1=5m$

(iv)Given : Relative density of lead - $11.3$
We know that : density of water $=1gc{m}^{-3}$

Formula: Relative density of lead = $\frac{\text{density of lead}}{\text{density of water}}$

density of lead = relative density of lead$\times$ density of water
$=11.3\times 1gc{m}^{-3}=11.3gc{m}^{-3}$

In S.I. System: density of water $={10}^{3}kg{m}^{-3}$
density of lead = relative density of lead$\times$ density of water
$=11.3\times {10}^{3}kg{m}^{-3}$
$=1.3\times {10}^{4}kg{m}^{-3}$