Question

Give some incident to show that the man referred to in the extract used to insult shylor

Answer 1

UR QUESTON:-

an iron pillar consisted of a cylindrical portion of 2.8 m height and 20 cm in diameter and a cone of 42 cm. Find the weight of the pillar if 1 cm³ of iron weights 7.5 g

UR ANSWER✓

$\Large\bold\purple{given,}$

$.Height \:of \:cylindrical \:portion \:= 2.8 m$

$Diameter\: of \:cylindrical \:portion \:= 20 cm$

$Height \:of \:the\: cone\: = 42 cm$

$\Large\underline\bold{TO\:FIND,}$

$\sf\large\dashrightarrow \:weight\:of\:pillar$

$\Large\underline\bold{SOLUTION,}$

$\sf\implies \:volume\:of\:pillar=volume\:of\:cone+\:volume\:of\:cylinder$

$\Large\underline\bold{for\:cylinder:-}$

$radius= \dfrac{diameter}{2}$

$\sf\implies \frac{20}{2}$

$\sf\implies 10cm$

$\sf\therefore height\:=2.8m$

converstion m into centimeter

$\sf\implies 2.8 m \times 100$

$\sf\implies height=280cm$

$. Volume = \pi r^2h$

$\sf\implies \dfrac{22}{7} \times 10 \times 10 \times 280$

$\sf\implies \dfrac{22}{7} \times 28000$

$\sf\implies \dfrac{616000}{7}$

$\sf\implies \cancel \dfrac{616000}{7}$

$\sf\implies 88000cm^3$

$\large{\boxed{\sf{ 88000cm^3}}}$

$\Large\underline\bold{for\:cone:-}$

$\sf Radius = \frac{diameter}{2}$

$\sf\implies \frac{20}{2}$

$\sf\implies 10 cm$

$\sf\therefore height=42cm$

$\sf\therefore Volume = \dfrac{1}{3} \pi r^2h$

$\sf\implies dfrac{1}{3} \times \dfrac{22}{7} \times 10 \times 10 \times 42$

$\sf\implies dfrac{1}{3} \times \dfrac{22}{7} \times 100 \times 42$

$\sf\implies dfrac{1}{3} \times \dfrac{22}{7} \times 4200$

$\sf\implies \dfrac{22}{21} \times 4200$

$\sf\implies \dfrac{92400}{21}$

$\sf\implies \cancel \dfrac{92400}{21}$

$\large{\boxed{\sf{{4400m^3}}}$

$\Large\underline\bold{now,}$

$\sf\therefore volume\:of\:cylinder= 88000cm^3$

$\sf\therefore volume\:of\:cone= 4400cm^3$

$\sf\implies \:volume\:of\:pillar=volume\:of\:cone+\:volume\:of\:cylinder$

$\sf\implies \:volume\:of\:pillar=88000+4400$

$\sf\implies 92400cm^3$

$\large{\boxed{\sf{92400cm^3}}}$

Total weight of pillar at a weight of 7.5 g per 1cm³

= 92400 × 7.5

$\sf {\fbox {answer\:=\:693000gms}}$

$\sf\implies conversion$

=$\sf \implies \dfrac{693000}{1000}kg$

$\sf\implies \cancel \frac{693000}{1000}kg$

$\sf {\fbox {answer\:=\:693kg }}$

__________________

ADDITIONAL INFORMATION,

$\Large\underline\bold{1)diagram\:of\:cylinder}$

$\Large\underline\bold{2)diagram\:of\: cone}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(5,6)\qbezier(18,0)(10,5)(0,0)\qbezier(18,0)(10,-5)(0,0)\qbezier(0,-27)(8,-33)(18,-27)\qbezier(0,-27)(8,-22)(18,-27)\put(0,-27){\line(0,1){27}}\put(18,-27){\line(0,1){27}}\put(9,-27){\circle*{0.6}}\multiput(9,-27)(0,2){14}{\line(0,1){1}}\put(9,0){\circle*{0.6}}\multiput(9,-27)(2,0){5}{\line(1,0){1}}\put(10,-29){10cm}\put(10,-13){2.8m}\end{picture}$

$\begin{lgathered}\setlength{\unitlength}{0.99cm}\begin{picture}(6, 4)\linethickness{0.26mm}\qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{42cm}}\put(3,2){\line(0,2){4.5}}\put(1.5,1.7){\sf{10cm}}\qbezier(.2,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(1,4){\sf l}\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}\\\end{lgathered}$