An open wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5 cm thick wood. Find (i) the capacity of the box, (ii) volume of wood used and (iii) weight of the box, it being given that 100 cm³ of wood weight 8 g.
Answers
- Here the concept of volume of cuboid has been used with the given dimensions we can find the volume of the box. Then we can divide the volume of 100 cm³ and multiply with 8 to find the total weight of box.
An open wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5 cm thick wood. Find
(i) the capacity of the box,
(ii) volume of wood used and
(iii) weight of the box, it being given that 100 cm³ of wood weight 8 g.
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Answer:
Here the concept of volume of cuboid has been used with the given dimensions we can find the volume of the box. Then we can divide the volume of 100 cm³ and multiply with 8 to find the total weight of box.
\huge\underline{\underline{\texttt{{Formula\:Used:-}}}}
FormulaUsed:-
\star{\boxed{\sf{volume \: of \: cuboid \: = lenght \times breath \times height}}}⋆
volumeofcuboid=lenght×breath×height
\star{\boxed{\sf{total \: weight \: of \: box \: = \: \frac{difference \: in \: volume \: of \: box}{100} \times 8}}}⋆
totalweightofbox=
100
differenceinvolumeofbox
×8
\huge\underline{\underline{\texttt{{Question:}}}}
Question:
An open wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5 cm thick wood. Find
(i) the capacity of the box,
(ii) volume of wood used and
(iii) weight of the box, it being given that 100 cm³ of wood weight 8 g.
\huge\underline{\underline{\texttt{{Solution:}}}}
Solution:
\begin{gathered}\small\bold\orange{external \: length \: of \: the \: box = 80 \: cm} \\ \end{gathered}
externallengthofthebox=80cm
\begin{gathered}\small\bold\orange{external \: breadth \: of \: the \: box = 65 \: cm.} \\ \end{gathered}
externalbreadthofthebox=65cm.
\begin{gathered}\small\bold\orange{external \: height \: of \: the \: box = 45 \: cm.} \\ \end{gathered}
externalheightofthebox=45cm.
\begin{gathered}\small\bold{external \: volume \: of \: the \: box = (80 \times 65 \times 45)cu \: cm} \\ = 234000 {cm}^{3} \end{gathered}
externalvolumeofthebox=(80×65×45)cucm
=234000cm
3
\begin{gathered}\small\bold{internal \: lenght \: of \: the \: box = (80 - (2.5 \times 2))cm = (80 - 5)cm} \\ = 75 \: cm.\end{gathered}
internallenghtofthebox=(80−(2.5×2))cm=(80−5)cm
=75cm.
\begin{gathered}\small\bold{internal \: breadth \: of \: the \: box = (65 - (2.5 \times 2))cm = (65 - 5)cm} \\ = 60 \: cm.\end{gathered}
internalbreadthofthebox=(65−(2.5×2))cm=(65−5)cm
=60cm.
\small\bold{internal \: height \: of \: the \: box = (45 - 2.5)cm = 42.5 \: cm.}internalheightofthebox=(45−2.5)cm=42.5cm.
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\begin{gathered}\small\bold\pink{(i) \: capacity \: of \: the \: box = internal \: volume \: of \: the \: box} \\ \: \: \: \: \: \: \: \: \: \small\bold\pink{ =(75 \times 60 \times 42.5) {cm}^{3} = 191250 {cm}^{3} }\end{gathered}
(i)capacityofthebox=internalvolumeofthebox
=(75×60×42.5)cm
3
=191250cm
3
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\begin{gathered}\small\bold\red{(ii) \: volume \: of \: wood \: used} \\\end{gathered}
(ii)volumeofwoodused
\begin{gathered}\small\bold\red{ = (external \: volume) - (internal \: volume)} \\ \end{gathered}
=(externalvolume)−(internalvolume)
\begin{gathered}\small\bold\red{ = (234000 - 191250) {cm}^{3} = 42750 {cm}^{3}. } \\ \end{gathered}
=(234000−191250)cm
3
=42750cm
3
.
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\begin{gathered}\small\bold\blue{(iii) \: weight \: of \: 100 \: {cm}^{3} \: of \: wood = 8g. } \\ \end{gathered}
(iii)weightof100cm
3
ofwood=8g.
\begin{gathered}\small\bold\blue{weight \: of \: 42750 \: {cm}^{3}of \: wood = ( \frac{8}{100} \times 42750)g = 3420 \: g } \\ \small\bold\blue{ = 3 \: kg \: 420 \: g.}\end{gathered}
weightof42750cm
3
ofwood=(
100
8
×42750)g=3420g
=3kg420g.
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\huge\underline{\underline{\texttt{{More\:to\:know:}}}}
Moretoknow:
\small\bold\green{volume \: of \: cube \: = {(side)}^{3} }volumeofcube=(side)
3
\small\bold\green{volume \: of \: cylinder = \ {\pi \: r}^{2}h }volumeofcylinder= πr
2
h
\small\bold\green{volume \: of \: cone = \frac{1}{3} \times \pi \: rh}volumeofcone=
3
1
×πrh
\begin{gathered}\small\bold\green{volume \: of \: hemisphere = \frac{2}{3} \times {\pi \: rh}^{2} } \\ \end{gathered}
volumeofhemisphere=
3
2
×πrh
2
\begin{gathered}\small\bold\green{TSA \: of \: cuboid \: = 2(lb + bh + lh)} \\ \end{gathered}
TSAofcuboid=2(lb+bh+lh)
\begin{gathered}\small\bold\green{CSA \: of \: cuboid = 2 \times (l + b) \times h} \\ \end{gathered}
CSAofcuboid=2×(l+b)×h
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