RD Sharma solution class 8 chapter 21 Mensuration II(Volume and Surface Area of Cuboid and a Cube) Exercise 21.2

Exercise 21.2

Page-21.15

Question 1:

Find the volume in cubic metre (cu. m) of each of the cuboids whose dimensions are:
(i) length = 12 m, breadth = 10 m, height = 4.5 cm
(ii) length = 4 m, breadth = 2.5 m, height = 50 cm.
(iii) length = 10 m, breadth = 25 dm, height = 50 cm.

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Length}=12 \mathrm{m} \\ \mathrm{Breadth}=10 \mathrm{m} \\ \mathrm{Height}=4.5 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cubo}\mathrm{id}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =12\times 10\times 4.5 \\ =540 {\mathrm{m}}^3 \\ \left(\mathrm{ii}\right) \\ \mathrm{Length}=4 \mathrm{m} \\ \mathrm{Breadth}=2.5 \mathrm{m} \\ \mathrm{Height}=50 \mathrm{cm} \\ =\frac{50}{100}\mathrm{m} (\because 1 \mathrm{m} = 100 \mathrm{cm} ) \\ =0.5 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cub}\mathrm{oid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =4\times 2.5\times 0.5 \\ =5 {\mathrm{m}}^3 \\ \left(\mathrm{iii}\right) \\ \mathrm{Length}=10 \mathrm{m} \\ \mathrm{Breadth}=25 \mathrm{dm} \\ =\frac{25}{10}\mathrm{m} (\because 10 \mathrm{dm}= 1\mathrm{m}) \\ =2.5 \mathrm{m} \\ \mathrm{Height}=25 \mathrm{cm}=\frac{25}{100}\mathrm{m}=0.25 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cub}\mathrm{oid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =10\times 2.5\times 0.25 \\ =6.25 {\mathrm{m}}^3 \end{gathered}$$

Question 2:

Find the volume in cubic decimetre of each of the cubes whose side is
(i) 1.5 m
(ii) 75 cm
(iii) 2 dm 5 cm

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{cube}=1.5 \mathrm{m} \\ =1.5\times 10 \mathrm{dm} (\because 1 \mathrm{m}= 10 \mathrm{dm}) \\ =15 \mathrm{dm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=(15{)}^3=3375 {\mathrm{dm}}^3 \\ \left(\mathrm{ii}\right) \\ \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{cube}=75 \mathrm{cm} \\ =75\times \frac{1}{10} \mathrm{dm} (\because 1 \mathrm{dm}=10 \mathrm{cm}) \\ =7.5 \mathrm{dm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=(7.5{)}^3=421.875 {\mathrm{dm}}^3 \\ \left(\mathrm{iii}\right) \\ \mathrm{Side} \mathrm{of} \mathrm{the} \mathrm{cube} =2 \mathrm{dm} 5 \mathrm{cm} \\ =2 \mathrm{dm}+5\times \frac{1}{10} \mathrm{dm} (\because 1 \mathrm{dm}=10 \mathrm{cm}) \\ =2 \mathrm{dm}+0.5 \mathrm{dm} \\ =2.5 \mathrm{dm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=(2.5{)}^3=15.625 {\mathrm{dm}}^3 \end{gathered}$$

Question 3:

How much clay is dug out in digging a well measuring 3 m by 2 m by 5 m?

Answer 3:

$$\begin{gathered} \mathrm{The} \mathrm{measure} \mathrm{of} \mathrm{well} \mathrm{is} 3 \mathrm{m}\times 2 \mathrm{m}\times 5 \mathrm{m}. \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{clay} \mathrm{dug} \mathrm{out}=(3\times 2\times 5) {\mathrm{m}}^3=30 {\mathrm{m}}^3 \end{gathered}$$

Question 4:

What will be the height of a cuboid of volume 168 m³, if the area of its base is 28 m²?

Answer 4:

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=168 {\mathrm{m}}^3 \\ \mathrm{Area} \mathrm{of} \mathrm{its} \mathrm{base}=28 {\mathrm{m}}^2 \\ \mathrm{Let} h \mathrm{m} \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid}. \\ \mathrm{Now}, \mathrm{we} \mathrm{have} \mathrm{the} \mathrm{following}: \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{base}=\mathrm{lenght}\times \mathrm{breadth} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{lenght}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=(\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base})\times \mathrm{height} \\ \Rightarrow 168=28\times h \\ \Rightarrow h=\frac{168}{28}=6 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{is} 6 \mathrm{m}. \end{gathered}$$

Question 5:

A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain?

Answer 5:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{tank}=8 \mathrm{m} \\ \mathrm{Breadth}=6 \mathrm{m} \\ \mathrm{Height}=2 \mathrm{m} \\ \therefore \mathrm{Its} \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=(8\times 6\times 2) {\mathrm{m}}^3=96 {\mathrm{m}}^3 \\ \mathrm{We} \mathrm{know} \mathrm{that} 1{\mathrm{m}}^3=1000 \mathrm{L} \\ \mathrm{Now}, 96 {\mathrm{m}}^3=96\times 1000 \mathrm{L}=96000 \mathrm{L} \\ \therefore \mathrm{The} \mathrm{tank} \mathrm{can} \mathrm{store} 96000 \mathrm{L} \mathrm{of} \mathrm{water}. \end{gathered}$$

Question 6:

The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its height and length are 10 m and 2.5 m respectively.

Answer 6:

$$\begin{gathered} \mathrm{Capacity} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{tank}=50000 \mathrm{L} \\ 1000 \mathrm{L}=1 {\mathrm{m}}^3 \\ \mathrm{i}.\mathrm{e}., 50000 \mathrm{L}=50\times 1000 \mathrm{litres}=50 {\mathrm{m}}^3 \\ \therefore \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{tank} \mathrm{is} 50 {\mathrm{m}}^3. \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{tank} \mathrm{is} 10 \mathrm{m}. \\ \mathrm{Height} =2.5 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{tank} \mathrm{is} b \mathrm{m}. \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 50=10\times b\times 2.5 \\ \Rightarrow 50=25\times b \\ \Rightarrow b=\frac{50}{25}=2 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{tank} \mathrm{is} 2 \mathrm{m}. \end{gathered}$$

Question 7:

A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold?

Answer 7:

$$\begin{gathered} \mathrm{Lenght} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{diesel} \mathrm{tanker}=2 \mathrm{m} \\ \mathrm{Breadth}=2 \mathrm{m} \\ \mathrm{Height}=40 \mathrm{cm} \\ =40\times \frac{1}{100}\mathrm{m} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =0.4 \mathrm{m} \\ \mathrm{So}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{tanker}=\mathrm{lenght}\times \mathrm{breadth}\times \mathrm{height} \\ =2\times 2\times 0.4 \\ =1.6 {\mathrm{m}}^3 \\ \mathrm{We} \mathrm{konw} \mathrm{that} 1 {\mathrm{m}}^3=1000 \mathrm{L} \\ \mathrm{i}.\mathrm{e}., 1.6 {\mathrm{m}}^3=1.6\times 1000 \mathrm{L}=1600 \mathrm{L} \\ \therefore \mathrm{The} \mathrm{tanker} \mathrm{can} \mathrm{hold} 1600 \mathrm{L} \mathrm{of} \mathrm{diesel}. \end{gathered}$$

Question 8:

The length , breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains.

Answer 8:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{room}=5 \mathrm{m} \\ \mathrm{Breadth}=4.5 \mathrm{m} \\ \mathrm{Height}=3 \mathrm{m} \\ \mathrm{Now}, \mathrm{volume}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =5\times 4.5\times 3 \\ =67.5 {\mathrm{m}}^3 \\ \therefore \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{in} \mathrm{the} \mathrm{room} \mathrm{is} 67.5 {\mathrm{m}}^3\mathrm{.} \end{gathered}$$

Question 9:

A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold?

Answer 9:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{tank}=3 \mathrm{m} \\ \mathrm{Breadth}=2 \mathrm{m} \\ \mathrm{Height}=1 \mathrm{m} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{tank}=3\times 2\times 1=6 {\mathrm{m}}^3 \\ \mathrm{We} \mathrm{know} \mathrm{that} 1 {\mathrm{m}}^3=1000 \mathrm{L} \\ \mathrm{i}.\mathrm{e}., 6 {\mathrm{m}}^3=6\times 1000 \mathrm{L}=6000 \mathrm{L} \\ \therefore \mathrm{The} \mathrm{water} \mathrm{tank} \mathrm{can} \mathrm{hold} 6000 \mathrm{L} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{it}. \end{gathered}$$

Question 10:

How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick?

Answer 10:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{wooden} \mathrm{block}=6 \mathrm{m} \\ =6\times 100 \mathrm{cm} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =600 \mathrm{cm} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{block}=75 \mathrm{cm} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{block}=45 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{block}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ =600\times 75\times 45 \\ =2025000 {\mathrm{cm}}^3 \\ \mathrm{Again}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{a} \mathrm{plank}=3 \mathrm{m} \\ =3\times 100 \mathrm{cm} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =300 \mathrm{cm} \\ \mathrm{Breadth}=15 \mathrm{cm}, \\ \mathrm{Height}=5 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{plank}=\mathrm{leng}\mathrm{th}\times \mathrm{breadth}\times \mathrm{height} \\ =300\times 15\times 5=22500 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{such} \mathrm{planks}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wooden} \mathrm{block}}{\mathrm{voume} \mathrm{of} \mathrm{a} \mathrm{plank}}=\frac{2025000 {\mathrm{cm}}^3}{22500 {\mathrm{cm}}^3}=90 \end{gathered}$$

Question 11:

How many bricks each of size 25 cm × 10 cm × 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick, assuming that the volume of sand and cement used in the construction is negligible?

Answer 11:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{a} \mathrm{brick}=25 \mathrm{cm}\times 10 \mathrm{cm}\times 8 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{brick}=25 \mathrm{cm}\times 10 \mathrm{cm}\times 8 \mathrm{cm} \\ =2000 {\mathrm{cm}}^3 \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{wall} \mathrm{is} 5 \mathrm{m} \\ =5\times 100 \mathrm{cm} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =500 \mathrm{cm} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{wall}=3 \mathrm{m} \\ =3\times 100 \mathrm{cm} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =300 \mathrm{cm} \\ \mathrm{It} \mathrm{is} 16 \mathrm{cm} \mathrm{thick}, \mathrm{i}.\mathrm{e}., \mathrm{breadth} =16 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{wall}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=500\times 300\times 16=2400000 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{bricks} \mathrm{needed} \mathrm{to} \mathrm{build} \mathrm{the} \mathrm{wall}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wall}}{\mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{brick}}=\frac{2400000 {\mathrm{cm}}^3}{2000 {\mathrm{cm}}^3}=1200 \end{gathered}$$

Question 12:

A village, having a population of 4000, requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days will the water of this tank last?

Answer 12:

$$\begin{gathered} \mathrm{A} \mathrm{village} \mathrm{has} \mathrm{population} \mathrm{of} 4000 \mathrm{and} \mathrm{every} \mathrm{person} \mathrm{needs} 150 \mathrm{L} \mathrm{of} \mathrm{water} \mathrm{a} \mathrm{day}. \\ \mathrm{So}, \mathrm{the} \mathrm{total} \mathrm{requirement} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{a} \mathrm{day}=4000\times 150 \mathrm{L}=600000 \mathrm{L} \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{tank} \mathrm{is} 20 \mathrm{m}. \\ \mathrm{Breadth}=15 \mathrm{m} \\ \mathrm{Height}=6 \mathrm{m} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{tank}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=20\times 15\times 6=1800 {\mathrm{m}}^3 \\ \mathrm{Now},\mathit{ }1 {\mathrm{m}}^3=1000 \mathrm{L} \\ \mathrm{i}.\mathrm{e}., 1800 {\mathrm{m}}^3=1800\times 1000 \mathrm{L}=1800000 \mathrm{L} \\ \mathrm{The} \mathrm{tank} \mathrm{has} 1800000 \mathrm{L} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{it} \mathrm{and} \mathrm{the} \mathrm{whole} \mathrm{village} \mathrm{need} 600000 \mathrm{L} \mathrm{per} \mathrm{day}. \\ \therefore \mathrm{The} \mathrm{water} \mathrm{in} \mathrm{the} \mathrm{tank} \mathrm{will} \mathrm{last} \mathrm{for} \frac{1800000}{600000} \mathrm{days}, \mathrm{i}.\mathrm{e}., 3 \mathrm{days}. \end{gathered}$$

Question 13:

A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m × 8 m × 6 m is dug outside the field and the earth dug-out from this well is spread evenly on the field. How much will the earth level rise?

Answer 13:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{the} \mathrm{well} = 14 \mathrm{m}\times 8 \mathrm{m}\times 6 \mathrm{m} \\ \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{dug}-\mathrm{out} \mathrm{earth} =14\times 8\times 6=672 {\mathrm{m}}^3 \\ \mathrm{Now}, \mathrm{we} \mathrm{will} \mathrm{spread} \mathrm{this} \mathrm{dug}-\mathrm{out} \mathrm{earth} \mathrm{on} \mathrm{a} \mathrm{field} \mathrm{whose} \mathrm{length}, \mathrm{breadth} \mathrm{and} \mathrm{height} \mathrm{are} 70 \mathrm{m}, 60 \mathrm{m} \mathrm{and} h \mathrm{m}, \mathrm{respectively}. \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{dug}-\mathrm{out} \mathrm{earth} =\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}=70\times 60\times h \\ \Rightarrow 672=4200\times h \\ \Rightarrow h=\frac{672}{4200}=0.16 \mathrm{m} \\ \mathrm{We} \mathrm{know} \mathrm{that} 1 \mathrm{m}= 100 \mathrm{cm} \\ \therefore \mathrm{The} \mathrm{earth} \mathrm{level} \mathrm{will} \mathrm{rise} \mathrm{by} 0.16 \mathrm{m}=0.16\times 100 \mathrm{cm}=16 \mathrm{cm}. \end{gathered}$$

Question 14:

A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water.

Answer 14:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{pool}=250 \mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{pool}=130 \mathrm{m} \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} 3250 {\mathrm{m}}^3 \mathrm{of} \mathrm{water} \mathrm{is} \mathrm{poured} \mathrm{into} \mathrm{it}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{the} \mathrm{pool}=3250 {\mathrm{m}}^3 \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{level} \mathrm{is} h \mathrm{m}. \\ \mathrm{Then}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{water}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 3250=250\times 130\times h \\ \Rightarrow 3250=32500\times h \\ \Rightarrow h=\frac{3250}{32500}=0.1 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{water} \mathrm{level} \mathrm{in} \mathrm{the} \mathrm{tank} \mathrm{will} \mathrm{rise} \mathrm{by} 0.1 \mathrm{m}. \end{gathered}$$

Question 15:

A beam 5 m long and 40 cm wide contains 0.6 cubic metre of wood. How thick is the beam?

Answer 15:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{beam}=5\mathrm{m} \\ \mathrm{Breadth}=40 \mathrm{cm} \\ =40\times \frac{1}{100}\mathrm{m} (\because 100 \mathrm{cm}=1 \mathrm{m}) \\ =0.4 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{beam} \mathrm{is} h \mathrm{m}. \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{beam} \mathrm{contains} 0.6 \mathrm{cubic} \mathrm{metre} \mathrm{of} \mathrm{wood}. \\ \mathrm{i}.\mathrm{e}., \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{beam}=0.6 {\mathrm{m}}^3 \\ \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{beam}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 0.6=5\times 0.4\times h \\ \Rightarrow 0.6=2\times h \\ \Rightarrow h=\frac{0.6}{2}=0.3 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{beam} \mathrm{is} 0.3 \mathrm{m} \mathrm{thick}. \end{gathered}$$

Question 16:

The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day?

Answer 16:

$$\begin{gathered} \mathrm{The} \mathrm{rainfall} \mathrm{on} \mathrm{a} \mathrm{certain} \mathrm{day}=6 \mathrm{cm} \\ =6\times \frac{1}{100}\mathrm{m} (\because 1 \mathrm{m} = 100 \mathrm{cm}) \\ =0.06 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{field}=3 \mathrm{hectares} \\ \mathrm{We} \mathrm{know} \mathrm{that} 1 \mathrm{hectare}=10000 {\mathrm{m}}^2 \\ \mathrm{i}.\mathrm{e}., 3 \mathrm{hectares}=3\times 10000 {\mathrm{m}}^2=30000 {\mathrm{m}}^2 \\ \mathrm{Thus}, \mathrm{volume} \mathrm{of} \mathrm{rain} \mathrm{water} \mathrm{that} \mathrm{fell} \mathrm{in} \mathrm{the} \mathrm{field}=(\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{field})\times (\mathrm{height} \mathrm{of} \mathrm{rainfall})=30000\times 0.06=1800 {\mathrm{m}}^3 \\ \mathrm{Since} 1 {\mathrm{m}}^3=1000 \mathrm{L}, \mathrm{we} \mathrm{have}: \\ 1800 {\mathrm{m}}^3=1800\times 1000 \mathrm{L}=1800000 \mathrm{L}=18\times 100000 \mathrm{L}=18\times {10}^5 \mathrm{L} \\ \therefore \mathrm{On} \mathrm{that} \mathrm{day}, 18\times {10}^5 \mathrm{L} \mathrm{of} \mathrm{rain} \mathrm{water} \mathrm{fell} \mathrm{on} \mathrm{the} \mathrm{field}. \end{gathered}$$

Question 17:

An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge.

Answer 17:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{wooden} \mathrm{beam}=8 \mathrm{m} \\ \mathrm{Width} =0.5 \mathrm{m} \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{beam} \mathrm{is} h \mathrm{m}. \\ \mathrm{Then}, \mathrm{its} \mathrm{volume} =\mathrm{length}\times \mathrm{width}\times \mathrm{height}=8\times 0.5\times \mathrm{h}=4\times \mathrm{h} {\mathrm{m}}^3 \\ \mathrm{Also}, \mathrm{it} \mathrm{produces} 4000 \mathrm{cubes}, \mathrm{each} \mathrm{of} \mathrm{edge} 1 \mathrm{cm}=1\times \frac{1}{100}\mathrm{m}=0.01 \mathrm{m} (100 \mathrm{cm} = 1 \mathrm{m}) \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3=(0.01{)}^3=0.000001 {\mathrm{m}}^3 \\ \therefore \mathrm{Volume} \mathrm{of} 4000 \mathrm{cubes}=4000\times 0.000001=0.004 {\mathrm{m}}^3 \\ \mathrm{Since} \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{wastage} \mathrm{of} \mathrm{wood} \mathrm{in} \mathrm{preparing} \mathrm{cubes}, \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{the} 4000 \mathrm{cubes} \mathrm{will} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{beam}. \\ \mathrm{i}.\mathrm{e}., \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{beam}=\mathrm{volume} \mathrm{of} 4000 \mathrm{cubes} \\ \Rightarrow 4\times h=0.004 \\ \Rightarrow h=\frac{0.004}{4}=0.001 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{third} \mathrm{edge} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{wooden} \mathrm{beam} \mathrm{is} 0.001 \mathrm{m}. \end{gathered}$$

Question 18:

The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed?

Answer 18:

$$\begin{gathered} \mathrm{Dimension} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{block} \mathrm{is} 2.25 \mathrm{m}\times 1.5 \mathrm{m}\times 27 \mathrm{cm}, \mathrm{i}.\mathrm{e}., 225 \mathrm{cm}\times 150 \mathrm{cm}\times 27 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}). \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{block}=225\times 150\times 27=911250 {\mathrm{cm}}^3 \\ \mathrm{This} \mathrm{metal} \mathrm{block} \mathrm{is} \mathrm{melted} \mathrm{and} \mathrm{recast} \mathrm{into} \mathrm{cubes} \mathrm{each} \mathrm{of} \mathrm{side} 45 \mathrm{cm}. \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3={45}^3=91125 {\mathrm{cm}}^3 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{such} \mathrm{cubes} \mathrm{formed} \mathrm{from} \mathrm{the} \mathrm{metal} \mathrm{block}=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{metal} \mathrm{block}}{\mathrm{volume} \mathrm{of} a \mathrm{metal} \mathrm{cube}}=\frac{911250 {\mathrm{cm}}^3}{91125 {\mathrm{cm}}^3}=10 \end{gathered}$$

Question 19:

A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 cm³ of iron weighs 8 gm.

Answer 19:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{an} \mathrm{iron} \mathrm{piece} \mathrm{is} 6 \mathrm{m}\times 6 \mathrm{cm}\times 2 \mathrm{cm}, \mathrm{i}.\mathrm{e}., 600 \mathrm{cm}\times 6 \mathrm{cm}\times 2 \mathrm{cm} (\because 1 \mathrm{m} = 100 \mathrm{cm}). \\ \mathrm{Its} \mathrm{volume}=600\times 6\times 2=7200 {\mathrm{cm}}^3 \\ \mathrm{Now}, 1 {\mathrm{cm}}^3=8 \mathrm{gm} \\ \mathrm{i}.\mathrm{e}., 7200 {\mathrm{cm}}^3=7200\times 8 \mathrm{gm}=57600 \mathrm{gm} \\ \therefore \mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{iron} \mathrm{piece}=57600 \mathrm{gm} \\ =57600\times \frac{1}{1000}\mathrm{kg} (\because 1 \mathrm{Kg}= 1000 \mathrm{gm}) \\ =57.6 \mathrm{kg} \end{gathered}$$

Page-21.16

Question 20:

Fill in the blanks in each of the following so as to make the statement true:
(i) 1 m³ = .........cm³
(ii) 1 litre = ....... cubic decimetre
(iii) 1 kl = ....... m³
(iv) The volume of a cube of side 8 cm is ........
(v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm³. The height of the cuboid is ........ cm.
(vi) 1 cu.dm = ........ cu. mm
(vii) 1 cu. km = ........ cu. m
(viii) 1 litre = ........ cu. cm
(ix) 1 ml = ........ cu. cm
(x) 1 kl = ........ cu. dm = ........ cu. cm.

Answer 20:

$$\begin{gathered} \left(\mathrm{i}\right) \\ 1 {\mathrm{m}}^3=1 \mathrm{m}\times 1 \mathrm{m}\times 1 \mathrm{m} \\ =100 \mathrm{cm}\times 100 \mathrm{cm}\times 100 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}) \\ =1000000 {\mathrm{cm}}^3 \\ ={10}^6 {\mathrm{cm}}^3 \\ \left(\mathrm{ii}\right) \\ 1 \mathrm{L}=\frac{1}{1000}{\mathrm{m}}^3\mathrm{ } \\ \mathrm{=}\frac{1}{1000}1 \mathrm{m}\times 1 \mathrm{m}\times 1 \mathrm{m} \\ \mathrm{=}\frac{1}{1000}\times 10 \mathrm{dm}\times 10 \mathrm{dm}\times 10 \mathrm{dm} \\ =1 {\mathrm{dm}}^3 \\ \left(\mathrm{iii}\right) \\ 1 \mathrm{kL}=1000 \mathrm{L} \\ =1 {\mathrm{m}}^3\mathrm{ } (1000 \mathrm{L}=1 {\mathrm{m}}^3) \\ \left(\mathrm{iv}\right) \\ \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{cube} \mathrm{of} \mathrm{side} 8 \mathrm{cm}=(\mathrm{side}{)}^3=8^3=512 {\mathrm{cm}}^3 \\ (\mathrm{v}) \\ \mathrm{Lenght} \mathrm{of} \mathrm{the} \mathrm{wooden} \mathrm{cuboid}=10 \mathrm{cm} \\ \mathrm{Breadth}= 8 \mathrm{cm} \\ \mathrm{Its} \mathrm{volume}=4000 {\mathrm{cm}}^3 \\ \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{is} h \mathrm{cm}. \\ \mathrm{Then}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height} \\ \Rightarrow 4000=10\times 8\times h \\ \Rightarrow 4000=80\times h \\ \Rightarrow h=\frac{4000}{80}=50 \mathrm{cm} \\ (\mathrm{vi}) \\ 1 \mathrm{cu} \mathrm{dm}=1 \mathrm{dm}\times 1 \mathrm{dm}\times 1 \mathrm{dm} \\ =100 \mathrm{mm}\times 100 \mathrm{mm}\times 100 \mathrm{mm} \\ =1000000 {\mathrm{mm}}^3 \\ ={10}^6 \mathrm{cu} \mathrm{mm} \\ (\mathrm{vii}) \\ 1 \mathrm{cu} \mathrm{km}=1 \mathrm{km}\times 1 \mathrm{km}\times 1 \mathrm{km} \\ =1000 \mathrm{m}\times 1000 \mathrm{m}\times 1000 \mathrm{m} (\because 1 \mathrm{km}=1000 \mathrm{m}) \\ =1000000000 {\mathrm{m}}^3 \\ ={10}^9 \mathrm{cu} \mathrm{m} \\ (\mathrm{viii}) \\ 1 \mathrm{L}=\frac{1}{1000}{\mathrm{m}}^3\mathrm{ } \\ \mathrm{=}\frac{1}{1000}\times 1 \mathrm{m}\times 1 \mathrm{m}\times 1 \mathrm{m} \\ \mathrm{=}\frac{1}{1000}\times 100 \mathrm{cm}\times 100 \mathrm{cm}\times 100 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}) \\ =1000 {\mathrm{cm}}^3 \\ ={10}^3 \mathrm{cu} \mathrm{cm} \\ (\mathrm{ix}) \\ 1 \mathrm{mL}=\frac{1}{1000}\times 1 \mathrm{L}=\frac{1}{1000}\mathrm{\times }\frac{1}{1000}{\mathrm{m}}^3\mathrm{ } \\ \mathrm{=}\frac{1}{1000}\mathrm{\times }\frac{1}{1000}\times 1 \mathrm{m}\times 1 \mathrm{m}\times 1 \mathrm{m} \\ \mathrm{=}\frac{1}{1000}\mathrm{\times }\frac{1}{1000}\times 100 \mathrm{cm}\times 100 \mathrm{cm}\times 100 \mathrm{cm} (\because 1 \mathrm{m}=100 \mathrm{cm}) \\ =1 \mathrm{cu} \mathrm{cm} \\ (\mathrm{x}) \\ 1 \mathrm{kL}=1000 \mathrm{L}=1000\times \frac{1}{1000}{\mathrm{m}}^3=1 {\mathrm{m}}^3 \\ =1 \mathrm{m}\times 1 \mathrm{m}\times 1 \mathrm{m} \\ =10 \mathrm{dm}\times 10 \mathrm{dm}\times 10 \mathrm{dm} (\because 1 \mathrm{m}=10 \mathrm{dm}) \\ =1000 \mathrm{cu} \mathrm{dm} \\ =1000\times 10 \mathrm{cm}\times 10 \mathrm{cm}\times 10 \mathrm{cm} (\because 1 \mathrm{dm}=10 \mathrm{cm}) \\ =1000000 {\mathrm{cm}}^3 \\ ={10}^6 \mathrm{cu} \mathrm{cm} \end{gathered}$$