Surface Areas and Volumes : Exercise 13.5 (Mathematics NCERT Class 9th)


Q.1     A matchbox measures 4 cm × 2.5 cm × 1.5 cm . What will be the volume of a packet containing 12 such boxes?
Sol.

Here , l = 4 cm, b = 2.5 cm and h = 1.5 cm
Therefore Volume of one matchbox  = \ell \times b \times h\,c{m^3}
                                                               = \left( {4 \times 2.5 \times 1.5} \right)c{m^3} = 15\,c{m^3}
Therefore Volume of a packet containing 12 such boxes  = \left( {12 \times 15} \right)c{m^3} = 180\,c{m^3}


Q.2     A cuboid water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold? (1\,{m^3}=1000\,\ell ).
Sol. 

Here , l = 6 m , b = 5 m and h = 4.5 m
Therefore Volume of the tank  = \ell bh\,{m^3}
                                                    = \left( {6 \times 5 \times 4.5} \right){m^3} = 135\,{m^3}
Therefore , the tank can hold = 135 × 1000 litres          [Since 1\,{m^3} = 1000\,litres]
                                                  = 135000 litres of water.


Q.3    A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
Sol.

Here, Length = 10 m , Breadth = 8 m and Volume = 380\,{m^3}
Volume of cuboid = Length x Breadth x Height
Height\, = {{Volume\,of\,cuboid} \over {Length\, \times \,Breadth}} 
                   \, = {{380} \over {10\, \times \,8}}m
                   = 4.75\,m


Q.4     Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3m deep at the rate of Rs 30 per {m^3}.
Sol.

Here, l = 8 m, b = 6 m and h = 3 m
Volume of the pit  = \ell bh{\mkern 1mu} {m^3}
                               =  \left( {8 \times 6 \times 3} \right){m^3} = 144\,{m^3}

Rate of digging is Rs 30 per {m^3}
Therefore Cost of digging the pit = Rs (144 × 30) = Rs 4320.


Q.5    The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Sol.

Here, length = 2.5 m, depth = 10 m and volume = 50000 litres
 = \left( {50000 \times {1 \over {1000}}} \right){m^3}\,\,\,\left[ {\,\,1{m^3} = 1000\,litres} \right]
 = 50\,{m^3}
Breadth = {{Volume\,of\,cuboid} \over {Length\, \times \,Depth}} = \left( {{{50} \over {25 \times 10}}} \right)m = 2\,m


Q.6     A village having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m ×6 m . For how many days will the water of this tank last?
Sol.

Here , l = 20 m , b = 15 m and h = 6 m
Therefore Capacity of the tank  = \ell bh\,\,{m^3}
                                                      = \left( {20 \times 15 \times 6} \right){m^3} = 1800\,{m^3}
Water requirement per person per day  = 150\,litres
Water required for 4000 person per day  = \left( {4000 \times 150} \right)\ell
                                                                      = \left( {{{4000 \times 150} \over {1000}}} \right){m^3} = 600\,{m^3}
Number of days the water will last  = {{Capacity\,of\,\tan k} \over {Total\,water\,required\,per\,day}}
                                                            = \left( {{{1800} \over {600}}} \right) = 3
Thus, the water will last for 30 days.


Q.7     A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Sol. 

Volume of the godown  = \left( {60\, \times 25 \times 10} \right){m^3}  = 15000\,{m^3}
Volume of 1 crates  = \left( {1.5 \times 1.25 \times 0.5} \right){m^3}  = 0.9375\,\,{m^3}
Number of crates that can be stored in the godown
 = {{Volume\,of\,the\,godown} \over {Volum\,of\,1\,crate}}
 = {{15000} \over {0.9375}} = 16000


Q.8     A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Sol. 

Let {V_1} be volume of the cube of edge 12 cm . So , length = breadth = height = 12 cm
Volume of the cube  = \left( {12 \times 12 \times 12} \right)c{m^3}
and {V_2} = Volume of the cube cut out of the first one
      
 = {1 \over 8} \times {V_1} = \left( {{1 \over 8} \times 12 \times 12 \times 12} \right)c{m^3}
        = \left( {6 \times 6 \times 6} \right)c{m^3}
Therefore , side of the new cube = 6 cm
Ratio of their surface areas  = {{6{{\left( {side} \right)}^2}} \over {6{{\left( {side} \right)}^2}}} = {{6 \times 12 \times 12} \over {6 \times 6 \times 6}}
                                                = {4 \over 1}i.e.,\,4\,:1


Q.9     A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Sol. 

Since the water flows at the rate of 2 km per hour, the water from 2 km of river flows into the sea in one hour.
Therefore The volume of water flowing into the sea in one hour = Volume of the cuboid
                                                                                                             = \ell \times b \times h\,{m^3}
                                                                                                             = \left( {2000 \times 40 \times 3} \right){m^3}
Therefore , the volume of water flowing into the sea in one minute
 = \left( {{{2000 \times 40 \times 3} \over {60}}} \right){m^3}

 = 4000\,{m^3}