The document RD Sharma Solutions -Ex-18.2 (Part - 2), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q16) A box with lid is made of 2cm thick wood. Its external length, breadth and height are 25cm, 18cm and 15cm respectively. How much cubic cm of a fluid can be placed in it? Also, find the volume of the wood used in it.**

**Solution:**

Given,

The external dimensions of cuboid are as follows

Length (l) = 25 cm

Breadth (b) = 18 cm

Heigth (h) = 15 cm

External volume of the case with cover (cuboid) = l*b*h cm^{3}

= 25*18*15 cm^{3}

= 6750 cm^{3}

Now, the internal dimensions of the cuboid is as follows

Length (l) = 25-(2*2) = 21 cm

Breadth (b) = 18-(2*2) = 14 cm

Height (h) = 15-(2*2) = 11cm

Now, Internal volume of the case with cover (cuboid) = l*b*h cm^{3}

= 21*14*11 cm^{3}

= 3234 cm^{3}

Therefore, Volume of the fluid that can be placed = 3234 cm^{3}

Now, volume of the wood utilized = External volume â€“ Internal volume

= 3516 cm^{3}

**Q17) The external dimensions of a closed wooden box are 48cm, 36cm, 30cm. The box is made of 1.5cm thick wood. How many bricks of size 6cm x 3cm x 0.75cm can be put in this box?**

**Solution:**

Given that,

The external dimensions of the wooden box are as follows:

Length (l) = 48cm, Breadth (b) = 36cm, Heigth (h) = 30cm

Now, the internal dimensions of the wooden box are as follows:

Length (l) = 48-(2*1.5) = 45cm

Breadth (b) = 36-(2*1.5) = 33cm

Height (h) = 30-(2*1.5) = 27cm

Internal volume of the wooden box = l*b*h cm^{3}

= 45*33*27 cm^{3}

= 40095 cm^{3}

Volume of the brick = 6*3*0.75 = 13.5 cm^{3}

Therefore,Number of bricks = = 2970 bricks

Therefore,2970 bricks can be kept inside the wooden box.

**Q18) How many cubic centimeters of iron are there in an open box whose external dimensions are 36cm, 25cm and 16.5cm, the iron being 1.5cm thick throughout? If 1 cubic cm of iron weighs 15gms. Find the weight of the empty box in kg.**

**Solution:**

Given,

Outer dimensions of iron:

Length (l) = 36cm

Breadth (b) = 25cm

Heigth (h) = 16.5cm

Inner dimensions of iron:

Length (l) = 36-(2*1.5) = 33cm

Breadth (b) = 25-(2*1.5) = 22cm

Height (h) = 16.5-1.5 = 15cm

Volume of Iron = Outer volume â€“ Inner volume

= (36*25*16.5) â€“ (33*22*15)

= 3960 cm3

Weight of Iron = 3960*15 = 59400 grams = 59.4 kgs

**Q19) A cube of 9cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15cm and 12cm, find the rise in water level in the vessel.**

**Solution:**

Volume of the cube = S3 = 93 = 729cm3

Area of the base = l*b = 15*12 = 180cm2

Risein water level =

= 4.05cm

**Q20) A rectangular container, whose base is a square of side 5cm, stands on a horizontal table, and holds water up to 1cm from the top. When a solid cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.**

**Solution:**

Let the length of each edge of the cube be â€˜xâ€™cm

Then, volume of the cube = Volume of water inside the tank + Volume of water that overflowed

x^{3 }= (5âˆ—5âˆ—1)+2

x^{3} = 27

x = 3cm

Hence, volume of the cube = 27cm^{3}

And edge of the cube = 3cm

**Q21) A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7m deep and the earth taken out is spread evenly on the field. By how many meters is the level of the field raised? Give the answer to the second place of decimal.**

**Solution:**

Volume of the earth dug out = 50*40*7 = 14000 m^{3}

Let â€˜hâ€™ be the rise in the height of the field

Therefore, volume of the field (cuboidal) = Volume of the earth dug out

â‡’ 200âˆ—150âˆ—h = 14000

â‡’ h = = 0.47m

**Q22) A field is in the form of a rectangular length 18m and width 15m. A pit 7.5m long, 6m broad and 0.8m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.**

**Solution:**

Let â€˜hâ€™metres be the rise in the level of field

Volume of earth taken out from the pit = 7.5*6*0.8 = 36 m^{3}

Area of the field on which the earth taken out is to be spread = 18*15 â€“ 7.5*6 = 225m^{2}

Now, Area of the field * h = Volume of the earth taken out from the pit

â‡’ 225âˆ—h = 7.5âˆ—6âˆ—0.8

â‡’ h = = 0.16m = 16cm

**Q23) A rectangular tank is 80m long and 25m broad. Water flows into it through a pipe whose cross-section is 25cm ^{2}, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes?**

**Solution:**

Consider â€˜hâ€™ be the rise in water level.

Volume of water in rectangular tank = 8000*2500*h cm^{2}

Cross-sectional area of the pipe = 25 cm^{2}

Water coming out of the pipe forms a cuboid of base area 25 cm^{2} and length equal to the distance travelled in 45 minutes with the speed 16 km/hour

i.e., length = Length = 16000âˆ—100âˆ—cm

Therefore, The Volume of water coming out pipe in 45 minutes = 25*16000*100* (45/60)

Now, volume of water in the tank = Volume of water coming out of the pipe in 45 minutes

â‡’ 8000âˆ—2500âˆ—h = 16000âˆ—100âˆ—4560âˆ—25

= 1.5cm

**Q24) Water in a rectangular reservoir having base 80 m by 60 m is 6.5m deep. In what time can the water be pumped by a pipe of which the cross-section is a square of side 20cm if the water runs through the pipe at the rate of 15km/hr.**

**Solution:**

Flow of water = 15km/hr

= 15000 m/hr

Volume of water coming out of the pipe in one hour,

âˆ—15000 = 600m^{3}

Volume of the tank = 80*60*6.5

= 31200 m^{3}

Timetakentoemptythetank =

= 52hours

**Q25) A village having a population of 4000 requires 150 liters of water per head per day. It has a tank measuring 20m x 15m x 6m. For how many days will the water of this tank last?**

**Solution:**

Given that,

Length of the cuboidal tank (l) = 20m

Breadth of the cuboidal tank (b) = 15m

Height of the cuboidal tank (h) = 6m

Capacity of the tank = l*b*h = 20*15*6

= 1800 m^{3}

= 1800000 litres

Water consumed by the people of village in one day = 4000*150 litres

= 600000 litres

Let water of this tank last for â€˜nâ€™ days

Therefore, water consumed by all people of village in n days = capacity of the tank

= n*600000 = 1800000

= n = = 3

Thus, the water will last for 3 days in the tank.

**Q26) A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in Fig. 18.12. If the edge of each cube is 3cm, find the volume of the structure built by the child.**

**Solution:**

Volume of each cube = edge*edge*edge

= 3*3*3

= 27 cm3

Number of cubes in the structure = 15

Therefore, volume of the structure = 27*15

= 405 cm^{3}

**Q27) A godown measures 40m x 25m x 10m. Find the maximum number of wooden crates each measuring 1.5m x 1.25m x 0.5m that can be stored in the godown.**

**Solution:**

Given,

Godown length (l_{1}) = 40m

Godown breadth (b_{1}) = 25m

Godown height (h_{1}) = 10m

Volume of the godown = l_{1*} b_{1*} h_{1 } = 40*25*10

= 10000 m^{3}

Wooden crate length (l_{2}) = 1.5m

Wooden crate breadth (b_{2}) = 1.25m

Wooden crate height (h_{2}) = 0.5m

Volume of the wooden crate = l_{2}*b_{2}*h_{2 } = 1.5*1.25*0.5

= 0.9375 m^{3}

The number of wooden crates stored in the godown is taken as â€˜nâ€™

Volume of â€˜nâ€™ wooden crates = Volume of godown

= 0.9375n = 10000

= n = = 10666.66

Therefore, the number of wooden crates that can be stored in the godown is 10666.66.

**Q28) A wall of length 10m was to be built across an open ground. The height of the wall is 4m and thickness of the wall is 24cm. If this wall is to be built up with bricks whose dimensions are 24cm x 12cm x 8cm, how many bricks would be required?**

**Solution:**

Given that,

The wall with all its bricks makes up space occupies by it, we need to find the volume of the wall, which is nothing but cuboid.

Here, length = 10m = 1000cm

Thickness = 24cm

Height = 4m = 400cm

Therefore, volume of the wall = l*b*h

= 1000*24*400 cm^{3}

Now, each brick is a cuboid with length = 24cm

Breadth = 12cm

Height = 8cm

So, volume of each brick = l*b*h = 24*12*8 = 2304 cm^{3}

The number of bricks required is given by,

= 4166.6 bricks

So, the wall requires 4167 bricks.

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