# VOLUME AND SURFACE AREA

Ans .

868 $$cm^2$$

1. Explanation :

Volume = (16 x 14 x 7) $$m^3$$ = 1568 $$m^3$$ Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] $$cm^2$$ = (2 x 434) $$cm^2$$= 868 $$cm^2$$.

Ans .

17 m.

1. Explanation :

Length of longest pole = Length of the diagonal of the room

= $$\sqrt{12^{2}+8^{2}+9^{2}}$$= .$$\sqrt{289}$$= 17 m.

Ans .

40 cm

1. Explanation :

Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

x * 5x * 40x = 12.8 --> $$x^3$$=12.8/200 = 128/2000 = 64/1000

So, x = (4/10) m =((4/10)*100)cm = 40 cm

Ans .

45000.

1. Explanation :

Volume of the wall = (2400 x 800 x 60) cu. cm.

Volume of bricks = 90% of the volume of the wall

=((90/100)*2400 *800 * 60)cu.cm.

Volume of 1 brick = (24 x 12 x 8) cu. cm.

Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.

Ans .

96min

1. Explanation :

Volume required in the tank = (200 x 150 x 2) $$x^3$$ = 60000 $$x^3$$

Length of water column flown in1 min =(20*1000)/60 m =1000/3 m

Volume flown per minute = 1.5 * 1.25 * (1000/3) $$x^3$$ = 625 $$x^3$$

Required time = (60000/625)min = 96min

Ans .

8.04 kg.

1. Explanation :

Volume of the metal used in the box = External Volume - Internal Volume

= [(50 * 40 * 23) - (44 * 34 * 20)]$$cm^3$$

= 16080 $$cm^3$$

Weight of the metal =((16080*0.5)/1000) kg = 8.04 kg.

Ans .

216 $$cm^2$$

1. Explanation :

Let the edge of the cube be a.

$$\sqrt{3}$$a = 6../3 _ a = 6.

So,Volume = $$a^3$$ = (6 x 6 x 6) cm3 = 216 $$cm^3$$

Surface area = 6$$a^2$$ = (6 x 6 x 6) $$cm^2$$== 216 $$cm^2$$

Ans .

4913 $$cm^3$$.

1. Explanation :

Let the edge of the cube bea. Then,

6$$a^2$$ = 1734 $$a^2$$ = 289 => a = 17 cm.

Volume = $$a^3$$= (17)3 $$cm^3$$= 4913 $$cm^3$$.

Ans .

40.

1. Explanation :

Volume of the block = (6 x 12 x 15)$$cm^3$$. = 1080$$cm^3$$.

Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.

Volume of this cube = (3 x 3 x 3) $$cm^3$$. = 27 $$cm^3$$..

Number of cubes = 1080/27 = 40.

Ans .

11.25 cm

1. Explanation :

Increase in volume = Volume of the cube = (15 x 15 x 15)$$cm^3$$.

Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.

Ans .

486 $$cm^2$$

1. Explanation :

Volume of new cube = ($$1^3$$ +$$6^3$$ + $$8^3$$) cm = 729 $$cm^3$$

Edge of new cube = $$\sqrt[3]{729}$$cm = 9 cm.

Surface area of the new cube = (6 x 9 x 9) $$cm^2$$ = 486 $$cm^2$$.

Ans .

125%

1. Explanation :

Let original length of each edge = a.

Then, original surface area = 6$$a$$ .

New edge = (150% of a) = (150a/100) = 3a/2

New surface area = 6x $${(3a/2)}^2$$ = 27 $$a^2$$/2

Increase percent in surface area =( $$\frac{15a^2}{2}$$ x $$\frac{1}{6a^2}$$ x 100)% = 125%

Ans .

1:9

1. Explanation :

Let their edges be a and b. Then,

$$a^3$$./$$b^3$$= 1/27 (or) $${(a/b)}^3$$ = $${(1/3)}^3$$ (or) (a/b) = (1/3).

Ratio of their surface area = 6$$a^2$$/6$$b^2$$ = $$a^2$$/$$b^2$$ = $${(a/b)}^2$$ = 1/9, i.e. 1:9.

Ans .

957 $$cm^2$$

1. Explanation :

Volume = $$\prod r^{2}$$h = ((22/7)x(7/2)x(7/2)x40) = 1540 $$cm^3$$. .

Curved surface area = 2$$\prod$$rh = (2x(22/7)x(7/2)x40)= 880 $$cm^2$$.

Total surface area = 2$$\prod$$rh + 2$$\prod r^{2}$$ = 2$$\prod$$r (h + r)

= (2 x (22/7) x (7/2) x (40+3.5)) $$cm^2$$= 957 $$cm^2$$.

Ans .

12 m

1. Explanation :

Let the depth of the tank be h metres. Then,

$$\prod$$ x $$7^2$$ x h = 1848 --> h = (1848 x (7/22) x (1/49) = 12 m

Ans .

112 m.

1. Explanation :

Let the length of the wire be h metres. Then,

$$\prod {(0.50/(2 x 100))}^2$$ x h = 2.2/1000

h = ( (2.2/1000) x (100 x 100)/(0.25 x 0.25) x (7/22) ) = 112 m.

Ans .

400

1. Explanation :

Volume of 1 rod = (( 22/7) x (1/100) x (1/100) x 7 ) cu.m = 11/5000 cu.m

Volume of iron = 0.88 cu. m.

Number of rods = (0.88 x 5000/11) = 400.

Ans .

2.5

1. Explanation :

Let the radii of the cylinders be 3x, 5x and their heights be 2y, 3y respectively. Then

Ratio of their curved surface area = $$\frac{2\prod X 3x X 2y}{2\prod X 5x X 3y}$$ = 2/5 = 2.5

Ans .

26.4 kg.

1. Explanation :

Inner radius = (3/2) cm = 1.5 cm, Outer radius = (1.5 + 1) = 2.5 cm.

Volume of iron = [$$\prod$$x (2.5)2 x 100 -$$\prod$$ x (1.5)2 x 100] $$cm^{3}$$

= (22/7) x 100 x [(2.5)2 - (1.5)2] $$cm^{3}$$

= (8800/7) $$cm^{3}$$

Weight of the pipe = ((8800/7) x (21/1000))kg = 26.4 kg.

Ans .

35cm

1. Explanation :

Here, r = 21 cm and h = 28 cm.

Slant height, l = $$\sqrt{r^{2}+h^{2}}$$ = $$\sqrt{21^{2}+28^{2}}$$ =$$\sqrt{1225}$$ = 35cm

Ans .

440 m.

1. Explanation :

Here, r = 7m and h = 24 m.

So,l = $$\sqrt{r^{2}+h^{2}}$$ =$$\sqrt{7^{2}+24^{2}}$$ =$$\sqrt{625}$$ = 25 m.

Area of canvas =$$\prod$$rl=((22/7)*7*25) $$m^{2}$$= 550 $$m^{2}$$.

Length of canvas = (Area/Width) = (550/1.25) m = 440 m.

Ans .

9 : 32

1. Explanation :

Let the radii of their bases be r and R and their heights be h and 2h respectively.

Then,(2$$\prod$$r/2$$\prod$$R)=(3/4)--> R=(4/3)r.

Ratio of volumes = (((1/3)$$\prod$$ $$r^{2}$$h)/((1/3)$$\prod$$$${4/3r}^2$$(2h)))=9 : 32.

Ans .

9 : 8.

1. Explanation :

Let the radii of the cylinder and the cone be 3r and 4r and their heights be 2h and

3h respectively.

$$\frac{Volume of cylinder}{Volume of cone}$$ =$$\frac{\prod3 r^{2}* 2h}{(1/3)\prod r^{2}* 3h}$$= 9/8 = 9 : 8.

Ans .

24 cm

1. Explanation :

Volume of the liquid in the cylindrical vessel

= Volume of the conical vessel

= ((1/3)* (22/7)* 12 * 12 * 50) )$$cm^{3}$$ = (22 *4 *12 * 50)/7 $$cm^{3}$$

Let the height of the liquid in the vessel be h.

Then (22/7)*10*10*h =(22*4*12*50)/7 or h = (4*12*50)/100 = 24 cm

Ans .

4851 $$cm^{3}$$,1386 $$cm^{2}$$

1. Explanation :

Volume = (4/3)?$$r^{3}$$ =(4/3)*(22/7)*(21/2)*(21/2)*(21/2) $$cm^{3}$$ = 4851 $$cm^{3}$$.

Surface area = 4$$\prod r^{2}$$ =(4*(22/7)*(21/2)*(21/2)) $$cm^{2}$$ = 1386 $$cm^{2}$$

Ans .

237.5%, 125%.

1. Explanation :

Original volume = (4/3)$$\prod R^{3}$$, New volume = (4/3)$$\prod {3R/2}^{3}$$ =($$\prod R^{3}$$/2)

Increase % in volume=((19/6)$$\prod R^{3}$$)*(3/4$$\prod R^{3}$$)*100))% = 237.5%

Original surface area =4$$\prod R^{2}$$. New surface area = 4$$\prod {3R/2}^{2}$$=9($$\prod R^{2}$$)

Increase % in surface area =(5$$\prod R^{2}$$/4$$\prod R^{2}$$) * 100) % = 125%.

Ans .

1728.7

1. Explanation :

Volume of larger sphere = (4/3)$$\prod$$*6*6*6) $$cm^{3}$$ = 288$$\prod$$ $$cm^{3}$$

Volume of 1 small lead ball = ((4/3)$$\prod$$*(1/2)*(1/2)*(1/2)) $$cm^{3}$$ = $$\prod$$/6 $$cm^{3}$$

Number of lead balls = (288$$\prod$$*(6/$$\prod$$)) = 1728.7

Ans .

1792

1. Explanation :

Volume of cylinder = ($$\prod$$ x 6 x 6 x 28 ) $$cm^{3}$$= ( 9$$\prod$$/16) $$cm^{3}$$.

Number of bullet =$$\frac{Volume of cylinder}{Volume of each bullet}$$ = [(36 x 28)$$\prod$$ x 16] /9$$\prod$$ = 1792.

Ans .

243m

1. Explanation :

Volume of sphere = ((4$$\prod$$/3) x 9 x 9 x 9 ) $$cm^{3}$$ = 972$$\prod$$$$cm^{3}$$

Volume of sphere = ($$\prod$$ x 0.2 x 0.2 x h ) $$cm^{3}$$

972$$\prod$$= $$\prod$$ x (2/10) x (2/10) x h --> h = (972 x 5 x 5 )cm = [( 972 x 5 x5 )/100 ] m

= 243m

Ans .

4.2.cm

1. Explanation :

Volume of sphere = Volume of 2 cones

= ($$\frac{1}{3}\prod$$x (2.102) x 4.1 + $$\frac{1}{3} \prod x {2.1}^2$$x 4.3)

Let the radius of sphere be R

(4/3)$$\prod R^{3}$$ = (1/3)$$\prod {2.1}^{3}$$ or R = 2.1cm

Hence , diameter of the sphere = 4.2.cm

Ans .

1:2

1. Explanation :

Let radius of each be R and height of the cone be H.

Then, (4/3) $$\prod R^{3}$$ = (1/3) $$\prod R^{2}$$H (or) R/H =1/4 (or) 2R/H = 2/4 =1/2

Required ratio = 1:2.

Ans .

1039.5 $$cm^{2}$$

1. Explanation :

Volume = (2 $$\prod r^{3}$$/3) = ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))$$cm^{3}$$

= 2425.5 $$cm^{3}$$

Curved surface area = 2$$\prod r^{3}$$ = (2 x (22/7) x (21/2) x (21/2))$$cm^{2}$$

=693 $$cm^{2}$$

Total surface area = 3$$\prod r^{3}$$ = (3 x (22/7) x (21/2) x (21/2))$$cm^{2}$$

= 1039.5 $$cm^{2}$$.

Ans .

54

1. Explanation :

Volume of bowl = ((2($$\prod$$/3) x 9 x 9 x 9 ) $$cm^{3}$$= 486($$\prod$$$$cm^{3}$$.

Volume of 1 bottle = ($$\prod$$ x (3/2) x (3/2) x 4 ) $$cm^{3}$$ = 9($$\prod$$$$cm^{3}$$

Number of bottles = (486($$\prod$$/9($$\prod$$) = 54.

Ans .

1:2:3

1. Explanation :

Let R be the radius of each

Height of the hemisphere = Its radius = R.

Height of each = R.

Ratio of volumes = (1/3)$$\prod R^{2}$$ x R : (2/3)$$\prod R^{3}$$ : $$\prod R^{2}$$ x R = 1:2:3