Ans .
868 \(cm^2\)
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.
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
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.
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
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.
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 \(cm^2\)
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 \(cm^3\).
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.
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
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 \(cm^2\)
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%
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
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 \(cm^2\)
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
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.
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
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
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.
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
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.
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
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.
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
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 \(cm^{2}\)
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%.
Let original radius = R. Then, new radius = (150/100)R=(3R/2)
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
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
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
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
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
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 \(cm^{2}\)
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
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
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