Given that, the height and radius of a right circular ,etal cone (solid) are 8 cm and 2 cm, respectively.
i.e., h = 8 cm and r = 2 cm
Let the radius of the sphere is R
Then, by condition,
1/3 ? r2 h = 4/3 ? R3 h
? 4 x 8 = 4 R3
? R3 = (2)3 = ? R = 2
? Radius odf the sphere = 2 cm
Let the breadth and height of room be m and h m, respectively.
Then, according to the question, l x b = n x Area occupied by one patient
? 14 x b = 56 x 22
? b = (56 x 22)/14 = 8.8 m3
? h x Area of occupied by one patient = 8.8
h = 8.8/2.2
h = 4 m
Volume of water in the tank = 100 x 500 x 2 = 30000 m3
Speed of water = 20 x 1000/60 = 1000/3 m/min
water flow per minute = (15/10) x (125/100) x (1000/3) = 625 m3
Time taken = 30000/625 = 48 min
Volume of the earth dugout as a tunnel = ?r2h = (22/7) x 2 x 2 x 56 = 704 m3
Volume of the ditch = (48 x 33)/(2 x 4) = 24 x 33 x 4 = 3168 m3
? Part required = 704/3168 = 2/9
r = Radius of the cone = 14/2 = 7 cm
h = Height of the cone = 12 x 2 = 24 cm
? Slant height (l) = ?r2 + h2
= ?(7)2 + (24)2
= ?49 + 576
= ?625
= 25 cm
Area of the sheet = Total surface area
= (?rl + ?r2)
= ?r(l + r)
= (22/7) x 7 x (25 + 7) = 704 sq cm
Let length of the wire be h According to the question,
Volume of sphere = Volume of wire
(4/3) x ? x 18 x 18 x 18 = ? x (2/10) x (2/10) x h
? h = (100 x 1944) cm
= (100 x 1944)/100 m = 1944 m [? 1 cm = 1/100 m]
Given that, Height = 1 m
Diameter = 140 cm or 140/100 m = 1.40 m
? r = 1.40/2 = 0.7 m
Now,
Required sheet = Total surface area = 2?r( h + r)
= (22/7) x 1.4 x (1 + 0.7) = 7.48 sq m
? Sheet required = 7.48 sq m
Height of the cylinder = 10 x Diameter of each ball
= 10 x 10 = 100 cm
? Required empty space
= ? (5)2 (100) - 4/3 ? (5)3 x 10
= ? (5)2 x 10 [10 - 20/3]
= ? (5)2 x 10 [30 - 20/3] = 2500/3 ? cm3
Let h be the height,
Volume of the cone = V
and curved surface area of the cone = c
? 1/3 ?r2h = V, ?rl = c
i.e., ?r?r2 + h2 = c
?2r2(r2 + h2) = c2
Consider 3?Vh3 - c2h2 + 9V2
= 3? x 1/3?r2h x h3 - ?2r2(r2 + h2)h + 9 x 1/9 x ?2r4 h2
= ?2r2h4 - ?2r4h2 - ?2r2h4 + ?2r4 = 0
Let the side of the cube be 3k, 4k and 5k, respectively.
So, volumes of these cubes are 27 k3, 64k3, 125 k3 respectively.
? Volume of the new bigger cube = 27 k3 + 64k3 + 125 k3
= 216k3
So, side of the new cube = 6k
Since, diagonal of the cube = ?(6k)2 + (6k)2 + (6k)2
= 12?3
? 108k2 = 432
?k = 2
So, sides of the three cubes were 6 cm, 8 cm and 10 cm respectively.
External radius = 2.5 cm,
length = 100 cm
? External volume = [? x (2.25)2 x 100] cm2
internal radius = 1.5 cm
? internal volume = [? x (1.5)2 x 100]cm3
Volume of metal
= [? x (2.5)2 x 100 - ? x (1.5)2 x 100] cm3
= ? x 100 x [(2.5)2 - (1.5)2] cm3
= (22/7 x 100 x 4 x 1 x 21/1000) kg
= 26.4 kg.
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