If 1 cubic cm of cast iron weights 21 gm, then the weight of a cast iron pipe of length 1 m with a bore of 3 cm and in which the thickness of the metal is 1 cm, is?

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.

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

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

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

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

Curved surface area of cylinder = 2?rh cm²
Total surface area of cylinder = 2?r(h + r) cm²
Now, According to the question
2?rh/2?r(h + r) = 1/2
? h/(h + r) = 1/2
? 2h = h + r
? Now, Total surface area = 616 cm²
? 2? (h + r) = 616
? 2?r (r + r) = 616 (? h = r)
? 2?r(2r) = 616
? 4?r² = 616
? r = ?style="text-decoration:overline">616/4?
= ?616 x 7/4 x 22
= ?7 x 7 =
? r = h = 7 cm
? Volume = ? r² h
= (22/7) x (7 x 7) x 7 = 1078 cm³

If 10 circular plates each of thickness 3 cm, each are placed one above the other, then it forms a cylinder with height (3 x 10 = 30 cm) and a hemisphere of radius 6 cm is placed on the top just to cover the cylinder that means its radius is 6 cm also
? Radius of hemisphere (R) = 6 cm
Radius of cylinder (r) = 6 cm
and height of cylinder (h) = 30 cm
? Volume of the solid = Volume of cylinder + Volume of hemisphere
= ?r²h + (2/3)?R³
= ?(6)² x 30 + (2/3)?(6)³
= ? x 36 x 30 + (2/3)? x 216
= 1080? + 2? x 72 = 1080? + 144?
= 1224? cm³
Which is the required volume of solid.

Given series is a consecutive prime number series. Therefore, the next term will be 23.

Series Pattern
+6, +9, +12, +15....
? Missing term = 6 + 6 = 12

(1 - 2) = -1
(3 - 4) = -1
(5 - 6) = -1
and so on.
? Total terms of -1 will be 50.
? Answer = -1 x 50 = -50