Question

Mathematics teacher of a school took her 10^th standard students to show Red fort. It was a part of their educational trip. The teacher had interest in history as well. She narrated the facts of Red fort to students. Then, the teacher said in this monument one can find combination of solid figures. There ae 2 pillars which are cylindrical in shape. Also, 2 domes at the corners which are hemispherical. 7 smaller domes at the center. Flag hoisting ceremony on Independence Day takes place near those domes.

(a) How much cloth material will be required to cover 2 big domes each of radius 2.5 m? (Take π = \(\frac{22}{7}\)) 

I. 75 m² 

II. 78.57 m² 

III. 87.47 m² 

IV. 25.8 m²

(b) Find the lateral surface area of two pillars if height of the pillar is 7m and radius of the base is 1.4m. 

I. 112.3 cm² 

II. 123.2 m² 

III. 90 m² 

IV. 345.2 cm²

(c) How much is the volume of a hemisphere if the radius of the base is 3.5m? 

I. 85.9 m³ 

II. 80 m³ 

III. 98 m³ 

IV. 89.83 m³

(d) A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the volume of wood in the toy? 

I. 205.33 cm³ 

II. 301 cm³ 

III. 420 cm³ 

IV. 420.23 cm³

(e) A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. the radius of the base of the cone is : 

I. 3 cm 

II. 4 cm 

III. 2.4 cm 

IV. 3.5 cm

Answer 1

(a) Option : (ii) 

Domes are in hemispherical shape of radius r = 2.5m. 

∴ Cloth material required to cover 2 big domes 

= 2 × curved surface area of 1 hemisphere 

= 2 × 2πr² 

= 2 × 2 × \(\frac{22}{7}\) × 2.5 × 2.5 

= 4 × \(\frac{22}{7}\)× \(\frac{625}{100}\)

= \(\frac{22}{7}\) × 25 = 550 7 = 78.57m². 

∴ Cloth material required to cover 2 big domes is 78.57 m². 

Hence, 

Option (ii) is correct.

(b) Option : (ii) 

Height of cylindrical pillar is h = 7m. 

And radius of the base of cylindrical pillar is r = 1.4 m. 

∴ Lateral surface area of two pillars = 2× surface area of 1 cylindrical pillar

= 2×2πrh 

= 4 x \(\frac{22}{7}\) x 1.4 × 7 

= 88 ×1.4 

= 123.2 m².

Hence,

Option (ii) is correct.

(c) Option : (iv)

The radius of the hemisphere is r = 13.5 m. 

∴ Volume of the hemisphere = \(\frac{2}{3}\)πr³ 

= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 3.5 

= \(\frac{269.5}{3}\) 

= 89.83 m². 

Hence,

Option (iv) is correct.

(d) Option : (i) 

The height of the cylinder is h = 10 cm. 

The radius of the base of cylinder = The radius of hemisphere = r = 3.5cm. 

The volume of wood in the toy = The volume of cylinder – 2× volume of hemisphere.

Hence,

The volume of wood in the toy is 205.33 cm³. 

Hence,

Option (i) is correct.

(e) Option : (iii) 

Let the radius of the base of the cone is R cm. 

Given that height of the cone is h = 75 cm. 

And radius of hemisphere is r = 6 cm. 

Since, 

Hemisphere is cast into the right circular cone. 

∴ The volume of cone = The volume of hemisphere.

Hence, 

The radius of the base of the cone is 2.4 cm. 

Hence,

Option (iii) is correct.