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Given:

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l² + b² + h²)

Floor diagonal = √(l² + b²)

Calculations:

⇒ l² + b² + h² = 14² = 196

⇒ l² + b² = (6√5)² = 36 × 5 = 180

⇒ h² = 196 − 180 = 16

⇒ h = √16 = 4

∴ h = 4 m

Given:

PQ = 14, PR = 6√5, h = 4 (from previous)​

Concept Used:

PQ is the space diagonal; its horizontal projection on the floor is PR and the vertical component is h. So the angle α between PQ and PR has adjacent = PR and hypotenuse = PQ. Hence

cos α = adjacent/hypotenuse = PR / PQ

Calculation:

⇒ cos α = (6√5) / 14

⇒ cos α = (3√5) / 7

∴ cos α = (3√5) / 7.

Alternate Method

Given: 

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l2 + b2 + h2)

Floor diagonal = √(l2 + b2)

Calculations:

⇒ l2 + b2 + h2 = 142 = 196

⇒ l2 + b2 = (6√5)2 = 36 × 5 = 180

⇒ h2 = 196 − 180 = 16

⇒ h = √16 = 4

Using dot product to find cos α

PQ = (l, b, h), PR = (l, b, 0)

PQ · PR = l² + b² + 0 = 180

|PQ| = 14, |PR| = 6√5

cos α = (PQ · PR) / (|PQ| × |PR|)

cos α = 180 / (14 × 6√5)

cos α = 180 / (84√5)

cos α = 15 / (7√5)

∴ cos α = 15 / (7√5)

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

∴ Weight of material = 730 gm.

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

Weight of material = 730 gm.

Volume of water = Volumeinner = 14 × 10 × 8 = 1120 cm3 = 1120 × 10^-6 m³

Weight of water = 1000 × 1120 × 10-6 kg = 1.12 kg

Total weight of the box with water = 0.73 kg + 1.12 kg = 1.850 kg

∴ The correct answer is option 1.

Given:

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC² = BC² - AB²

AC = √(100 - 64)

AC = √36 = 6 cm  

AD = (AB × AC)/(BC)

AD = (6 × 8)/10 = 4.8 cm

∴ The correct answer is option 1.

Given:

 

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC2 = BC2 - AB2

AC = √(100 - 64)

AC = √36 = 6 cm  

ΔADB ~ ΔADB

So, $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{6^2}{8^2}$

⇒ $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{36}{64}$ 

⇒ $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{9}{16}$ 

∴ The correct answer is option 2.

Given:

Annual rainfall = 40 cm = 0.40 m

Area = 1 hectare = 10,000 m²

Formula used:

Volume = Area × Rainfall depth

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m³

∴ Volume of rainfall = 4,000 m³ per hectare (annually).

Given:

Weight of water = 1 metric tonne per m³

Formula used:

Volume = Area × Rainfall depth

Weight = Volume × Density (tonnes/m3)

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m3

⇒ Weight = 4,000 × 1

⇒ Weight = 4,000 metric tonnes

∴ Weight of water = 4,000 metric tonnes per hectare (annually).

Calculation:

In Δ AOB,

sin 60° = p/h = OB / AB

⇒ √3/2 = r / l

⇒ r : l = √3 : 2

Hence, the correct answer is option (3).

Given:

Angle at the vertex of the conical body = 120°

Sum of slant height, height, and radius = 9 + 3√3 cm

Formula used:

Volume of the cone = (1/3)πr²h

Relationship between slant height, radius, and height = l² = r² + h²

Calculations:

Let the height of the cone be h, radius be r, and slant height be l.

We know that angle at the vertex is 120° and tan(60°) = r/h, so r = √3 × h.

We are given that r + h + l = 9 + 3√3, substituting r = √3 × h and l = √(r² + h²) = 2h:

(√3h) + h + 2h = 9 + 3√3

⇒ (√3 + 3)h = 9 + 3√3

⇒ h = (9 + 3√3) / (√3 + 3)

Rationalize the denominator:

h = (9 + 3√3)(√3 - 3) / (3 - 9)

⇒ h = -18 / -6

⇒ h = 3 cm

Now, r = √3 × h = √3 × 3 = 3√3 cm.

Now, volume of the cone = (1/3)πr²h = (1/3)π(3√3)² × 3

⇒ Volume = (1/3)π × 27 × 3 = 27π cubic cm.

∴ The volume of the cone is 27π cubic cm.