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Combination and Conversion of Solids

Introduction

Mensuration में कई problems में solids को combine किया जाता है (जैसे एक cone को cylinder के ऊपर रखना) या एक solid को melt करके दूसरे solid में convert किया जाता है (melting और recasting)। यह pattern आपको सिखाता है कि volumes को कैसे add/subtract करना है और conversion में volume equality कैसे apply होती है - यह exams में बहुत common और high-yield topic है।

इसे master करने से आप tank capacities, metal casting, hollow objects और shaded/remaining volumes जैसे practical questions आसानी से solve कर पाएंगे।

Pattern: Combination and Conversion of Solids

Pattern

Key concept: Combined shapes में volumes (या areas) को add/subtract करते हैं। Conversion में total volume before = total volume after का rule लगाते हैं।

Useful relations:
• Volume add/subtract: Volume(composite) = अलग-अलग parts के volumes का sum - removed parts का volume।
• Conversion: Volume(before) = Volume(after) (mass और density constant होने पर)।
• Problems में commonly used formulas: cylinder πr²h, cone (1/3)πr²h, sphere (4/3)πr³, cuboid lbh.

Step-by-Step Example

Question

9 cm radius वाली एक solid metal sphere को melt करके identical cones में recast किया जाता है, जिनका base radius 3 cm और height 8 cm है। कितने full cones बन सकते हैं?

Solution

  1. Step 1: Sphere का volume निकालें।

    Sphere volume = (4/3)πR³. यहाँ R = 9 cm: V_sphere = (4/3)π × 9³ = (4/3)π × 729 = 972π.

  2. Step 2: एक cone का volume निकालें।

    Cone volume = (1/3)πr²h. यहाँ r = 3 cm, h = 8 cm: V_cone = (1/3)π × 3² × 8 = (1/3)π × 9 × 8 = 24π.

  3. Step 3: Conversion equality apply करें।

    Number of cones = V_sphere ÷ V_cone = (972π) ÷ (24π) = 972 ÷ 24 = 40.5.

  4. Step 4: Final Answer.

    केवल full cones count होते हैं → 40 full cones बनेंगे (half-cone बच जाएगा)।

  5. Quick Check:

    40 × 24π = 960π जो 972π से कम है; एक और cone बनाने के लिए 24π extra चाहिए जो sphere volume से ज्यादा है ✅

Quick Variations

1. Cylindrical cavity वाला cylinder (hollow pipe): outer cylinder volume - inner cylinder volume करें।

2. Cone + cylinder से बना solid: दोनों के volumes add करें।

3. कई छोटे solids का metal मिलाकर बड़ा solid बनाना: छोटे volumes को sum करके बड़े shape से equate करें।

4. Shaded region problems: outer solid/area से inner को subtract करके remaining material निकालें।

5. Conversion with loss/gain: अगर percentage loss दिया है तो पहले loss factor लगाएँ (जैसे usable volume = (1 - loss%) × initial volume) फिर equate करें।

Trick to Always Use

  • Step 1 → हर part का exact formula लिखें (शुरुआत में approximations न लें) और π को तब तक symbolic रखें जब तक cancellation न हो सके।
  • Step 2 → Conversion में π, 1/3 जैसे common factors को cancel कर दें-calculation बहुत आसान हो जाती है।
  • Step 3 → "How many" वाले questions में total available volume को single-item volume से divide करें और integer part (floor) लें।
  • Step 4 → Feasibility check करें (जैसे source volume ≥ required volume) और अगर material loss दिया हो तो उसे जरूर consider करें।

Summary

Summary

Combination और conversion problems का core है volume का सही bookkeeping:

  • Joined parts के volumes add करें; cavities या removed parts को subtract करें।
  • Recasting में initial और final total volume को equate करें (loss factor हो तो apply करें)।
  • Units consistent रखें और π का numeric use तब करें जब वह cancel न हो।
  • Integer counts में हमेशा division का floor लें; leftover हो तो बताएं।

Practice

(1/5)
1. A solid sphere of radius 6 cm is melted to form small spherical balls of radius 3 cm. How many such small spheres can be made?
easy
A. 8
B. 6
C. 4
D. 2

Solution

  1. Step 1: Use volume ratio for same-shape conversion.

    Number = (Volume of large sphere) ÷ (Volume of small sphere) = (R³) ÷ (r³).

  2. Step 2: Substitute radii.

    R = 6, r = 3 ⇒ Number = (6³) ÷ (3³) = 216 ÷ 27.

  3. Step 3: Compute.

    Number = 8.

  4. Final Answer:

    8 spheres → Option A.

  5. Quick Check:

    Ratio (R/r) = 2 → (2)³ = 8, so 8 small spheres fit exactly by volume ✅
Hint: For identical shapes: number = (linear scale)³ = (R/r)³.
Common Mistakes: Using diameters instead of radii when cubing.
2. A solid metal cone of radius 3 cm and height 12 cm is melted and recast into smaller cones each of radius 1 cm and height 3 cm. How many small cones are formed?
easy
A. 8
B. 36
C. 16
D. 18

Solution

  1. Step 1: Use cone volume formula (factor (1/3) cancels in ratio).

    Number = (R²·H) ÷ (r²·h).

  2. Step 2: Substitute values.

    Large: R²·H = 3² × 12 = 9 × 12 = 108. Small: r²·h = 1² × 3 = 3.

  3. Step 3: Compute.

    Number = 108 ÷ 3 = 36.

  4. Final Answer:

    36 cones → Option B.

  5. Quick Check:

    Scaling: (R/r)² × (H/h) = 9 × 4 = 36 ✅
Hint: For cones: number = (R/r)² × (H/h).
Common Mistakes: Treating cone like sphere (using cubic ratio) - cones use r²·h.
3. A solid cylinder of radius 7 cm and height 10 cm is melted and recast into spheres of radius 3.5 cm. How many complete spheres are made? (Use π = 22/7 if needed.)
easy
A. 2
B. 3
C. 8
D. 5

Solution

  1. Step 1: Compute cylinder volume.

    V_cyl = πR²H = (22/7) × 7² × 10 = 1540 cm³.

  2. Step 2: Compute one sphere volume.

    V_sph = (4/3)πr³ with r = 3.5 → using (22/7) gives V_sph ≈ 179.6667 cm³.

  3. Step 3: Divide total by one sphere and take integer part.

    1540 ÷ 179.6667 ≈ 8.57 → full spheres = 8.

  4. Final Answer:

    8 spheres → Option C.

  5. Quick Check:

    8 × 179.6667 ≈ 1437.33 < 1540, adding one more sphere would exceed the available volume ✅
Hint: Divide total volume by one-piece volume; take floor for full items.
Common Mistakes: Rounding too early - compute with adequate precision before flooring.
4. A cone of radius 7 cm and height 24 cm is placed on top of a solid cylinder of the same radius and height 10 cm (shared base). Find the total volume of the combined solid. (Use π = 22/7.)
medium
A. 2772 cm³
B. 3000 cm³
C. 2500 cm³
D. 3100 cm³

Solution

  1. Step 1: Cylinder volume.

    V_cyl = πr²h = (22/7) × 7² × 10 = 1540 cm³.

  2. Step 2: Cone volume.

    V_cone = (1/3)πr²h = (1/3) × (22/7) × 7² × 24 = 1232 cm³.

  3. Step 3: Add volumes.

    Total = 1540 + 1232 = 2772 cm³.

  4. Final Answer:

    2772 cm³ → Option A.

  5. Quick Check:

    Cone: (1/3) of π×49×24 = (1/3)×(22/7)×1176 = 1232; sum with 1540 gives 2772 ✅
Hint: Add individual volumes; keep π symbolic until cancellation if convenient.
Common Mistakes: Forgetting the 1/3 factor for the cone.
5. A hollow sphere has outer radius 10 cm and inner radius 6 cm. Find the volume of metal used. (Use π = 22/7.)
medium
A. 3285.33 cm³
B. 3300 cm³
C. 3150.50 cm³
D. 3400 cm³

Solution

  1. Step 1: Use hollow-sphere volume formula.

    V = (4/3)π(R³ - r³).

  2. Step 2: Substitute values.

    R³ - r³ = 1000 - 216 = 784. So V = (4/3) × (22/7) × 784.

  3. Step 3: Simplify and compute.

    784 ÷ 7 = 112 → V = (4/3) × 22 × 112 = 9856/3 ≈ 3285.3333 cm³.

  4. Final Answer:

    3285.33 cm³ → Option A.

  5. Quick Check:

    Outer sphere volume - inner sphere volume ≈ 4186.67 - 901.33 = 3285.34 (matches rounding) ✅
Hint: Compute R³ - r³ first, then multiply by (4/3)π; cancel 7 early if using 22/7.
Common Mistakes: Forgetting to cube radii or omitting the (4/3) factor.

Mock Test

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