491. Thousand solid metallic spheres of diameter 6 cm each are melted and recast into a new solid sphere. The diameter of the new sphere (in cm) is
     

1. 1. 30
      

   
   
   

   2. 90
      

   
   
   

   3. 45
      

   
   
   

   4. 60

   
   
   

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1. View Hint View Answer Discuss in Forum

   Volume of each smaller sphere =	4	πr³
   	3

   
   

   =	4	π × (3)³ = 36 &p; cu.cm.
   	3

   
   If the radius of larger sphere be R cm, then
   

   	4	πR³ = 1000 × 36π
   	3

   
   

   ⇒ R³ =	1000 × 36 × 3
   	4

   
   = 1000 × 3 × 3 × 3
   ∴ R = ³√1000 × 3 × 3 × 3
   = 10 × 3 = 30 cm.
   ∴ Its diameter = (2 × 30) cm. = 60 cm

   Correct Option: D

   Volume of each smaller sphere =	4	πr³
   	3

   
   

   =	4	π × (3)³ = 36 &p; cu.cm.
   	3

   
   If the radius of larger sphere be R cm, then
   

   	4	πR³ = 1000 × 36π
   	3

   
   

   ⇒ R³ =	1000 × 36 × 3
   	4

   
   = 1000 × 3 × 3 × 3
   ∴ R = ³√1000 × 3 × 3 × 3
   = 10 × 3 = 30 cm.
   ∴ Its diameter = (2 × 30) cm. = 60 cm

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492. A hemispherical bowl has internal radius of 6 cm. The internal surface area would be : (Takep π = 3.14)
     

1. 1. 225 cm²
      

   
   
   

   2. 400 cm²
      

   
   
   

   3. 289.75 cm²
      

   
   
   

   4. 226.08 cm²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Internal surface area of hemispherical bowl = 2πr²
   = (2 × 3.14 × 6 × 6) sq.cm. = 226.08 sq.cm.

   Correct Option: D

   Internal surface area of hemispherical bowl = 2πr²
   = (2 × 3.14 × 6 × 6) sq.cm. = 226.08 sq.cm.

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493. The diameter of a 120 cm long roller is 84 cm. It takes 500 complete revolutions of the roller to level a ground. The cost of levelling the ground at Rs. 1.50 per sq. m. is :
     

1. 1. Rs. 6000
      

   
   
   

   2. Rs. 3762
      

   
   
   

   3. Rs. 2376
      

   
   
   

   4. Rs. 5750

   
   
   

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1. View Hint View Answer Discuss in Forum

   Levelled area in one revolution of roller = 2πrh
   

   = 2 ×	22	× 42 × 120 = 31680 sq. cm.
   	7

   
   Area levelled in 500 revolutions = (31680 × 500) sq. cm. = 15840000 sq. cm.
   = 1584 sq. metre
   ∴ Required cost = Rs. (1584 × 1.5) = Rs. 2376

   Correct Option: C

   Levelled area in one revolution of roller = 2πrh
   

   = 2 ×	22	× 42 × 120 = 31680 sq. cm.
   	7

   
   Area levelled in 500 revolutions = (31680 × 500) sq. cm. = 15840000 sq. cm.
   = 1584 sq. metre
   ∴ Required cost = Rs. (1584 × 1.5) = Rs. 2376

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494. If the radius of a sphere is increased by 2 cm, then its surface area increases by 352 cm². The radius of the sphere initially was : (use π = 22/7)
     

1. 1. 4 cm
      

   
   
   

   2. 5 cm
      

   
   
   

   3. 3 cm
      

   
   
   

   4. 6 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Initial radius of sphere = r cm (let).
   According to the question, 4π(r + 2)² – 4πr² = 352
   ⇒ 4π ((r + 2)² – r²) = 352
   

   ⇒ r² + 4r + 4 – r² =	352
   	4π

   
   

   =	352
   	4 ×	22
   		7

   
   

   ⇒ 4r + 4 =	352 × 7	= 28
   	4 × 22

   
   ⇒ 4r = 28 – 4 = 24
   

   ⇒ r =	24	= 6 cm
   	4

   Correct Option: D

   Initial radius of sphere = r cm (let).
   According to the question, 4π(r + 2)² – 4πr² = 352
   ⇒ 4π ((r + 2)² – r²) = 352
   

   ⇒ r² + 4r + 4 – r² =	352
   	4π

   
   

   =	352
   	4 ×	22
   		7

   
   

   ⇒ 4r + 4 =	352 × 7	= 28
   	4 × 22

   
   ⇒ 4r = 28 – 4 = 24
   

   ⇒ r =	24	= 6 cm
   	4

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495. The total surface area of a right circular cylinder with radius of the base 7 cm and height 20 cm, is:
     

1. 1. 900 cm²
      

   
   
   

   2. 140 cm²
      

   
   
   

   3. 1000 cm²
      

   
   
   

   4. 1188 cm²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total surface area of right circular cylinder = 2πrh + 2πr² = 2πr (h + r)
   

   = 2 ×	22	× 7 (20 + 7)
   	7

   
   = 2 × 22 × 27 = 1188 sq. cm.

   Correct Option: D

   Total surface area of right circular cylinder = 2πrh + 2πr² = 2πr (h + r)
   

   = 2 ×	22	× 7 (20 + 7)
   	7

   
   = 2 × 22 × 27 = 1188 sq. cm.

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