Mensuration
- The volume of the metal of a cylindrical pipe is 748 cm3. The length of the pipe is 14cm and its external radius is 9 cm. Its thickness is (Take π = 22/7)
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Let the thickness of the pipe be x cm
∴ If the external radius = 9 cm then, in radius = (9 – x) cm
According to the question, π × 9² × 14 – π × 14 × (9 – x)² = 748
↠ π × 14 (81 – (81 + x² – 18x)) = 748
π p × 14 (–x² + 18x) = 748⇒ –x² + 18x = 784 = 784 × 7 π × 14 22 × 14
⇒ –x² + 18x = 17
⇒ x² – 18x + 17 = 0
⇒ x² – 17x – x + 17 = 0
⇒ x (x – 17) – 1 (x – 17) = 0
⇒ (x – 1) (x – 17) = 0
⇒ x = 1 or 17 but x = 17 is inadmissible
∴ x = 1 cmCorrect Option: A
Let the thickness of the pipe be x cm
∴ If the external radius = 9 cm then, in radius = (9 – x) cm
According to the question, π × 9² × 14 – π × 14 × (9 – x)² = 748
↠ π × 14 (81 – (81 + x² – 18x)) = 748
π p × 14 (–x² + 18x) = 748⇒ –x² + 18x = 784 = 784 × 7 π × 14 22 × 14
⇒ –x² + 18x = 17
⇒ x² – 18x + 17 = 0
⇒ x² – 17x – x + 17 = 0
⇒ x (x – 17) – 1 (x – 17) = 0
⇒ (x – 1) (x – 17) = 0
⇒ x = 1 or 17 but x = 17 is inadmissible
∴ x = 1 cm
- A hollow cylindrical tube 20 cm long, is made of iron and its external and internal diameters are 8 cm and 6 cm respectively. The volume of iron used in making the tube is ( π = 22/7)
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The volume of iron used = πr1²h1 - πr2²h2 = (r1² - r2²)
= 22 × 20(4² - 3²) 7 = 22 × 20 × 7 = 440 cu.cm. 7 Correct Option: C
The volume of iron used = πr1²h1 - πr2²h2 = (r1² - r2²)
= 22 × 20(4² - 3²) 7 = 22 × 20 × 7 = 440 cu.cm. 7
- Four equal sized maximum circular plates are cut off from a square paper sheet of area 784 sq.cm. The circumference of each plate is (Take π = 22/7)
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Using Rule 10 and 14,
Side of the square paper sheet = √784 = 28 cm.Obviously, radius of each circle = 28 = 7 cm 4
Circumference of each circular plate = 2πr= 2 × 22 × 7 = 44 cm 7 Correct Option: B
Using Rule 10 and 14,
Side of the square paper sheet = √784 = 28 cm.Obviously, radius of each circle = 28 = 7 cm 4
Circumference of each circular plate = 2πr= 2 × 22 × 7 = 44 cm 7
- If the perimeter of a semicircular field is 144m, then the diameter of the field is (take π = 22/7)
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Using Rule 14,
If the radius be r metre, then πr + 2r = 144
⇒ r (π + 2) = 144⇒ r = 144 = 144 = 144 × 7 = 28 π + 2 22 + 2 36 7
∴ Diameter = 2r = 2 × 28 = 56 metreCorrect Option: D
Using Rule 14,
If the radius be r metre, then πr + 2r = 144
⇒ r (π + 2) = 144⇒ r = 144 = 144 = 144 × 7 = 28 π + 2 22 + 2 36 7
∴ Diameter = 2r = 2 × 28 = 56 metre
- If the difference between the circumference and diameter of a circle is 30 cm, then the radius of the circle must be
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Using Rule 14,
Let the radius of the circle be r cm.
Then, 2πr – 2r = 30
2r (π – 1) = 30⇒ 2r × 22 - 7 = 30 7 ⇒ 2r ×15 = 30 × 7 ⇒ r = 30 × 7 ⇒ r = 7 cm 30 Correct Option: B
Using Rule 14,
Let the radius of the circle be r cm.
Then, 2πr – 2r = 30
2r (π – 1) = 30⇒ 2r × 22 - 7 = 30 7 ⇒ 2r ×15 = 30 × 7 ⇒ r = 30 × 7 ⇒ r = 7 cm 30
