Mensuration
- Assume that a drop of water is spherical and its diameter is onetenth of a cm. A conical glass has a height equal to the diameter of its rim. If 32,000 drops of water fill the glass completely, then the height of the glass (in cm) is
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Let Height of glass = h cm
then Radius = h cm. 2
Volume of glass = volume of 32000 drops∴ 1 π 
h 
² × h 3 2 = 4 π 1 ³ × 32000 3 20 ⇒ h³ = 4 × 1 × 32000 4 8000
⇒ h³ = 4³ ⇒ h = 4 cmCorrect Option: D
Let Height of glass = h cm
then Radius = h cm. 2
Volume of glass = volume of 32000 drops∴ 1 π h ² × h 3 2 = 4 π 1 ³ × 32000 3 20 ⇒ h³ = 4 × 1 × 32000 4 8000
⇒ h³ = 4³ ⇒ h = 4 cm
- A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is :
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Walls are 5 cm thick.
∴ Internal length = (330 – 2 × 5) cm = 320 cm.
Let the thickness of bottom be x Breadth = (260 – 10)cm
= 250 cm Height = (110 – x) cm
Here, the cistern is assumed to be open and x is the thickness of bottom.
∴ 320 × 250 × (110 – x) = 8000 litres
⇒ 320 × 250 × (110 – x) = 8000 × 1000 cm³⇒ (110 - x) = 8000000 320 × 250
⇒ 110 – x = 100
⇒ x = 10 cm or 1 dm.Correct Option: C
Walls are 5 cm thick.
∴ Internal length = (330 – 2 × 5) cm = 320 cm.
Let the thickness of bottom be x Breadth = (260 – 10)cm
= 250 cm Height = (110 – x) cm
Here, the cistern is assumed to be open and x is the thickness of bottom.
∴ 320 × 250 × (110 – x) = 8000 litres
⇒ 320 × 250 × (110 – x) = 8000 × 1000 cm³⇒ (110 - x) = 8000000 320 × 250
⇒ 110 – x = 100
⇒ x = 10 cm or 1 dm.
- A cone is cut at mid point of its height by a frustum parallel to its base. The ratio between the two parts of cone would be
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∆ADE ~ ∆ABC∴ AD = DE = 1 AB BC 2 AD = AB ; DE = 1 BC 2 Required ratio = 1 π(DE)² × AD 3 1 π BC² × AB - 1 π (DE)² × AD 3 3 = DE² × AD BC² × AB - DE² × AD = 1 BC² × 1 AB 4 2 BC² × AB - 1 BC² × AB 4 2 = 1 = 1 : 7 8 1 - 1 8 Correct Option: D
∆ADE ~ ∆ABC∴ AD = DE = 1 AB BC 2 AD = AB ; DE = 1 BC 2 Required ratio = 1 π(DE)² × AD 3 1 π BC² × AB - 1 π (DE)² × AD 3 3 = DE² × AD BC² × AB - DE² × AD = 1 BC² × 1 AB 4 2 BC² × AB - 1 BC² × AB 4 2 = 1 = 1 : 7 8 1 - 1 8
- The area of a circle of radius 5 is numerically what percent of its circum ference?
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We know that Circumference = 2πr
= 2 × 22 × 5 = 220 7 7 Area = πr² = 22 × 25 = 550 7 7
⇒ Required %= 550 × 7 × 100 = 250 % 7 220 Correct Option: D
We know that Circumference = 2πr
= 2 × 22 × 5 = 220 7 7 Area = πr² = 22 × 25 = 550 7 7
⇒ Required %= 550 × 7 × 100 = 250 % 7 220
- If the circumference and area of a circle are numerically equal, then the diameter is equal to :
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We know that The area of circle = πr²
πD² 4
[where D : diameter of circle] and Circumference of circle = 2πr = πD
Now, according to question,πD² = πD 4
D² = 4D
D² – 4D = 0
⇒ D(D – 4) = 0 ⇒ D = 4Correct Option: D
We know that The area of circle = πr²
πD² 4
[where D : diameter of circle] and Circumference of circle = 2πr = πD
Now, according to question,πD² = πD 4
D² = 4D
D² – 4D = 0
⇒ D(D – 4) = 0 ⇒ D = 4
