Mensuration
- The base of a right prism is an equilateral triangle. If the lateral surface area and volume is 120 cm² 40√3 cm³ respectively then the side of base of the prismis
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Lateral surface area of prism = 3 × side × height
∴ 3 × side × height = 120⇒ Side × height = 120 = 40 sq.cm.........(i) 3
Volume of prism = Area of base × height⇒ 40√3 = √3 × (side)² × height 4 ⇒ 40√3 × 4 × (side)² × height √3
⇒ (side)² × height = 160 cu.cm ...(ii)
Dividing equation (ii) by (i),Side = 160 = 4 cm 40 Correct Option: A
Lateral surface area of prism = 3 × side × height
∴ 3 × side × height = 120⇒ Side × height = 120 = 40 sq.cm.........(i) 3
Volume of prism = Area of base × height⇒ 40√3 = √3 × (side)² × height 4 ⇒ 40√3 × 4 × (side)² × height √3
⇒ (side)² × height = 160 cu.cm ...(ii)
Dividing equation (ii) by (i),Side = 160 = 4 cm 40
- A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given ³√2 = 1.2593] is approximately
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Volume of lead = 4 πr³ = 4 π × 2³ 3 3
If the thickness of gold be x cm, then Volume of gold= 4 π[(2 + x)³ - 2³]cu.cm. 3 ∴ 4 π[(2 + x)³ - 2³] 3 = 4 π × 2³ 3
7rArr; (2 + x )³ – 2³ = 2³
⇒ (2 + x )³ = 8 + 8 = 16
⇒ (2 + x )³ = 2³.2
⇒ 2 + x = 2 × ³√2
⇒ 2 + x = 2 × 1.259 = 2.518
∴ x = 2.518 – 2 = 0.518 cmCorrect Option: D
Volume of lead = 4 πr³ = 4 π × 2³ 3 3
If the thickness of gold be x cm, then Volume of gold= 4 π[(2 + x)³ - 2³]cu.cm. 3 ∴ 4 π[(2 + x)³ - 2³] 3 = 4 π × 2³ 3
7rArr; (2 + x )³ – 2³ = 2³
⇒ (2 + x )³ = 8 + 8 = 16
⇒ (2 + x )³ = 2³.2
⇒ 2 + x = 2 × ³√2
⇒ 2 + x = 2 × 1.259 = 2.518
∴ x = 2.518 – 2 = 0.518 cm
- A right triangle with sides 9 cm, 12 cm and 15 cm is rotated about the side of 9 cm to form a cone. The volume of the cone so formed is :
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Radius of cone so formed = 9 cm
Its height = 12 cm∴ Volume of cone = 1 πr²h 3 = 1 π × 9 × 9 × 12 3
= 324π cu. cm.Correct Option: D
Radius of cone so formed = 9 cm
Its height = 12 cm∴ Volume of cone = 1 πr²h 3 = 1 π × 9 × 9 × 12 3
= 324π cu. cm.
- The portion of a ditch 48m long, 16.5 m wide and 4 m deep that can be filled with stones and earth available during excavation of a tunnel, cylindrical in shape, of diameter 4 m and length 56 m is (Take π = 22/7)
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Volume of earth and stones taken out from the tunnel = πr²h
= 
22 × 2 × 2× 56 
cu.metre. 7
= 704 cu. metre
Volume of ditch = (48 × 16.5 × 4) cu. metre = 3168 cu. metre∴ Part of ditch filled = 704 = 2 parts 3168 9 Correct Option: C
Volume of earth and stones taken out from the tunnel = πr²h
= 22 × 2 × 2× 56 cu.metre. 7
= 704 cu. metre
Volume of ditch = (48 × 16.5 × 4) cu. metre = 3168 cu. metre∴ Part of ditch filled = 704 = 2 parts 3168 9
- If a hemisphere is melted and four spheres of equal volume are made, the radius of each sphere will be equal to
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Volume of hemisphere = 2 πR³ cu. units 3 Volume of newphere = 4 πr³ cu. units 3
According to the question,2 πR³ = 4 × 2 πr³ 3 3
⇒ R³ = 8r³
⇒ R = 2r units∴ r = 1 R Units 2 Correct Option: C
Volume of hemisphere = 2 πR³ cu. units 3 Volume of newphere = 4 πr³ cu. units 3
According to the question,2 πR³ = 4 × 2 πr³ 3 3
⇒ R³ = 8r³
⇒ R = 2r units∴ r = 1 R Units 2
