Mensuration
- 2 cm of rain has fallen on a square km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m × 10 m base, by what level would the water level in the pool have increased ?
-
View Hint View Answer Discuss in Forum
Volume of rain water = Area of base × height
= 1000000 × 2 = 20000 cu. metre 100
Water stored in pool = 10000 cu. metre∴ Required water level = 10000 = 10 metre 100 Correct Option: B
Volume of rain water = Area of base × height
= 1000000 × 2 = 20000 cu. metre 100
Water stored in pool = 10000 cu. metre∴ Required water level = 10000 = 10 metre 100
- A parallelopiped whose sides are in ratio 2 : 4 : 8 have the same volume as a cube. The ratio of their surface area is :
-
View Hint View Answer Discuss in Forum
Let the sides of the parallelopiped be 2x, 4x and 8x units respectively and the edge of cube be y units.
∴ 2x × 4x × 8x = y³
⇒ 8 × 8 x3 = y³
Taking cube roots,
⇒ 4x = y .....(i)
Surface area of parallelopiped = 2 (2x × 4x +4x ×8x + 8x × 2x)
= 2 (8x² + 32x² + 16x²) = 112x² sq. units.
Surface area of cube = 6y² sq. units.∴ Required ratio = 112x² = 112x² 6y² 6 × 16x² = 7 or 7 : 6 6 Correct Option: D
Let the sides of the parallelopiped be 2x, 4x and 8x units respectively and the edge of cube be y units.
∴ 2x × 4x × 8x = y³
⇒ 8 × 8 x3 = y³
Taking cube roots,
⇒ 4x = y .....(i)
Surface area of parallelopiped = 2 (2x × 4x +4x ×8x + 8x × 2x)
= 2 (8x² + 32x² + 16x²) = 112x² sq. units.
Surface area of cube = 6y² sq. units.∴ Required ratio = 112x² = 112x² 6y² 6 × 16x² = 7 or 7 : 6 6
- If two adjacent sides of a rectangular parallelopiped are 1 cm and 2 cm and the total surface area of the parallelopiped is 22 square cm, then the diagonal of the parallelopiped is
-
View Hint View Answer Discuss in Forum
Let the third side of the rectangular parallelopiped be x cm, then
2 (x × 1 + 1 × 2 + 2 × x) = 22
⇒ 3x + 2 = 11
⇒ 3x = 11 – 2 = 9
Diagonal = √l² + b² + h²
= √3² + 2² + 1²
= √9 + 4 + 1
= √14 cmCorrect Option: C
Let the third side of the rectangular parallelopiped be x cm, then
2 (x × 1 + 1 × 2 + 2 × x) = 22
⇒ 3x + 2 = 11
⇒ 3x = 11 – 2 = 9
Diagonal = √l² + b² + h²
= √3² + 2² + 1²
= √9 + 4 + 1
= √14 cm
- What part of a ditch, 48 metres long, 16.5 metres broad and 4 metres deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres ? (Use π = 22/7)
-
View Hint View Answer Discuss in Forum
Volume of earth taken out = πr²h
= 22 × 2 × 2 × 56 = 704 m³ 7
Volume of the ditch = 48 × 16.5 × 4 m³
= 3168 m³Part of the ditch filled = 704 = 2 3168 9 Correct Option: B
Volume of earth taken out = πr²h
= 22 × 2 × 2 × 56 = 704 m³ 7
Volume of the ditch = 48 × 16.5 × 4 m³
= 3168 m³Part of the ditch filled = 704 = 2 3168 9
- The perimeters of a circle, a square and an equilateral triangle are same and their areas are C, S and T respectively. Which of the following statement is true ?
-
View Hint View Answer Discuss in Forum
Radius of circle = x cm.
Side of square = y cm.
Side of equilateral triangle = z cm.
Circumference of circle = Perimeter of square = Perimeter of equilateral triangle
⇒ 2πx = 4y = 3z⇒ x = 4y = 2y z = 4y 2π π 3
Area of circle ‘C’ = πx²= π × 4 y² = 4 y² > y² π² π
Area of square ‘S’ = y²Area of triangle ‘T’ = √3 z² 4 = √3 × 4 × 4 y² 4 3 × 3 = 4 y² < y² 3√3
∴ T < S < CCorrect Option: B
Radius of circle = x cm.
Side of square = y cm.
Side of equilateral triangle = z cm.
Circumference of circle = Perimeter of square = Perimeter of equilateral triangle
⇒ 2πx = 4y = 3z⇒ x = 4y = 2y z = 4y 2π π 3
Area of circle ‘C’ = πx²= π × 4 y² = 4 y² > y² π² π
Area of square ‘S’ = y²Area of triangle ‘T’ = √3 z² 4 = √3 × 4 × 4 y² 4 3 × 3 = 4 y² < y² 3√3
∴ T < S < C
