2
–1
1
0
2. Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at the speed of 15 km/h. In what time will the water be 3 m deep?
100 hours
120 hours
140 hours
150 hours
3. If a solid cone of volume 27 πcm^3 is kept inside a hollow cylinder whose radius and height are that of the cone, then the volume of water needed to fill the empty space is:
3πcm^3
18πcm^3
54πcm^3
81πcm^3
4. Two 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 has increased?
1 km
10 m
11 m
1 m
5. The height of a circular cylinder is increased 6 times and the base area is decreased to 1/9 of its value. The factor by which the lateral surface of the cylinder increases is:
2
1/2
2/3
3/2
6. The height of a right prism with a square base is 15 cm. if the area of the total surfaces of the prism is 608 cm2, its volume is:
910 cm2
920 cm2
960 cm2
980 cm2
7. The volume of a solid hemisphere is 19404 cm3. Its total surface area is:
4158 cm3
2858 cm3
1738 cm3
2038 cm3
8. Base of right pyramid is a square, length of diagonal of the base is 24√2 m. If the volume of the pyramid is 1728 cu. m, its height is:
7 m
8 m
9 m
10 m
9. The height of a right circular cone and the radius of its circular base are 9 cm and 3 cm respectively. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e, the lower part) of the cone is 44 cm2. The radius of the upper circular surface of the frustum (taking π=22/7) is:
∛12 cm
∛13 cm
∛6 cm
∛20 cm
10. Base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is 108√3 cc, its height is:
9 cm
10 cm
11 cm
12 cm
Answers:
- d
Solution:Radius of the base of cone = r units ∴
Volume (v) =1/3 πr^2 h
Curved surface area =πr√(h^2+r^2 ) ∴
3πvh^3-c^2 h^2+9v^2=3π×1/3 πr^2 h×h^3
-π^2 r^2 (h^2+r^2 ) h^2+9×1/9 π^2 r^4 h^3
=π^2 r^2 h^4-π^2 r^2 h^4-π^2 r^4 h^2+π^2 r^4 h^2=0 - a
Solution:15 km/h = 15000 m/hour
Water flown in an hour =(2 ×1.5 ×15000 )/100=450〖 m〗^3
Volume of desired water in the tank =(150×100×3) m^3
∴ Time =(150 ×100 ×3)/450=100 hours - c
Solution:Let, the radius and height of the cone be r cm and h cm respectively.
Now, according to the question,
Required volume of water
=πr^2 h-1/3 πr^2 h=2/3 π^2 h
=2×(1/3 πr^2 h)
=2× volume of cone =(2×27π)=54π cm^3
Quicker Method:
Volume of required water = 2 × volume of cone
= 2×27π=54πcm^3 - b
Solution:Volume of rain water = Area of base × height =1000000×2/100=20000 m^3
Water stored in pool
=(50% of 20000)=10000 m^3 ∴
Required water level =(10000/1000)=10 m - a
Solution:Curved surface of cylinder =2πrh
Case II:
Radius =1/3 r and height = 6h
Curved surface =2π×1/3 r×6h=(2πrh)×2
∴ Increase in curved surface of cylinder will be twice. - c
Solution:Total surface area of prism = Curved surface area + 2 × Area of base
⇒ 608 = Perimeter of base × height + 2 × Area of base
⇒ 608 = 4x×15+2x^2
(When x = side of square)
⇒ x^2+30x-304=0
⇒ x^2+38x-8x-304=0
⇒ x(x +38)-8(x+38)=0
⇒(x-8)(x+38)=0 ⇒ x = 8
Volume of prism = Area of base × height
8 × 8 × 15 = 960 cm3 - a
Solution:2/3 πr^3=19404
⇒2/3×22/7×r^3=19404
⇒r^3=(19404 × 3 × 7)/(2 ×22)=9261
∴r=∛(21×21×21)=21 cm. ∴
Total surface area =3πr^2
=3×22/7×21×21 = 4158 cm2 - c
Solution:Area of the base =1/2×(diagonal)^2
=1/2×24√2×24√2=576 m2
∴ Volume of pyramid =1/3× height × area of base
=1728=1/3×h×576⇒h=(1728 ×3)/576=9m - b
- d
Solution:Area of the base =√3/4×side^2
=√3/4×6×6=9√3 cm2 ∴
Volume of the prism = Area of base × height
⇒108√3=9√3×h ⇒
h=(108√3)/(9√3)=12 cm.
