Volume and Surface Area Problems and Solutions
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15. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.
- 42 m sqaure
- 49 m sqaure
- 52 m sqaure
- 64 m sqaure
Answer :
Option B
Explanation:
Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square -
16. If a right circular cone of height 24 cm has a volume of 1232 cm cube, then the area of its curved surface is :
- \begin{aligned} 450 cm^2 \end{aligned}
- \begin{aligned} 550 cm^2 \end{aligned}
- \begin{aligned} 650 cm^2 \end{aligned}
- \begin{aligned} 750 cm^2 \end{aligned}
Answer :
Option B
Explanation:
Volume is given, we can calculate the radius from it, then by calculating slant height, we can get curved surface area.
\begin{aligned}
\frac{1}{3}*\pi *r^2*h = 1232 \\
\frac{1}{3}*\frac{22}{7}*r^2*24 = 1232 \\
r^2 = \frac{1232*7*3}{22*24} = 49 \\
r = 7 \\
\text{Now, r = 7cm and h = 24 cm } \\
l = \sqrt{r^2+h^2} \\
= \sqrt{7^2+24^2} = 25cm \\
\text{Curved surface area =}\pi rl\\
= \frac{22}{7}*7*25 = 550 cm^2
\end{aligned} -
17. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:
- 260
- 262
- 270
- 272
Answer :
Option C
Explanation:
Volume will be length * breadth * height, but in this case two heights are given so we will take average,
\begin{aligned}
Volume = \left(12*9*\left(\frac{1+4}{2}\right)\right)m^3 \\
12*9*2.5 m^3 = 270 m^3
\end{aligned} -
18. The volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is:
- \begin{aligned} 79.8 cm^3 \end{aligned}
- \begin{aligned} 79.4 cm^3 \end{aligned}
- \begin{aligned} 89.8 cm^3 \end{aligned}
- \begin{aligned} 89.4 cm^3 \end{aligned}
Answer :
Option C
Explanation:
Volume of the largest cone = Volume of the cone with diameter of base 7 and height 7 cm
\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
= \frac{1}{3}*\frac{22}{7}*3.5*3.5*7 \\
= \frac{269.5}{3}cm^3 \\
= 89.8 cm^3
\end{aligned}
Note: radius is taken as 3.5, as diameter is 7 cm -
19. A circular well with a diameter of 2 meters, is dug to a depth of 14 meters. What is the volume of the earth dug out.
- \begin{aligned} 40 m^3 \end{aligned}
- \begin{aligned} 42 m^3 \end{aligned}
- \begin{aligned} 44 m^3 \end{aligned}
- \begin{aligned} 46 m^3 \end{aligned}
Answer :
Option C
Explanation:
\begin{aligned}
Volume = \pi r^2h \\
Volume = \left(\frac{22}{7}*1*1*14\right)m^3 \\
= 44 m^3
\end{aligned} -
20. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :
- \begin{aligned} 2:1 \end{aligned}
- \begin{aligned} 1:\sqrt{2} \end{aligned}
- \begin{aligned} \sqrt{2}:1 \end{aligned}
- \begin{aligned} \sqrt{3}:1 \end{aligned}
Answer :
Option C
Explanation:
Let the radius of hemisphere and cone be R,
Height of hemisphere H = R.
So the height of the cone = height of the hemisphere = R
Slant height of the cone
\begin{aligned}
= \sqrt{R^2+R^2} \\
= \sqrt{2}R \\
\frac{\text{Hemisphere Curved surface area}}{\text{Cone Curved surface area}} = \\
\frac{2\pi R^2}{\pi *R*\sqrt{2}R} \\
= \sqrt{2}:1
\end{aligned} -
21. The maximum length of a pencil that can he kept is a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is
- \begin{aligned} 2\sqrt{17} \end{aligned}
- \begin{aligned} 2\sqrt{16} \end{aligned}
- \begin{aligned} 2\sqrt{26} \end{aligned}
- \begin{aligned} 2\sqrt{24} \end{aligned}
Answer :
Option C
Explanation:
In this question we need to calculate the diagonal of cuboid,
which is =
\begin{aligned}
\sqrt{l^2+b^2+h^2} \\
= \sqrt{8^2+6^2+2^2} \\
= \sqrt{104} \\
= 2\sqrt{26}
\end{aligned}