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Question Detail
The perimeter of one face of a cube is 20 cm. Its volume will be:
- \begin{aligned} 125 cm^3 \end{aligned}
- \begin{aligned} 400 cm^3 \end{aligned}
- \begin{aligned} 250 cm^3 \end{aligned}
- \begin{aligned} 625 cm^3 \end{aligned}
Answer: Option A
Explanation:
Edge of cude = 20/4 = 5 cm
Volume = a*a*a = 5*5*5 = 125 cm cube
1. 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}
2. 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}
3. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8g/cm cube, then find the weight of the pipe.
- 3.696 kg
- 3.686 kg
- 2.696 kg
- 2.686 kg
Answer: Option A
Explanation:
In this type of question, we need to subtract external radius and internal radius to get the answer using the volume formula as the pipe is hollow. Oh! line become a bit complicated, sorry for that, lets solve it.
External radius = 4 cm
Internal radius = 3 cm [because thickness of pipe is 1 cm]
\begin{aligned}
\text{Volume of iron =}\pi r^2h\\
= \frac{22}{7}*[4^2 - 3^2]*21 cm^3\\
= \frac{22}{7}*1*21 cm^3\\
= 462 cm^3 \\
\end{aligned}
Weight of iron = 462*8 = 3696 gm
= 3.696 kg
4. 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
5. The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is:
- \begin{aligned} 116 \pi cm^2 \end{aligned}
- \begin{aligned} 122 \pi cm^2 \end{aligned}
- \begin{aligned} 124 \pi cm^2 \end{aligned}
- \begin{aligned} 136 \pi cm^2 \end{aligned}
Answer: Option D
Explanation:
\begin{aligned}
\text{Curved surface area of cone=}\pi rl\\
l = \sqrt{r^2+h^2} \\
l = \sqrt{8^2+15^2} = 17cm \\
\text{Curved surface area =}\pi rl\\
= \pi *8*17 = 136 \pi cm^2
\end{aligned}
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