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Question Detail
A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets into it. The mass of the man is :
- 50 kg
- 60 kg
- 70 kg
- 80 kg
Answer: Option B
Explanation:
In this type of question, first we will calculate the volume of water displaces then will multiply with the density of water.
Volume of water displaced = 3*2*0.01 = 0.06 m cube
Mass of Man = Volume of water displaced * Density of water
= 0.06 * 1000 = 60 kg
1. 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
2. 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}
3. 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}
4. 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}
5. 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
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