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Question Detail
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}
1. 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
2. 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}
3. The radii of two cones are in ratio 2:1, their volumes are equal. Find the ratio of their heights.
- 1:4
- 1:3
- 1:2
- 1:5
Answer: Option A
Explanation:
Let their radii be 2x, x and their heights be h and H resp.
Then,
\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
\frac{\frac{1}{3}*\pi *{(2x)}^2*h}{\frac{1}{3}*\pi *{x}^2*H} \\
=> \frac{h}{H} = \frac{1}{4} \\
=> h:H = 1:4
\end{aligned}
4. 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
5. 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}
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