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Question Detail
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}
1. 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
2. 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}
3. There are bricks with 24 cm x 12 cm x 8 cm dimensions. Find the total number of bricks required to construct a wall 24 m long, 8 m high and 60 m thick with 10% of wall filled with mortar.
- 35000
- 40000
- 45000
- 50000
Answer: Option C
Explanation:
So as per question,
\begin{aligned}
\text{Volume of wall} = (2400 * 800 * 60 ) cm^3 \\
\text{Volume of bricks} = \text { 90% of volume of wall } \\
= [ \frac{90}{100} * 2400 * 800 * 60 ] cm^3 \\
\text{ Volume of 1 brick = } (24 * 12 * 8) cm^3 \\
\text{Number of Bricks required} \\
= \frac{ (\frac{90}{100}) * (2400 * 800 * 60)}{24 * 12 * 8} \\
= 45000
\end{aligned}
4. 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
5. A hollow spherical metallic ball has an external diameter 6 cm and is 1/2 cm thick. The volume of metal used in the metal is:
- \begin{aligned} 47\frac{1}{5} cm^3 \end{aligned}
- \begin{aligned} 47\frac{3}{5} cm^3 \end{aligned}
- \begin{aligned} 47\frac{7}{5} cm^3 \end{aligned}
- \begin{aligned} 47\frac{9}{5} cm^3 \end{aligned}
Answer: Option B
Explanation:
Please note we are talking about "Hollow" ball. Do not ignore this word in this type of question in a hurry to solve this question.
If we are given with external radius and thickness, we can get the internal radius by subtracting them. Then the volume of metal can be obtained by its formula as,
External radius = 3 cm,
Internal radius = (3-0.5) cm = 2.5 cm
\begin{aligned}
\text{Volume of sphere =}\frac{4}{3}\pi r^3 \\
= \frac{4}{3}*\frac{22}{7}*[3^2 - 2.5^2]cm^3 \\
= \frac{4}{3}*\frac{22}{7}*\frac{91}{8}cm^3 \\
= \frac{143}{3} cm^3 \\
= 47\frac{2}{3}cm^3
\end{aligned}
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