Question Detail
Find the surface area of a 10cm*4cm*3cm brick.
- 154 cm square
- 156 cm square
- 160 cm square
- 164 cm square
Answer: Option D
Explanation:
Surface area of a cuboid = 2(lb+bh+hl) cm square
So,
Surface area of a brick = 2(10*4+4*3+3*10) cm square
= 2(82) cm square = 164 cm square
1. 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
2. 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}
3. 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}
4. A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by:
- \begin{aligned} 11\frac{3}{7} cm \end{aligned}
- \begin{aligned} 11\frac{2}{7} cm \end{aligned}
- \begin{aligned} 11\frac{1}{7} cm\end{aligned}
- \begin{aligned} 11 cm\end{aligned}
Answer: Option A
Explanation:
Let the drop in the water level be h cm, then,
\begin{aligned}
\text{Volume of cylinder= }\pi r^2h \\
=> \frac{22}{7}*\frac{35}{2}*\frac{35}{2}*h = 11000 \\
=> h = \frac{11000*7*4}{22*35*35}cm\\
= \frac{80}{7}cm\\
= 11\frac{3}{7} cm
\end{aligned}
5. 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}