Question Detail

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

1. If the volume of two cubes are in the ratio 27:1, the ratio of their edges is:

Answer: Option A

Explanation:

Let the edges be a and b of two cubes, then

\begin{aligned}
\frac{a^3}{b^3} = \frac{27}{1} \\
=> \left( \frac{a}{b} \right)^3 = \left( \frac{3}{1} \right)^3 \\
\frac{a}{b}=\frac{3}{1} \\
=> a:b = 3:1
\end{aligned}

2. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length if the wire in meters will be:

76 m
80 m
84 m
88 m

Answer: Option C

Explanation:

Let the length of the wire be h
\begin{aligned}
Radius = \frac{1}{2}mm = \frac{1}{20}cm\\
\pi r^2h = 66 \\
\frac{22}{7}*\frac{1}{20}*\frac{1}{20}*h = 66 \\
=> h = \frac{66*20*20*7}{22} \\
= 8400 cm \\
= 84 m
\end{aligned}

3. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:

Answer: Option C

Explanation:

Volume will be length * breadth * height, but in this case two heights are given so we will take average,

\begin{aligned}
Volume = \left(12*9*\left(\frac{1+4}{2}\right)\right)m^3 \\
12*9*2.5 m^3 = 270 m^3
\end{aligned}

4. 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}

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}