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 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}

2. 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}

3. 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}

4. 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}

5. 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}