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 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}

2. Two right circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.

\begin{aligned} \sqrt{3}:1 \end{aligned}
\begin{aligned} \sqrt{7}:1 \end{aligned}
\begin{aligned} \sqrt{2}:1 \end{aligned}
\begin{aligned} 2:1 \end{aligned}

Answer: Option C

Explanation:

Let their heights be h and 2h and radii be r and R respectively then.

\begin{aligned}
\pi r^2h = \pi R^2(2h) \\
=> \frac{r^2}{R^2} = \frac{2h}{h} \\
= \frac{2}{1} \\
=> \frac{r}{R} = \frac{\sqrt{2}}{1} \\
=> r:R = \sqrt{2}:1 \\
\end{aligned}

3. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

4 cm
6 cm
8 cm
10 cm

Answer: Option B

Explanation:

\begin{aligned}
\text{Curved surface area of sphere =}\\
\frac{4}\pi r^2 \\
\text{Surface area of cylinder =} \\
2\pi rh \\
=> \frac{4}\pi r^2 = 2\pi rh \\
=> r^2 = \frac{6*12}{2} \\
=> r^2 = 36 \\
=> r = 6
\end{aligned}

Note: Diameter of cylinder is 12 so radius is taken as 6.

4. 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}

5. 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