Question Detail

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}

1. 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

2. The cost of the paint is Rs. 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint outside of a cube having 8 feet each side.

Rs. 850
Rs. 860
Rs. 876
Rs. 886

Answer: Option C

Explanation:

We will first calculate the Surface area of cube, then we will calculate the quantity of paint required to get answer.
Here we go,

\begin{aligned}
\text{Surface area =}6a^2 \\
= 6 * 8^2 = 384 \text{sq feet} \\
\text{Quantity required =}\frac{384}{16} \\
= 24 kg\\
\text{Cost of painting =} 36.50*24 \\
= Rs. 876
\end{aligned}

3. 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

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 radii of two cones are in ratio 2:1, their volumes are equal. Find the ratio of their heights.

Answer: Option A

Explanation:

Let their radii be 2x, x and their heights be h and H resp.
Then,
\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
\frac{\frac{1}{3}*\pi *{(2x)}^2*h}{\frac{1}{3}*\pi *{x}^2*H} \\
=> \frac{h}{H} = \frac{1}{4} \\
=> h:H = 1:4
\end{aligned}