Question Detail

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}

1. 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}

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 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

4. 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}

5. A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. The percentage of the wood wasted is :

Answer: Option D

Explanation:

We will first subtract the cone volume from wood volume to get the wood wasted.
Then we can calculate its percentage.

\begin{aligned}
\text{Sphere Volume =}\frac{4}{3}\pi r^3 \\
\text{Cone Volume =}\frac{1}{3}\pi r^2h\\

\text{Volume of wood wasted =}\\
\left(\frac{4}{3}\pi *9*9*9\right)-\left(\frac{1}{3}\pi *9*9*9\right) \\
= \pi *9*9*9 cm^3 \\
\text{Required Percentage =} \\
\frac{\pi *9*9*9}{\frac{4}{3}\pi *9*9*9}*100 \% \\
= \frac{3}{4}*100 \% \\
= 75\%
\end{aligned}