Question Detail

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}

1. 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}

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 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 volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is:

\begin{aligned} 79.8 cm^3 \end{aligned}
\begin{aligned} 79.4 cm^3 \end{aligned}
\begin{aligned} 89.8 cm^3 \end{aligned}
\begin{aligned} 89.4 cm^3 \end{aligned}

Answer: Option C

Explanation:

Volume of the largest cone = Volume of the cone with diameter of base 7 and height 7 cm

\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
= \frac{1}{3}*\frac{22}{7}*3.5*3.5*7 \\
= \frac{269.5}{3}cm^3 \\
= 89.8 cm^3
\end{aligned}

Note: radius is taken as 3.5, as diameter is 7 cm

5. A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by:

\begin{aligned} 11\frac{3}{7} cm \end{aligned}
\begin{aligned} 11\frac{2}{7} cm \end{aligned}
\begin{aligned} 11\frac{1}{7} cm\end{aligned}
\begin{aligned} 11 cm\end{aligned}

Answer: Option A

Explanation:

Let the drop in the water level be h cm, then,

\begin{aligned}
\text{Volume of cylinder= }\pi r^2h \\
=> \frac{22}{7}*\frac{35}{2}*\frac{35}{2}*h = 11000 \\
=> h = \frac{11000*7*4}{22*35*35}cm\\
= \frac{80}{7}cm\\
= 11\frac{3}{7} cm
\end{aligned}