Question Detail

A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets into it. The mass of the man is :

50 kg
60 kg
70 kg
80 kg

Answer: Option B

Explanation:

In this type of question, first we will calculate the volume of water displaces then will multiply with the density of water.
Volume of water displaced = 3*2*0.01 = 0.06 m cube

Mass of Man = Volume of water displaced * Density of water

= 0.06 * 1000 = 60 kg

1. 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}

2. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge.

10000 cubes
1000 cubes
100 cubes
50 cubes

Answer: Option B

Explanation:

\begin{aligned}
\text{Number of cubes =}\frac{100*100*100}{10*10*10} \\
= 1000
\end{aligned}

Note: 1 m = 100 cm

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 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}

5. 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}