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 :

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

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}

Answer: Option D

Explanation:

l = (48 - 16)m = 32 m, [because 8+8 = 16]
b = (36 -16)m = 20 m,
h = 8 m.

Volume of the box = (32 x 20 x 8) m cube
= 5120 m cube.

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}

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}