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. 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. Two right circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.

\begin{aligned} \sqrt{3}:1 \end{aligned}
\begin{aligned} \sqrt{7}:1 \end{aligned}
\begin{aligned} \sqrt{2}:1 \end{aligned}
\begin{aligned} 2:1 \end{aligned}

Answer: Option C

Explanation:

Let their heights be h and 2h and radii be r and R respectively then.

\begin{aligned}
\pi r^2h = \pi R^2(2h) \\
=> \frac{r^2}{R^2} = \frac{2h}{h} \\
= \frac{2}{1} \\
=> \frac{r}{R} = \frac{\sqrt{2}}{1} \\
=> r:R = \sqrt{2}:1 \\
\end{aligned}

3. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km square. The height of mountain is :

2.3 km
2.4 km
2.5 km
2.6 km

Answer: Option B

Explanation:

Let the radius of the base be r km. Then,
\begin{aligned}
\pi r^2 = 1.54 \\
r^2 = \frac{1.54*7}{22} = 0.49\\
= 0.7 km \\
\text{Now l=2.5 km, r = 0.7 km} \\
h = \sqrt{2.5^2 - 0.7^2} km \\
=\sqrt{6.25 - 0.49}\\
=\sqrt{5.76} km \\
= 2.4 km
\end{aligned}

4. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

4 cm
6 cm
8 cm
10 cm

Answer: Option B

Explanation:

\begin{aligned}
\text{Curved surface area of sphere =}\\
\frac{4}\pi r^2 \\
\text{Surface area of cylinder =} \\
2\pi rh \\
=> \frac{4}\pi r^2 = 2\pi rh \\
=> r^2 = \frac{6*12}{2} \\
=> r^2 = 36 \\
=> r = 6
\end{aligned}

Note: Diameter of cylinder is 12 so radius is taken as 6.

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}