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.

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}

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}

Answer: Option C

Explanation:

Let the length of the wire be h
\begin{aligned}
Radius = \frac{1}{2}mm = \frac{1}{20}cm\\
\pi r^2h = 66 \\
\frac{22}{7}*\frac{1}{20}*\frac{1}{20}*h = 66 \\
=> h = \frac{66*20*20*7}{22} \\
= 8400 cm \\
= 84 m
\end{aligned}

Answer: Option B

Explanation:

Volume is given, we can calculate the radius from it, then by calculating slant height, we can get curved surface area.
\begin{aligned}
\frac{1}{3}*\pi *r^2*h = 1232 \\
\frac{1}{3}*\frac{22}{7}*r^2*24 = 1232 \\
r^2 = \frac{1232*7*3}{22*24} = 49 \\
r = 7 \\
\text{Now, r = 7cm and h = 24 cm } \\
l = \sqrt{r^2+h^2} \\
= \sqrt{7^2+24^2} = 25cm \\
\text{Curved surface area =}\pi rl\\
= \frac{22}{7}*7*25 = 550 cm^2
\end{aligned}

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}