The maximum length of a pencil that can he kept is a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

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 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:

\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.

Answer: Option A

Explanation:

In this type of question, just equate the two volumes to get the answer as,
\begin{aligned}
\text{Volume of cylinder =}\pi r^2h\\
\text{Volume of sphere =} \frac{4}{3}\pi r^3\\
=> 12*\frac{4}{3}\pi r^3 = \pi r^2h \\
=> 12*\frac{4}{3}\pi r^3 = \pi *8*8*2 \\
=> r^3 = \frac{8*8*2*3}{12*4} \\
=> r^3 = 8 \\
=> r = 2 cm \\
=> \text{Diameter =}2*2 = 4 cm
\end{aligned}

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