From the question we know that,
The Distance covered at 22 \(\frac{km}{hr}\) in 30 hours = 22 × 30 = 660 km = 66 × 106 \(cm\)
The radius of a cycle wheel = 35 cm \((\frac{70}{2})\ cm\) as 70 is diameter
In one revolution, the distance covered by the wheel will be equal to the Circumference of cycle wheel
Circumference of cycle wheel = \(2 \pi r\)
= \(2 \times \frac{22}{7} \times 35 = 220\ cm\)
Now, The Number of revolutions in the total journey = \(\frac{66 \times 10^6}{220}\) = 3 × 105
= 300000
The correct option is (B): 3 lakh
On the day of her examination, Riya sharpened her pencil from both ends as shown below. 
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.