Step 1: Volume of the cube Let the side of the cube be $a$. The volume of the cube is: \[ V_{\text{cube}} = a^3. \] Step 2: Volume of the sphere The sphere that fits exactly inside the cube will have a diameter equal to the side of the cube, $a$. The radius of the sphere is: \[ r = \frac{a}{2}. \] The volume of the sphere is: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{a}{2}\right)^3. \] Simplify: \[ V_{\text{sphere}} = \frac{4}{3} \pi \cdot \frac{a^3}{8} = \frac{\pi a^3}{6}. \] Step 3: Find the ratio The ratio of the volume of the cube to the volume of the sphere is: \[ \text{Ratio} = \frac{V_{\text{cube}}}{V_{\text{sphere}}} = \frac{a^3}{\frac{\pi a^3}{6}} = \frac{6}{\pi}. \] Correct Answer: The ratio is $6 : \pi$.
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.
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