We have 2 rectangular sheets of paper, M and N, of dimensions 6 cm × 1 cm each. Sheet M is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet N is cut into equal square patches and assembled to form the largest possible closed cube. Assuming the ends of the cylinder are closed, the ratio of the volume of the cylinder to that of the cube is _________.**(A)** π/2**(B)** 3/π**(C)** 9/π**(D)** 3π**Answer:** **(C)****Explanation:** Sheet first M is 6×1 size, in which cylinder will be formed by bringing the short edges of the sheet together. Given that ends of the cylinder are closed.

So, volume of cylinder (V_{cylinder}) = π.r^{2}.h

Here, circumference = 2πr = 6

So, r = 3/π

and h = 1

Substitute in above equation,

= volume of cylinder (V_{cylinder})

= π.(3/π)^{2}.1

= π.(9/π^{2}).1

= 9/π

Now, second sheet N is 6×1 size, which cut into equal square patches and assembled to form the largest possible closed cube.

So, there would be 6 unit size square which formed a cube which would have unit length (i.e.,) of each side.

So, volume of cube (V_{cube})

= a^{3}

= 1.1.1

= 1

Therefore, the ratio of the volume of the cylinder to that of the cube is

= (9/π) / 1

= 9/π

Option (C) is correct.

Quiz of this Question