C

Along one edge, the number of small cubes that can be cut
= [latex]\frac{100}{10}[/latex] = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height).
Total number of small cubes that can be cut = 10 [latex]\times [/latex] 10 [latex]\times [/latex] 10 = 1000

B

In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.
= 500 [latex]\times [/latex] 2 [latex]\times [/latex] [latex]\frac{22}{7}[/latex] [latex]\times [/latex] 22.4
= 70400 cm = 704 m

A

The volume of the cone = [latex](\frac{1}{3}) \pi {r}^{2}h[/latex]
Only radius (r) and height (h) are varying.
Hence, [latex](\frac{1}{3}) \pi[/latex] may be ignored.
[latex]\frac {{V}_{1}}{{V}_{2}} [/latex] = [latex]\frac {{{r}_{1}}^{2} {h}_{1}}{{{r}_{2}}^{2} {h}_{2}} = \Rightarrow \frac{1}{10} = \frac {{(1)}^{2} {h}_{1}}{{(2)}^{2} {h}_{2}}[/latex]
[latex] \frac{{h}_{1}}{{h}_{2}} = \frac {2}{5}[/latex]
i.e. [latex]{h}_{1} : {h}_{2}[/latex] = 2 : 5

C

Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
[latex]\pi {(16)}^{2} \times h = (\frac {4}{3}) \pi {(12)}^{3} \Rightarrow h = 9 cm[/latex]

A

Ratio of the sides = [latex]\sqrt [3] {729}[/latex] : [latex]\sqrt [3] {1331}[/latex] = 9 : 11
Ratio of surface areas = [latex] {(9)}^{2} : {(11)}^{2}[/latex] = 81 : 121