Note: Volume of earth dug out `= (4 xx 5/2 xx 3/2) m^3` `= 15 m^3.` Area over which earth is spread `= (31 xx 10 - 4 xx 5/2) m^2` `= 300 m^2` Rise in Level `= (Volume/Area)` `= (15/300 xx 100) cm` `= 5 cm.`

Note: Let original edge`= a` Then, volume`= a^3` New edge`= 2a.` So, new volume `= (2a)^3 = 8a^3.` :. Volume becomes `8` times.

Note: Let original edge = a. Then, surface area`= 6a^2` New edge`= 125/100 a = (5a/4).` New surface area `= 6 xx ((5a)/(4))^2` `= (75a^2)/8.` Increase in surface area `= ((75a^2)/(8) - 6a^2)` `= (27a^2)/8.` Increase`% = ((27a^2)/8 xx 1/(6a^2) xx 100)%` `= 56.25%.`

Note: Let their edges be a. Then, `a^3/b^3 = 8/27 ⇔ (a/b)^3` `= (2/3)^3 ⇔ a/b = 2/3` `⇔ a^2/b^2 ⇔ 6a^2/6b^2 = 4/9.`

Note: `4π R^2 = 6a^2 ⇒ R^2/a^2` `= 3/(2π) ⇒ R/a` `=(√3)/(√2π).` `Volume of sphere/Volume of cube` `= (4/3 πR^3)/a^3` `= (4)/(3) π.(R/a)^3` `= 4/3π . (3√3)/(2π√2π)` `= (2√3)/(√2π)` `= √12/√2π` `= (√6)/(√π).`

Note: Let the edge of the cube be a. Then,volume of the cube `= a^3.` Radius of the sphere `= (a/2).` Volume of the sphere `= 4/3 r (a/2)^3` `= (ra^3)/6.` :. Requied Ratio `= a^3 : (ra^3/6` `= 6 : r.`

Note: New cuboid has length`= 12 cm,` breadth`= 6 cm` & height `= 6 cm` :. Its surface area`= 2(12 xx 6 xx 6 xx 6 + 12 xx 6) cm^2` ` = 360 cm^2.`

Note: Let length` = 1`breadth`= 6 cm` & height`= h.` Then, lb`= x, bh`= y, and lh`= z.` On multiplying, we get `(lbh)^2 = xyz` `or lbh =√xyz.` :. Volume`= √xyz.`

Note: `1/v = 1/s xx s/v` `= (2 (ab + bc + ca))/(s xx abc)` `= 2/s (1/a + 1/b + 1/c)`

Note: Number of cubes `= (Volume of bigger cube)/(Volume of smaller cube)` `= ((6 xx 6 xx 6)/(2 xx 2 xx 2))` `= 27.`