Note: Let their radii be R and r. Then, `(4/3 π R^3)/(4/3 π r^3)` `1/8 ⇒ (R/r)^3` `= 1/8 = (1/2)^3.` So, R/r = 1/2.` Ratio of surface areas `= (4 πR^2)/(4πr^2)` `= (R/r)^2` `= 1/4.`

Note: Let their radii be R and r. Then, `(πR^2)/(4πr^2)` `= 4/25 ⇒ (R/r)^2` `= (2/5)^2 or R/r` `2/5.` :. Ratio of their volumes `= (4/3 πR^3)/(4/3 πr^3)` `= (R/r)^3` `= (2/5)^3` `= 8/125`

Note: The total surface area `= 3 π R^2` `= (3 xx 22/7 xx 7 xx 7) cm^2` `= 462 cm^2.`

Note: Volume flown in conical vessel `= 1/3 π xx (2o)^2 xx 24` `= 3200 π` Volume flown in `1` min `= (π xx 2.5/10 xx 2.5/10 xx 1000)` `= 62.5 π.` :. Time taken `= (3200π)/(62.5π)` `51 min. 12 sec.`

Note: Let R be the radius of each Height of hemi - sphere = its radius = R. :. Height of each = R Ratio of their volumes `= 1/3 πR^2 xx R : 2/3 πR^3 : πR^2 xx R` `= 1 : 2 : 3`.

Note: volume of ball = volume of water displaced by it `:. 4/3 π R^3` `= π xx 12 xx 12 xx 6.75 ⇒ R^3` `= 9 xx 9 xx 9 or R = 9 cm.`

Note: Volume of bowl `= (2/3 π xx 9 xx 9 xx 9) cm^3` `= 486 π cm^3` Volume of `1` bottle `= (π xx 3/2 xx 3/2 xx 4) cm^3` `= 9 π cm^3. Number of bottles `= ((486 π)/(9π))` `= 54.`

Note: Volulme of the block `= (10 xx 5 xx 2) cm^3` `= 100 cm^3` Volume of the cone carved out `= (1/3 xx 22/7 xx 3 xx 3 xx 7) cm^3` `= 66 cm.^3` Wood wasted `= (100 - 66)%` `= 34%`

Note: Let the original height = h. Then, volume `= 1/3 π r^2 h.` New height`= 2h.` So, volume `=1/3 π r^2 xx 92h)` `= 2(1/3 π r^2 h)` :. Increase `= ((1/3 π r^2h)/(1/3 π r^2h) xx 100)%` `= 100%`.

Note: Let their heights be h and 2h & let their radii be R and r. Then, `π R^2h` `= π r^2 (2h) ⇒ R^2/r^2` `= 2/1 ⇒ R/r` `= √2/1`