This question was reported to be quite unpopular among the candidates. Majority of those who attempted it were also reported to have performed poorly. Candidates were expected to recall that . If volume of sphere = V = r3, = 4pr2. Hence, dV = 4pr2dr. = = .
But = 6%. This implies that = 6 i.e = = 2%.
Surface area of a sphere = A = 4pr2. = 8pr. Therefore, dA = 8prdr. = = . Hence, = = 2 ´ 2 = 4%. i.e. The area is increasing by 4%.
. If volume of sphere = V = 
r3, 
= 4pr2. Hence, dV = 4pr2dr. 
= 
= 
.
= 6%. This implies that 
= 6 i.e 
= 
= 2%.
= 8pr. Therefore, dA = 8prdr. 
= 
= 
. Hence, 
= 
= 2 ´ 2 = 4%. i.e. The area is increasing by 4%
.