Correct Answer - B

Let water equivalent be x

`therefore` `(10+x)c((DeltaT)/(DeltaT))= -k(T_(1)+T_(2))/2 - T_(0))`

Where c is the specific heat of water

`DeltaT` = Change in temperature

`Deltat` = time at which temperature falls

`therefore` (10+x)c(20-15)/(15) = -k(20+15)/2 -T_(0) ................(i)

`(20+x)c(20-15)/15 = -k((20+15)/2)= -k((20+15)/2)-T_(0))`.....................(ii)

From eqs. (i) and (ii), we get

`(10+x)/10 = (20+x)/15`

`150+15x = 200 + 10x`

`5x =50 rArr x=10g`