A number of tiny drops of water all of the same radius r cm combine to form a single drop of radius R cm. The rise in temperature of water is given by (if `sigma` = S.T. specific heat= 1 cal/g) :

Options

A. `(sigma)/J(1/r-1/R)`
B. `(3sigma)/J(1/r-1/R)`
C. `(3sigma)/J(1/(r^(2))-1/(R^(2)))`
D. `(sigma)/J(1/(r^(2))-1/(R^(2)))`

Answer

B

Explanation

Here Energy released =`4pi(nr^(2)-R^(2)).sigma`
`therefore` Heat produced=`W/J=(4sigmapi)/J(nr^(2)-R^(2))`
Heat produced in drop=`4/3piR^(3)xx1xxtheta`
`therefore(4sigmapi)/J(nr^(2)-R^(2))=4/3piR^(3)theta`
or `theta=(3sigma)/J(1/r-1/R)`
`therefore` Correct choice is (b).