Correct Answer - (a) (i) Let extension in each spring is `x` then `F_(1) = K_(1)x_(1)`
`F_(2) = K_(2)x_(1), F_(3) = K_(3)x = …… = K_(n)x`
`F_(1) + F_(2) + F_(3) + ….. + F_(n) = F`
`rArr K_(1)x + K_(2)x + K_(3)x + ...... + K_(n)x = K_(eq)x`
`rArr K_(eq) = K_(1) + K_(2) + K_(3) + .... + K_(n)`
(ii)
Let `F` be the force in each spring and `x_(1), x_(2), x_(3)......x_(n)`
are the extensions in different springs
then `F = K_(1)x_(1) = K_(2)x_(2) = K_(3)x_(3) = ....... = K_(n)x_(n)`
`x_(1) + x_(2) + x_(3) .... + x_(n) = x`
`rArr (F)/(K_(1)) + (F)/(K_(1)) + (F)/(K_(3)) + ....... + (F)/(K_(n)) = (F)/(K_(eq))`
`rArr (F)/(K_(eq)) = (F)/(K_(1)) + (F)/(K_(2)) + (F)/(K_(3)) +.........+ (F)/(K_(n))`