A density bottle weighs 120 g and 100 g when filled completely with oil ad water, respectively. If the weight of a empty density bottle is 40g, then arrange the following steps in sequence meant to solve the problem to get the density of oil
(A). The density of the oil
`D=("Mass of the oil")/("Weight of the water")=("Mass of the oil")/("Mass of the water")`
`=("Weighht of oil")/("Weight of the water")=(W_(3)-W_(1))/(W_(2)-W_(1))`
(B). Let the weight of the bottle +oil=`W_(3)=120g`
and the weight of the bottle+water =`W_(2)`
100g, where bottle is completely filled with liquid, i.e., oil or water.
(C). Let the weight of the empty bottle`=W_(1)=40g`
(D). The the weight of the oil ad water would be equal to `(W_(3)-W_(1)) and (W_(2)-W_(1))`, respectively.

Options

A. CBDA
B. ABCD
C. BADC
D. DCBA

Answer

A

Explanation

Consider the weight of the empty density bottle as, `W_1= 40 g.` Let the weight of the bottle +oil be `W_3= 120 g` and weight of the bottle + water as `W_2=100 g.` Here bottle is completely filled with given liquids, i.e., either oil or water. From above, the weight of the oil and water would be equal to `(W_3 -W_1) and (W_2-W_1),` respectively. The density of the oil `=("mass of oil "(M_e))/("mass of oil "(V_(e))=("mass of oil")/("volume of water")`
`=("weight of the oil")/("mass of the water")=("weight of the oil")/("weight of the water")`
`rArr D=(W_(3)-W_(1))/(W_(2)-W_(1))`
(A) (Here density of water in CGS system is `"1 g cm"^(-3)`.