If two liquids are mixed in equal volumes, their resultant density is 5 grcc and mixed in equal masses their resultant density is 4.8 g/ec. The ratio of their densities is 2/k . Then k is
Answers
Answer:
k = 3
Explanation:
Let say Theit Densities are D1 & D2
Density = Mass/Volume
Case 1 : Equal Volume
=> M1 = D1 * V
& M2 = D2 * V
Density combined = (M1 + M2)/2V = (D1 + D2)/2 = 5
=> D1 + D2 = 10
Case 2 : Equal Masses
=> V1 = M/D1
=> V2 = M/D2
Density combined = 2M /(M/D1 + M/D2) = 2D1D2/(D 1 + D2) = 4.8
=> D1 * D2 /10 = 2.4
=> D1 * D2 = 24
=> D1(10-D1) = 24
=> -D1² + 10D1 = 24
=> D1² - 10D1 + 24 = 0
=> D1² -6D1 - 4D1 + 24 = 0
=> (D1 -6)(D1-4) = 0
=> D1 = 6 or 4
] D2 are 4 & 6
D1/D2 = 6/4 = 3/2 is not of form 2/k
while D1/D2 = 4/6 = 2/3 = 2/k => k = 3
Answer:
Correct option is B)
Let the densities be D
1
and D
2
Density=
Volume
Mass
Case 1 : Equal Volume
⟹M
1
=D
1
×V
and
M
2
=D
2
×V
Density combined =
2V
(M
1
+M
2
)
=
2
(D
1
+D
2
)
=5
⟹D
1
+D
2
=10
Case 2 : Equal Masses
⟹V
1
=
D
1
M
⟹V
2
=
D
2
M
Density combined =
(M/D
1
+M/D
2
)
2M
=
(D+1+D
2
)
2D
1
D
2
=4.8
10
D
1
×D
2
=2.4
⟹D
1
×D
2
=24
impliesD
1
(10−D
1
)=24
⟹−D
1
2
+10D
1
=24
⟹D
1
2
−10D
1
+24=0
⟹D
1
2
−6D
1
−4D
1
+24=0
⟹(D
1
−6)(D
1
−4)=0
Then,
D
1
=6 or 4
D
2
are 4&6
D
2
D
1
=
4
6
=
2
3
is not of form
k
2
while
D
2
D
1
=
6
4
=
3
2
=
k
2
⟹k=3