Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
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Mixtures and Alligation Questions

Multiple choice

When n fluids of masses $m_1,m_2...m_n$ and densities $p_1,p_2...p_n$ respectively are mixed together then resultant density of mixture is :
  1. $\dfrac{{\displaystyle\sum\limits_{i = 1}^n {{m_i}} }}{{\displaystyle\sum\limits_{i = 1}^n {{p_i}} }}$
  2. $\dfrac{{\displaystyle\sum\limits_{i = 1}^n {{m_i}{p_i}} }}{{\displaystyle\sum\limits_{i = 1}^n {{m_i}} }}$
  3. $\dfrac{{\displaystyle\sum\limits_{i = 1}^n {{m_i}} }}{{\displaystyle\sum\limits_{i = 1}^n {\dfrac{{{m_i}}}{{{p_i}}}} }}$

  4. infinity
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Density = Total Mass / Total Volume. Total mass = sum(m_i). Total volume = sum(m_i / p_i). Therefore, density = sum(m_i) / sum(m_i / p_i).

Multiple choice

If two liquids are mixed in equal volumes, their resultant density is 5 gr/cc and mixed in equal masses their resultant density is 4.8 g/cc . The ratio of their densities is $\frac{2}{k}$ .Then K is

  1. 1

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let densities be d1, d2. Equal volumes: (d1+d2)/2 = 5 => d1+d2 = 10. Equal masses: 2/(1/d1 + 1/d2) = 4.8 => 2*d1*d2 / (d1+d2) = 4.8 => d1*d2 = 24. Roots of x^2 - 10x + 24 = 0 are 4 and 6. Ratio d1/d2 = 4/6 = 2/3 or 6/4 = 3/2. Given ratio is 2/k. If 2/3, k=3. If 3/2, k=4/3.

Multiple choice

Water of density $4 kg/m^3$ and ice of density $3 kg/m^3$ are mixed together. If their masses are equal then the density of mixture is
  1. $\dfrac{8}{3}kg\,/ m^3$
  2. $\dfrac{7}{3}kg\,/ m^3$
  3. $\dfrac{24}{7}kg\,/ m^3$
  4. $\dfrac{2}{3}kg\,/ m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Density = Total Mass / Total Volume. Let mass of each be m. Volume_water = m/4, Volume_ice = m/3. Total mass = 2m. Total volume = m/4 + m/3 = 7m/12. Density = 2m / (7m/12) = 24/7.

Multiple choice

Two liquids of masses $M_1$ and $M_2$ and specific heats $S_1$ and $S_2$ respectively are mixed. The specific heat of the mixture is ____________.
  1. $\cfrac { { { M }_{ 1 }S }_{ 1 }+{ { M }_{ 2 }S }_{ 2 } }{ M_{ 1 }+M_{ 2 } } $
  2. $\cfrac { { { M }_{ 1 }S }_{ 1 }+{ { M }_{ 2 }S }_{ 2 } }{ 2(M_{ 1 }+M_{ 2 }) } $
  3. $\cfrac { { 2({ M }_{ 1 }S }_{ 1 }+{ { M }_{ 2 }S }_{ 2 }) }{ M_{ 1 }+M_{ 2 } } $
  4. $\cfrac { { { M }_{ 1 }S }_{ 1 }-{ { M }_{ 2 }S }_{ 2 } }{ M_{ 1 }+M_{ 2 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The specific heat of a mixture is the weighted average of the specific heats of the components based on mass: S_mix = (M1*S1 + M2*S2) / (M1 + M2).

Multiple choice

In an insulated vessel, $0.05\ kg$ steam at $373\ K$ and $0.45\ kg$ of ice at $253\ K$ are mixed. Then, find the final temperature of the mixture. Given, $L_{fusion} = 80 cal/g = 336 J/g$, $L_{vaporisation} = 540 cal/g = 2268 J/g$, $S_{ice} = 2100 J/kg K = 0.5 cal/gK$ and $S_{water} = 4200 J/kg K = 1 cal /gK$
  1. $273\ K$
  2. $373\ K$
  3. $100\ K$
  4. $0\ K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat lost by steam (cooling to 100C + condensing + cooling to 0C) vs heat gained by ice (warming to 0C + melting). 0.05kg steam releases 0.05*2268000 + 0.05*4200*100 = 113400 + 21000 = 134400 J. 0.45kg ice requires 0.45*2100*20 + 0.45*336000 = 18900 + 151200 = 170100 J. Since heat required to melt all ice is greater than heat released by steam, the final state is a mix of ice and water at 273 K.

Multiple choice

Two liquids $L_{1}$ and $L_{2}$ are at $20^{\circ}C$ and $40^{\circ}C$ respectively. When equal masses of them are mixed together the temperature of the mixture becomes $35^{\circ}C$. The ratio of specific heat capacities of $L_{1}$ and $L_{2}$ is:
  1. $1 : 1$
  2. $3 : 1$
  3. $1 : 3$
  4. $2 : 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the principle of calorimetry, m * c1 * (35 - 20) = m * c2 * (40 - 35). This simplifies to 15 * c1 = 5 * c2, so c1 / c2 = 5 / 15 = 1 / 3.

Multiple choice

$20g$ of ice and $20 g$ of hot water are mixed, when the ice is melted the temperature of the mixture was found to be $0^oC$. The temperature of hot water taken should be ($L_{ice}=$80 cal/g)
  1. $40^{0}C$
  2. $72^{0}C$
  3. $80^{0}C$
  4. $96^{0}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat lost by hot water = Heat gained by ice. Heat gained by ice = m*L = 20 * 80 = 1600 cal. Heat lost by water = m*c*deltaT = 20 * 1 * (T - 0) = 20T. 20T = 1600, so T = 80 degrees Celsius.

Multiple choice

Three liquids $A,B\ and\ C$ of masses $400 gm,\ 600 gm\ and\ 800 gm$ are at $30^{0}C$, $40^{0}C$ and $50^{0}C$ respectively. When $A$ and $B$ are mixed, resultant temperature is $36^{0}C$ and when $B$ and $C$ are mixed, resultant temperature is $44^{0}C$. The ratio of their specific heats are :
  1. $2:1:1$
  2. $3:2:1$
  3. $2:2:1$
  4. $1:4:9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat lost = Heat gained. For A and B: 400*sA*(36-30) = 600*sB*(40-36) => 2400*sA = 2400*sB => sA = sB. For B and C: 600*sB*(44-40) = 800*sC*(50-44) => 2400*sB = 4800*sC => sB = 2*sC. Thus sA:sB:sC = 2:2:1.

Multiple choice

$1$ g of ice at $0^{\circ}C$ is mixed $1$ g of stream at $100^{\circ}C$. After thermal equilibrium is achieved, the temperature of the mixture is
  1. $100^{\circ}C$
  2. $55^{\circ}C$
  3. $75^{\circ}C$
  4. $0^{\circ}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1g of steam at 100C releases 540 calories to condense into water at 100C. 1g of ice at 0C requires 80 calories to melt into water at 0C. The remaining heat (540 - 80 = 460 calories) will raise the temperature of the 2g of water. Since 460 calories is more than enough to raise 2g of water to 100C (which takes 2 * 100 = 200 calories), the mixture will reach 100C.

Multiple choice

A closed container of volume 0.02 m$^3$ contains a mixture of neon and argon gases at a temperature of 27$^{0}$ C and at a pressure of $1\times 10^{5}N/m^{2}$ . The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

  1. 16 gm, 12 gm

  2. 4 gm, 24 gm

  3. 6 gm, 22 gm

  4. 12 gm, 16 gm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using PV = nRT, n = PV/RT = (10^5 * 0.02) / (8.314 * 300) = 0.801 moles. Let m1 be mass of Neon (20g/mol) and m2 be mass of Argon (40g/mol). m1 + m2 = 28 and m1/20 + m2/40 = 0.801. Solving gives m1 = 4g and m2 = 24g.

Multiple choice

$1\;kg$ ice at $-20^{\circ}C$ is mixed with $1\;kg$ steam at $200^{\circ}C$. Then find equilibrium temperature and mixture content.
  1. $T=100^{\circ}C$, $\dfrac{20}{27}$ kg steam and $\dfrac{34}{27}$ kg water
  2. $T=0^{\circ}C$, $\dfrac{20}{27}$ kg steam and $\dfrac{34}{27}$ kg water
  3. $T=100^{\circ}C$, $\dfrac{34}{27}$ kg steam and $\dfrac{20}{27}$ kg water
  4. $T=0^{\circ}C$, $\dfrac{10}{27}$ kg steam and $\dfrac{38}{27}$ kg water
Reveal answer Fill a bubble to check yourself
A Correct answer