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**Hint:**When the two liquids are mixed together, heat rejected from the liquid with the higher temperature will be absorbed by the liquid with the lower temperature. So we can say that the energy is conserved during the mixing of these two liquids. The amount of heat rejected or absorbed will be the product of the mass of the liquid, its specific heat and its change in temperature.

**Formula used:**

The amount of heat rejected or absorbed by a liquid is given by, $\Delta Q = ms\left( {{T_f} - {T_i}} \right)$ where $m$ is the mass of the liquid, $s$ is its specific heat, ${T_i}$ and ${T_f}$ are its initial and final temperatures.

**Complete step by step solution:**

List the known parameters of the two liquids.

Let the two liquids involved be named as A and B.

The temperature of liquid A is given to be ${T_A} = 60^\circ {\text{C}}$ and the temperature of liquid B is given to be ${T_B} = 20^\circ {\text{C}}$ .

Let ${m_A}$ and ${m_B}$ be the masses of the liquids A and B respectively.

Also, let ${s_A}$ and ${s_B}$ be the specific heats of the liquids A and B respectively.

Then it is given that $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{3}{4}$ and $\dfrac{{{s_A}}}{{{s_B}}} = \dfrac{4}{5}$ .

Let $T$ be the resultant temperature of the mixture of the two liquids.

Express the amount of heat rejected by liquid A and the amount of heat absorbed by liquid B.

The amount of heat rejected by liquid A can be expressed as $\Delta {Q_A} = {m_A}{s_A}\left( {{T_A} - T} \right)$ ------- (1)

The amount of heat absorbed by liquid B can be expressed as $\Delta {Q_B} = {m_B}{s_B}\left( {T - {T_B}} \right)$ ------- (2)

Apply the conservation of heat energy during mixing to obtain the resultant temperature.

While mixing, the amount of heat lost by liquid A will be the amount of heat gained by liquid B.

$ \Rightarrow \Delta {Q_A} = \Delta {Q_B}$ ------- (3)

Substituting equations (1) and (2) in equation (3) we get, ${m_A}{s_A}\left( {{T_A} - T} \right) = {m_B}{s_B}\left( {T - {T_B}} \right)$

$ \Rightarrow \dfrac{{{m_A}}}{{{m_B}}} \times \dfrac{{{s_A}}}{{{s_B}}} \times \left( {{T_A} - T} \right) = \left( {T - {T_B}} \right)$ ------- (4)

Substituting for ${T_A} = 60^\circ {\text{C}}$ , ${T_B} = 20^\circ {\text{C}}$ , $\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{3}{4}$ and $\dfrac{{{s_A}}}{{{s_B}}} = \dfrac{4}{5}$ in equation (4) we get, $\dfrac{3}{4} \times \dfrac{4}{5} \times \left( {60 - T} \right) = \left( {T - 20} \right)$

$ \Rightarrow 180 - 3T = 5T - 100 \Rightarrow 280 = 2T$

$ \Rightarrow T = 35^\circ {\text{C}}$

$\therefore $ the resultant temperature is obtained to be $T = 35^\circ {\text{C}}$ .

**Hence the correct option is D.**

**Note:**When liquid A of higher temperature and liquid B of a lower temperature are mixed together, the resultant temperature $T$ of the mixture will have a temperature less than the higher temperature of liquid A but higher than the lower temperature of liquid B i.e., ${T_A} > T > {T_B}$ . So while expressing the heat lost by liquid A, we take the temperature change to be ${T_A} - T$ in equation (1) and while expressing the heat gained by liquid B, we take the temperature change to be $T - {T_B}$ in equation (2).

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