A cup of coffee cools from \(90^{\circ}\text{C}\) $\mathrm{to}$ \(80^{\circ}\text{C}\) in \(t\) minutes, when the room temperature is \(20^{\circ}\text{C}\)$\mathrm{}$. The time taken by a similar cup of coffee to cool from \(80^{\circ}\text{C}\) $\mathrm{to}$ \(60^{\circ}\text{C}\) at room temperature same at \(20^{\circ}\text{C}\) is:
1. \(\dfrac{10}{13}t\) 
2. \(\dfrac{5}{13}t\)
3. \(\dfrac{13}{10}t\)
4. \(\dfrac{13}{5}t\)

An object kept in a large room having an air temperature of \(25^\circ \text{C}\) takes \(12\) minutes to cool from \(80^\circ \text{C}\) to \(70^\circ \text{C}.\) The time taken to cool for the same object from \(70^\circ \text{C}\) to \(60^\circ \text{C}\) would be nearly:
1. \(10\) min
2. \(12\) min
3. \(20\) min
4. \(15\) min

A body cools from a temperature of \(3T\) to \(2T\) in \(10\) minutes. The room temperature is \(T\). Assuming that Newton's law of cooling is applicable, the temperature of the body at the end of the next \(10\) minutes will be:

A certain quantity of water cools from \(70~^{\circ}\text{C}\) to \(60~^{\circ}\text{C}\) in the first \(5\) minutes and to \(54~^{\circ}\text{C}\) in the next \(5\) minutes. The temperature of the surroundings is:
1. \(45~^{\circ}\text{C}\)
2. \(20~^{\circ}\text{C}\)
3. \(42~^{\circ}\text{C}\)
4. \(10~^{\circ}\text{C}\)