Comprehension Passage:

The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWERmodetoreduce the temperature.

If the AC operates in REGULAR mode, then it brings down the temperature inside the room (called inside temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWERmode, then this is achieved in 30 minutes.

If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature outside at the time of switching off in 1 hour.

The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night. The following graph shows the inside temperature between 11 pm (23:00) and 2 am (2:00) that night.

The following facts are known about the AC operation that night:
• The ACwas turned on for the first time that night at 11 pm (23:00).
• The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the beginning of the hour or at 30 minutes after the hour.
• The ACwasusedinPOWERmodeforlongerduration than in REGULARmodeduring this 3-hour period.

What was the maximum difference between temperature outside and inside temperature, in degree Celsius, between 11:01 pm and 1:59 am?

The correct answer/option is 10.

To find the maximum difference between the temperature outside ($T_{\text{out}}$) and inside temperature ($T_{\text{in}}$) of the room between 11:01 pm and 1:59 am, we analyze the rates of temperature change described in the problem.

1. Inside Temperature Variation when the AC is Switched Off:
When the AC is switched off at a time $t$, the inside temperature rises at a constant rate such that it would reach the outside temperature at the time of switching off ($T_{\text{out}}\left(t\right)$) in exactly 1 hour.
From the graph, we can observe that the inside temperature rises linearly during two intervals:

• Between 0:00 (12:00 am) and 0:30 (12:30 am): rising from $26{°}^C$ to $31{°}^C$.
• Between 1:00 (1:00 am) and 1:30 (1:30 am): rising from $26{°}^C$ to $30{°}^C$.

This indicates that the AC was switched off at 0:00 and at 1:00.

Using the rate of rise:
For the interval 0:00 to 0:30 (duration of 30 minutes, or 0.5 hours):
$T_{\text{in}}\left(\mathrm{0:30}\right)-T_{\text{in}}\left(\mathrm{0:00}\right)=0.5\times \left(T_{\text{out}}\right(\mathrm{0:00})-T_{\text{in}}(\mathrm{0:00}\left)\right)$
Substituting the values from the graph:
$31-26=0.5\times \left(T_{\text{out}}\right(\mathrm{0:00})-26)$
$5=0.5\times \left(T_{\text{out}}\right(\mathrm{0:00})-26)$
$T_{\text{out}}\left(\mathrm{0:00}\right)=36{°}^C$

Similarly, for the interval 1:00 to 1:30:
$T_{\text{in}}\left(\mathrm{1:30}\right)-T_{\text{in}}\left(\mathrm{1:00}\right)=0.5\times \left(T_{\text{out}}\right(\mathrm{1:00})-T_{\text{in}}(\mathrm{1:00}\left)\right)$
Substituting the values from the graph:
$30-26=0.5\times \left(T_{\text{out}}\right(\mathrm{1:00})-26)$
$4=0.5\times \left(T_{\text{out}}\right(\mathrm{1:00})-26)$
$T_{\text{out}}\left(\mathrm{1:00}\right)=34{°}^C$

2. Outside Temperature Trend:
Since the outside temperature falls at a constant rate from 7 pm onward:
Rate of fall of outside temperature = $\frac{36{°}^C-34{°}^C}{\text{1 hour}}=2{°}^C/\text{hour}$
This yields the outside temperature at the boundary half-hour intervals:

• At 11:00 pm (23:00): $T_{\text{out}}=38{°}^C$
• At 11:30 pm (23:30): $T_{\text{out}}=37{°}^C$
• At 12:00 am (0:00): $T_{\text{out}}=36{°}^C$
• At 12:30 am (0:30): $T_{\text{out}}=35{°}^C$
• At 1:00 am (1:00): $T_{\text{out}}=34{°}^C$
• At 1:30 am (1:30): $T_{\text{out}}=33{°}^C$
• At 2:00 am (2:00): $T_{\text{out}}=32{°}^C$

3. Calculating the Temperature Difference ($T_{\text{out}}-T_{\text{in}}$):
Let us calculate the difference at each interval boundary point:

• At 11:00 pm (23:00): Difference = $38-38=0{°}^C$
• At 12:00 am (0:00): Difference = $36-26=10{°}^C$
• At 12:30 am (0:30): Difference = $35-31=4{°}^C$
• At 1:00 am (1:00): Difference = $34-26=8{°}^C$
• At 1:30 am (1:30): Difference = $33-30=3{°}^C$
• At 2:00 am (2:00): Difference = $32-28=4{°}^C$

Since all changes within the half-hour blocks are linear, the maximum difference occurs at one of these key milestones. The maximum difference is achieved at 12:00 am (0:00) with a value of 10.