Three taps A, B, C fill up a tank independent | vedclass

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(A) The rates of filling for taps $A, B, C$ are $\frac{1}{10}, \frac{1}{20}, \frac{1}{30}$ of the tank per hour,respectively.Let the pairs be $(A, B),

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The rates of filling for taps $A, B, C$ are $\frac{1}{10}, \frac{1}{20}, \frac{1}{30}$ of the tank per hour,respectively.Let the pairs be $(A, B), (B, C), (C, A)$. The combined rates are:$(A+B) = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} = 0.15$$(B+C) = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} \approx 0.0833$$(C+A) = \frac{1}{30} + \frac{1}{10} = \frac{4}{30} = \frac{2}{15} \approx 0.1333$To minimize time,we use the fastest pairs as much as possible. The rates in descending order are $(A+B) > (C+A) > (B+C)$.In $3 \, h$,using each pair once,the tank filled is $\frac{3}{20} + \frac{2}{15} + \frac{1}{12} = \frac{9+8+5}{60} = \frac{22}{60} = \frac{11}{30}$.Remaining part is $1 - \frac{11}{30} = \frac{19}{30}$.To fill the remaining $\frac{19}{30}$ as fast as possible,we use the fastest pair $(A+B)$ with rate $\frac{3}{20}$.Time required = $\frac{19/30}{3/20} = \frac{19}{30} 	imes \frac{20}{3} = \frac{38}{9} \approx 4.22 \, h$.Total time = $3 + 4.22 = 7.22 \, h$. Since we must use each pair for at least one hour,and the question implies integer hours,we check the options. Given the options,$8 \, h$ is the smallest integer satisfying the condition.

Process	Condition
$(I)$ Adiabatic	$(A) \Delta W = 0$
$(II)$ Isothermal	$(B) \Delta Q = 0$
$(III)$ Isochoric	$(C) \Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D) \Delta U = 0$

Three taps $A, B, C$ fill up a tank independently in $10 \, h$,$20 \, h$,and $30 \, h$,respectively. Initially,the tank is empty. Exactly one pair of taps is open during each hour,and every pair of taps is open for at least one hour. What is the minimum number of hours required to fill the tank?

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