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Sever21 [200]
3 years ago
15

In the long run, if the firm decides to keep output at its initial level, what will it likely do? Stay on SRATC3 but decrease to

the point touching LRATC Shift to operate on SRATC4 Shut down Shift to operate on SRATC2 True or False: The minimum efficient scale is achieved at the minimum point on each average total cost curve. False True
Engineering
1 answer:
romanna [79]3 years ago
8 0

Answer:

Stay on SRATC3 but decrease to the point touching LRATC.

False

Explanation:

The long run average total cost is the total cost for the firm to continue its operations. In the given scenario the firm decides to keep its level of out =put at initial level then it should stay in short run average total cost and then gradually moving towards long run average total cost.

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Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

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