A letter 'A' is constructed of a uniform wire with resistance 1.0$$\Omega$$ per cm. The sides of the letter are 20 cm and the cross piece in the middle is 10 cm long. The apex angle is 60°. The resistance between the ends of the legs is close to:

The letter 'A' is made of uniform wire with a resistance of 1.0 Ω per cm. The sides (legs) are each 20 cm long, and the cross piece is 10 cm long. The apex angle is 60°. We need to find the resistance between the ends of the legs, labeled as points A and B.

First, visualize the structure. The apex angle at the top is 60°, and the legs CA and CB are both 20 cm. The cross piece is horizontal and connects the midpoints of the legs. Let D be the midpoint of CA and E be the midpoint of CB. Since the legs are 20 cm, CD = DA = CE = EB = 10 cm. The cross piece DE is 10 cm long. Given the apex angle of 60° and CD = CE = 10 cm, triangle CDE is equilateral, so DE = 10 cm, which matches.

The circuit has the following segments with resistances (since resistance per cm is 1 Ω):

• AD (from A to D): 10 cm → 10 Ω
• DC (from D to C): 10 cm → 10 Ω
• CE (from C to E): 10 cm → 10 Ω
• EB (from E to B): 10 cm → 10 Ω
• DE (from D to E): 10 cm → 10 Ω

The circuit between A and B can be represented as:

• A is connected to D by 10 Ω.
• D is connected to E by 10 Ω (DE) and to C by 10 Ω (DC).
• E is connected to B by 10 Ω and to C by 10 Ω (CE).

To find the equivalent resistance between A and B, note that between D and E, there are two parallel paths:

1. Direct path DE: 10 Ω
2. Path via C: DC + CE = 10 Ω + 10 Ω = 20 Ω

The equivalent resistance between D and E is the parallel combination:

$$ \frac{1}{R_{DE}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} $$

$$ R_{DE} = \frac{20}{3} \Omega \approx 6.6667 \Omega $$

Now, the total path from A to B is: A to D (10 Ω), then D to E (20/3 Ω), then E to B (10 Ω). These are in series, so:

$$ R_{AB} = R_{AD} + R_{DE} + R_{EB} = 10 + \frac{20}{3} + 10 = 20 + \frac{20}{3} = \frac{60}{3} + \frac{20}{3} = \frac{80}{3} \Omega \approx 26.6667 \Omega $$

Comparing with the options:

• A. 50.0 Ω
• B. 10 Ω
• C. 36.7 Ω
• D. 26.7 Ω

The value 26.6667 Ω is closest to 26.7 Ω.

Hence, the correct answer is Option D.