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Grade: 10
        A rod of resistance 4 ohm is turned into half circle.What is the resistance along the diameter
11 months ago

## Answers : (2)

Arun
24739 Points
							 Let the resistance per unit length be $\dpi{80} \lambda$. Then, $\dpi{80} \lambda$ = 4/$\dpi{80} 2\pi r$=> resistance of the specimen wire connected along the diameter is:$\dpi{80} R = \lambda \times 2r = \frac{4}{\pi } \Omega$Now after the specimen wire is connected along the diameter, three resistance come in parallel:$\dpi{80} 2 \Omega$ (resistance of one half of the circumference),  $\dpi{80} 2 \Omega$ (resistance of the other half of the circumference) and R(resistance of the diameter). Solving, we get:$\dpi{80} {\color{DarkRed}R_{eq} = \frac{4}{4+\pi } \Omega }$

11 months ago
Rajat
213 Points
							Perimeter of the half circle is πr + 2r where r is radius of the circle.  Along the diameter the circle is divided into two unequal parts of length πr and 2r which are parallel to each other.Resistance of each part:1) 4*π/(π+2)2) 4*2/(π+2)Equivalent resistance is:R1*R2/(R1+R2)=[32π/(π+2)^2]/4=8π/(π+2)^2

11 months ago
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