A uniform metallic wire of length L is mounte | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b)

Let $a=$ side length of equilateral triangle, $r=$ radius of circle and $x=$ resistance per unit length of wire used. Then, $L=3 a=2 \pi r$ or

Vedclass · Accepted Answer

$9 / 8$

Vedclass · Answer

(b)

Let $a=$ side length of equilateral triangle, $r=$ radius of circle and $x=$ resistance per unit length of wire used. Then, $L=3 a=2 \pi r$ or $a=\frac{L}{3}$ and $r=\frac{L}{2 \pi}$ Now,

Equivalent resistance across $A B$ is

$R_{A B}=(a x \| 2 a x)=\frac{a x 	imes 2 a x}{a x+2 a x}$

$=\frac{2 a^2 x^2}{3 a x}=\frac{2}{3} a x$

$\Rightarrow \quad R_{A B}=\frac{2}{3} 	imes \frac{L}{3} 	imes x$

Power dissipated is

$P_1=\frac{V^2}{R_{A B}}=\frac{9 V^2}{2 L x} \quad \dots(i)$

$R_{P Q}=(\pi r x \| \pi r x)=\frac{\pi r x 	imes \pi r x}{\pi r x+\pi r x}=\frac{\pi^2 r^2 x^2}{2 \pi r x}$

$=\frac{1}{2} \pi r x=\frac{1}{2} \pi 	imes \frac{L}{2 \pi} x=\frac{L x}{4}$

So, power dissipated is

$P_2=\frac{V^2}{R_{P Q}}=\frac{4 V^2}{L x}$

Ratio of power dissipated in two cases is

$\frac{P_1}{P_2}=\frac{9 V^2 / 2 L x}{4 V^2 / L x}=\frac{9}{8}$

Similar Questions

An electric bulb is rated $220\, volt$ and $100\, watt$. Power consumed by it when operated on $110\, volt$ is ............. $watt$

Of the two bulbs in a house hold circuit, one glows brighter than the other, Which of the two bulbs has a large resistance?

Which of the following is not a correct statement

The heat produced by a $100\, watt$ heater in $2$ minute will be equal to

A battery is charged at a potential of $15\, V$ for $8$ hours when the current flowing is $10\, A$. The battery on discharge supplies a current of $5\, A$ for $15$ hours. The mean terminal voltage during discharge is $14\, V$. The "Watt-hour" efficiency of the battery is .............. $\%$