A letter'A' is constructed of a uniform wire with resistance 1.0 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

Correct answer: D. 26.7 For ADE 1 R^ =(1)/(2 x)+(1)/(10) or R^ =(20 x)/(10+2 x)R_ BC = 20 x 10+2 x +20-x+20-x (i) or 20 x 10+2 x +40=2 x Solving we get x=10 Putting the value of x=10 in equation (i) We get R_ BC =(20 × 10)/(10+2 × 10)+20-10+20-10=(80)/(3)=26.7

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Physics
Chapter: Current Electricity

BITSAT 2012 · Physics · Current Electricity

A letter'A' is constructed of a uniform wire with resistance 1.0...

A letter'A' is constructed of a uniform wire with resistance

1.0 \Omega

per

\mathrm{cm}

, The sides of the letter are

20 \mathrm{ cm}

and the cross piece in the middle is 10

\mathrm{cm}

long. The apex angle is

60.

The resistance between the ends of the legs is close to: A letter'A' is constructed of a uniform wire with resistance 1.0 per cm , The sides of the letter are 20 cm and the

A letter'A' is constructed of a uniform wire with resistance 1.0 per cm , The sides of the letter are 20 cm and the

Options

A $50.0 \Omega$
B $10 \Omega$
C $36.7 \Omega$
D $26.7 \Omega$

Correct answer

D. $26.7 \Omega$

Step-by-step solution

For ADE

\frac{1}{R^{\prime}}=\frac{1}{2 x}+\frac{1}{10} \quad

\quad R^{\prime}=\frac{20 x}{10+2 x}\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} \quad \ldots

(i) or

\quad \frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}

Solving we get

\mathrm{x}=10 \Omega

Putting the value of

x=10 \Omega

in equation (i) We get

\mathrm{R}_{\mathrm{BC}}=\frac{20 \times 10}{10+2 \times 10}+20-10+20-10=\frac{80}{3}=26.7 \Omega

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