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Current Electricity question

2012 · Q175
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Current Electricity question

2012 · Q175

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A ring is made of a wire having a resistance R0 = 12 Ω\OmegaΩ. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub circuit between these point is equal to 83Ω{8 \over 3}\Omega38​Ω.
AIPMT 2012 Prelims Physics - Current Electricity Question 88 English
  1. A
    l1l2=58{{{l_1}} \over {{l_2}}} = {5 \over 8}l2​l1​​=85​
  2. B
    l1l2=13{{{l_1}} \over {{l_2}}} = {1 \over 3}l2​l1​​=31​
  3. C
    l1l2=38{{{l_1}} \over {{l_2}}} = {3 \over 8}l2​l1​​=83​
  4. D
    l1l2=12{{{l_1}} \over {{l_2}}} = {1 \over 2}l2​l1​​=21​
View written solutionFree

Correct answer: D

Let x is the resistance per unit length then

AIPMT 2012 Prelims Physics - Current Electricity Question 88 English Explanation

equivalent resistance R=R1R2R1+R2R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}R=R1​+R2​R1​R2​​

(x1l1)(x2l2)xl1+xl2{{\left( {{x_1}{l_1}} \right)\left( {{x_2}{l_2}} \right)} \over {x{l_1} + x{l_2}}}xl1​+xl2​(x1​l1​)(x2​l2​)​

⇒83=xl1l1l2+1 \Rightarrow {8 \over 3} = x{{{l_1}} \over {{{{l_1}} \over {{l_2}}} + 1}}⇒38​=xl2​l1​​+1l1​​   ...(i)

also R0=xl1+xl2R_0 = xl_1 + xl_2R0​=xl1​+xl2​
12=x(l1+l2)12 = x(l_1 + l_2)12=x(l1​+l2​)

12=xl2(l1l2+1)12 = x{l_2}\left( {{{{l_1}} \over {{l_2}}} + 1} \right)12=xl2​(l2​l1​​+1)   ...(ii)

Now
(i)(ii)⇒83121=xl1(l1l2+1)xl2(l1l2+1){{\left( i \right)} \over {\left( {ii} \right)}} \Rightarrow {{{8 \over 3}} \over {{{12} \over 1}}} = {{{{x{l_1}} \over {\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}}} \over {x{l_2}\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}}(ii)(i)​⇒112​38​​=xl2​(l2​l1​​+1)(l2​l1​​+1)xl1​​​

=l1l2(l1l2+1)2 = {{{l_1}} \over {{l_2}{{\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}^2}}}=l2​(l2​l1​​+1)2l1​​

(l1l2+1)2×836=l1l2{\left( {{{{l_1}} \over {{l_2}}} + 1} \right)^2} \times {8 \over {36}} = {{{l_1}} \over {{l_2}}}(l2​l1​​+1)2×368​=l2​l1​​

(y2+1+2y)×836=y\left( {{y^2} + 1 + 2y} \right) \times {8 \over {36}} = y(y2+1+2y)×368​=y   (where y = l1l2{{{{l_1}} \over {{l_2}}}}l2​l1​​)

8y2 + 8 + 16y = 36y

⇒\Rightarrow⇒ 8y2 – 20y + 8 = 0

⇒\Rightarrow⇒ 2y2 – 5y + 2 = 0

⇒\Rightarrow⇒ 2y2 – 4y – y + 2 = 0

⇒\Rightarrow⇒ 2y (y – 2) – 1(y – 2) = 0

⇒\Rightarrow⇒ (2y – 1) (y – 2) = 0

⇒\Rightarrow⇒ y=l1l2=12y = {{{l_1}} \over {{l_2}}} = {1 \over 2}y=l2​l1​​=21​ or 2

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