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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Current Electricity

NEET 2013 · Physics · Current Electricity

A 12 ~cm wire is given a shape of a right angled triangle A B C...

A 12 cm12 \mathrm{~cm} wire is given a shape of a right angled triangle ABCA B C having sides 3 cm3 \mathrm{~cm}, 4 cm4 \mathrm{~cm} and 5 cm5 \mathrm{~cm} as shown in the figure. The resistance between two ends (AB, BC,CA)B C, C A) of the respective sides are measured one by one by a multi-meter. The resistance will be in the ratio: A 12 ~cm wire is given a shape of a right angled triangle A B C having sides 3 ~cm , 4 ~cm and 5 ~cm as shown in the

Options

  1. A9:16:259: 16: 25
  2. B27:32:3527: 32: 35
  3. C21:24:2521: 24: 25
  4. D3:4:53: 4: 5

Correct answer

B. 27:32:3527: 32: 35

Step-by-step solution

Resistance of side AB=R1=3ρAA B=R_1=\frac{3 \rho}{A} Resistance of side BC=R2=4ρAB C=R_2=\frac{4 \rho}{A} Resistance of side AC=P3=5ρA\mathrm{AC}=\mathrm{P}_3=\frac{5 \rho}{\mathrm{A}} (Where A=A= Area of cross-section and ρ=\rho= resistivity of wire) A 12 ~cm wire is given a shape of a right angled triangle A B C having sides 3 ~cm , 4 ~cm and 5 ~cm as shown in the
∴\therefore The resistance between AandB\mathrm{A} and \mathrm{B} isRAB=R1(R2+R3)R1+R2+R3=3ρA(4ρA+5ρA)3ρA+4ρA+5ρA=27ρ12A Similarly RBC=R2(R1+R3)R1+R2+R34ρA(3ρA+5ρA)3ρA+4ρA+5ρA=3212ρA And  RAC=R3(R1+R2)R1+R2+R3=5ρA(3ρA+3ρA)5ρA+3ρA+4ρA=3512ρA∴RAB:RBC:RAC=2712:3212:3512=27:32:35\begin{aligned}& \mathrm{R}_{\mathrm{AB}}=\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}=\frac{\frac{3 \rho}{A}\left(\frac{4 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}}\\ & =\frac{27 \rho}{12 A}\\ & \text{ Similarly }R_{B C}=\frac{R_2\left(R_1+R_3\right)}{R_1+R_2+R_3}\\ & \frac{\frac{4 \rho}{A}\left(\frac{3 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}}=\frac{32}{12} \frac{\rho}{A}\\ & \text{\text{ And } }R_{A C}=\frac{R_3\left(R_1+R_2\right)}{R_1+R_2+R_3}\\ & =\frac{\frac{5 \rho}{A}\left(\frac{3 \rho}{A}+\frac{3 \rho}{A}\right)}{\frac{5 \rho}{A}+\frac{3 \rho}{A}+\frac{4 \rho}{A}}=\frac{35}{12} \frac{\rho}{A}\\ & \therefore R_{A B}: R_{B C}: R_{A C}=\frac{27}{12}: \frac{32}{12}: \frac{35}{12}\\ & =27: 32: 35 \\ & \end{aligned}
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