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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Current Electricity

BITSAT 2024 · Paper 2 · Physics · Current Electricity

A wire of resistance 160 is melted and drawn in wire of one-fourth...

A wire of resistance 160Ω160 \Omega is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

Options

  1. A10Ω10 \Omega
  2. B640Ω640 \Omega
  3. C40Ω40 \Omega
  4. D16Ω16 \Omega

Correct answer

A. 10Ω10 \Omega

Step-by-step solution

Let Initial length =l1=l_1 Final length =l2=l_2 Initial area =A1=\mathrm{A}_1 Final area =A1∴=A_1\therefore Volume remains same ∴A1l1=A2l2⇒A1l1=A2l14⇒4A1=A2\therefore \mathrm{A}_1 \mathrm{l}_1=\mathrm{A}_2 \mathrm{l}_2 \Rightarrow \mathrm{ A}_1 \mathrm{l}_1=\mathrm{A}_2 \frac{\mathrm{l}_1}{4}\Rightarrow 4 \mathrm{ A}_1=\mathrm{A}_2 Initial resistance, R1=ρl1A1=160Ω\mathrm{R}_1=\frac{\rho l_1}{A_1}=160 \Omega (given) Final resistance, R2=ρl2A2∴R2R1=l2A1A2l1=l14A14A1l1⇒R2=116R1=116×160=10Ω\mathrm{R}_2=\frac{\mathrm{\rho l}_2}{\mathrm{ A}_2}\therefore \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{\mathrm{l}_2 \mathrm{ A}_1}{\mathrm{ A}_2 \mathrm{l}_1}=\frac{l_1}{4} \frac{\mathrm{ A}_1}{4 \mathrm{ A}_1 l_1}\Rightarrow R_2=\frac{1}{16} R_1=\frac{1}{16} \times 160=10 \Omega
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