Q147 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics

A cylindrical wire $P$ has resistance $10 \Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is

  • A
    $10 \Omega$
  • Correct answer
    B
    $20 \Omega$
  • C
    $5 \Omega$
  • D
    $\frac{5}{2} \Omega$

Solution

Given, length of wire, $P=l$ Length of wire, $Q=\frac{l}{2}$ Diameter of wire, $P=d$ Diameter of wire, $Q=\frac{d}{2}$ Resistance $R=\rho \cdot \frac{l}{A}=\rho \cdot \frac{l}{\pi d^2 / 4}$ $\therefore \frac{\text { Resistance of } Q}{\text { Resistance of } P}=\left(\frac{\rho \cdot \frac{l}{2}}{\frac{\pi}{4}\left(\frac{d}{2}\right)^2}\right) / \frac{\rho l}{\frac{\pi}{4} d^2}=\frac{\frac{1}{2}}{\frac{1}{4}}=\frac{4}{2}=\frac{2}{1}$ $\Rightarrow$ Resistance of Q = Resistance of P $\times 2$ $=10 \times 2=20 \Omega$

Hence, the correct answer is option: 'B'.

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