Q110 / 150 · MHT CET 2026 (15 April Shift 1) · Physics

When a resistance of $200\,\Omega$ is connected in series with a galvanometer of resistance G, its range is V. To triple its range, a resistance of $2000\,\Omega$ is connected in series. The value of G is

  • A
    $400\,\Omega$
  • B
    $500\,\Omega$
  • Correct answer
    C
    $700\,\Omega$
  • D
    $900\,\Omega$

Solution

Let $I_g$ be the current for full scale deflection of the galvanometer.

When a resistance of $200\,\Omega$ is connected in series, the range is $V$.

$V = I_g(G + 200)$

When a resistance of $2000\,\Omega$ is connected in series, the range becomes $3V$.

$3V = I_g(G + 2000)$

Dividing the second equation by the first equation:

$\dfrac{3V}{V} = \dfrac{I_g(G + 2000)}{I_g(G + 200)}$

$3 = \dfrac{G + 2000}{G + 200}$

$3G + 600 = G + 2000$

$2G = 1400$

$G = 700\,\Omega$

Answer: $700\,\Omega$

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

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