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MHT CET 2025 (25 Apr, Shift 1) · Reflection & Mirrors

Reflection & Mirrors question from MHT CET 2025 (25 Apr, Shift 1)

Q13MHT CET 2025 · 25 APR · SHIFT 1Difficult

Reflection & Mirrors

The image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 253 m\frac{25}{3}\text{ m} to 507 m\frac{50}{7}\text{ m} in 30 second. The speed of the object in km/hr\text{km/hr} is

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Answer: (d)

  1. Convex mirror: f=+10 mf = +10\text{ m}. Use 1u=1f−1v\frac{1}{u} = \frac{1}{f} - \frac{1}{v}.
  2. v=253v = \frac{25}{3}: 1u=110−325=−150⇒u=−50 m\frac{1}{u} = \frac{1}{10} - \frac{3}{25} = -\frac{1}{50} \Rightarrow u = -50\text{ m}
  3. v=507v = \frac{50}{7}: 1u=110−750=−125⇒u=−25 m\frac{1}{u} = \frac{1}{10} - \frac{7}{50} = -\frac{1}{25} \Rightarrow u = -25\text{ m}
  4. Speed =2530=56 m/s=56×185=3 km/hr= \frac{25}{30} = \frac{5}{6}\text{ m/s} = \frac{5}{6} \times \frac{18}{5} = 3\text{ km/hr}
  5. So the answer is (d) 3.
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