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A Metal Ball Cools from 64 °C to 50 °C in 10 Minutes and to 42 °C in Next 10 Minutes. the Ratio of Rates of Fall of Temperature During the Two Intervals is _______. - Physics

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Question

A metal ball cools from 64 °C to 50 °C in 10 minutes and to 42 °C in next 10 minutes. The ratio of rates of fall of temperature during the two intervals is _______.

Options
  • `4/7`

  • `7/4`

  • 2

  • 2.5

Solution

`7/4`

Rate of fall of temperature in first 10 minutes = R1 = `(64 - 50)/10`.°C/min = 1.4°C/min

Rate of fall of temperature in next 10 minutes = R2 = `(50 - 42)/10`.°C/min = 0.8°C/min

⇒ `R_1/R_2 = 7/4`

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Solution A Metal Ball Cools from 64 °C to 50 °C in 10 Minutes and to 42 °C in Next 10 Minutes. the Ratio of Rates of Fall of Temperature During the Two Intervals is _______. Concept: Temperature and Heat.
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