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The density of 400cm\(^3\) of palm oil was 0.9gcm-3 before frying. If the density of...

Physics
JAMB 2018

The density of 400cm\(^3\) of palm oil was 0.9gcm-3 before frying. If the density of the oil was 0.6gcm-3 after frying, assuming no loss of oil due to spilling, its new volume was?

  • A. 1360cm\(^3\)
  • B. 600cm\(^3\)
  • C. 240cm\(^3\)
  • D. 8000m\(^3\)
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Correct Answer: Option B
Explanation

\(Density= \frac{mass}{volume}\)

⇒ \(0.9gcm^{-3} = \frac{mass}{400cm^{3}} = 360g\)

without any loss in the frying process, we have

\(0.6gcm^{-3} = \frac{360g}{v}\)

\(v = \frac{360g}{0.6gcm^{-3}} = 600cm^{3}\) 

There is an explanation video available below.


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Explanation Video

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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free