Discussion
An aircraft in a steady climb suddenly experiences a 10% drop in thrust. After a new equilibrium is reached at the same speed, the new rate of climb is
Answer: B
The new rate of climb is lower by more than 10%.
Rate of climb depends on excess power: \(R/C = (T - D)V/W\). At the same speed and weight, drag is unchanged, so only the thrust term moves.
A 10% drop in thrust removes \(0.1T\) from the excess thrust \(T - D\). Because \(T - D\) is smaller than \(T\), losing \(0.1T\) is more than 10% of the excess. For example, if \(D = 0.6T\), the excess falls from \(0.4T\) to \(0.3T\), a 25% drop in rate of climb.
'Exactly 10%' would only be true if there were no drag, and 'less than 10%' gets the direction of the effect wrong.
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