IMU Question 4 The thrust \( T \) of a propeller varies jointly as the square of the number \( n \) of revolutions per second and the fourth power of its diameter \( d \). What is the effect on \( T \) if \( n \) is doubled and \( d \) is halved? ( 3 marks)

To solve this problem, we start with the relationship given by the formula for thrust \( T \): \[ T = k \cdot n^2 \cdot d^4 \] where \( k \) is a constant of proportionality. Now, if we double the number of revolutions per second \( n \) (so it becomes \( 2n \)) and halve the diameter \( d \) (so it becomes \( \frac{d}{2} \)), we can substitute these values into the thrust equation: \[ T' = k \cdot (2n)^2 \cdot \left(\frac{d}{2}\right)^4 \] Calculating this gives: \[ T' = k \cdot 4n^2 \cdot \frac{d^4}{16} \] Simplifying this, we have: \[ T' = k \cdot \frac{4}{16} \cdot n^2 \cdot d^4 = k \cdot \frac{1}{4} \cdot n^2 \cdot d^4 \] Thus, we find: \[ T' = \frac{1}{4} T \] This means that if \( n \) is doubled and \( d \) is halved, the thrust \( T' \) becomes one-quarter of the original thrust \( T \). So the effect on thrust is a reduction to just 25% of its original value.