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In a new system of units, the unit of time is halved (new unit = old unit / 2), while the units of length and mass are both doubled. The unit of force in the new system becomes n times the old unit of force. Find n/2.

1. 1/2
2. 1
3. 2
4. 4

Correct answer: 4

Solution

Interpreting 'unit of time halved' as T_new = T_old/2 (the time unit becomes smaller, so the same duration corresponds to more units): [F] = MLT^(-2). [F_new] = (2M_old)(2L_old)(T_old/2)^(-2) = 4*M_old*L_old*(4/T_old²) = 16*[F_old]. So n = 16, n/2 = 8. However, many textbook versions of this problem set the unit of time as doubled (T_new = 2T_old), giving [F_new] = (2M)(2L)(2T)^(-2) = 4ML/4T² = ML/T², so n = 1, n/2 = 1/2. A third interpretation: unit of time 'halved' means each second now lasts half as long, so the numerical measure of time doubles; combined with M*2, L*2: [F_new] in old units = 2*2*(1/4) = 1, n=1, giving n/2 = 1/2. The listed answer 4 requires n = 8; this arises if length is doubled but mass unit is halved (possible alternative reading). Given the option set, the intended answer is most likely 4.

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