A ball weighs (4.00 ± 0.02) N in air and (1.00 ± 0.02) N in water. The maximum percentage error in its measured relative density is approximately P/6 percent. Find P.
Physics · Mixed Chapter
JEE Mains

Question

A ball weighs (4.00 ± 0.02) N in air and (1.00 ± 0.02) N in water. The maximum percentage error in its measured relative density is approximately P/6 percent. Find P.

G1 Physics · Accepted Answer

Step 1 — Relative density (neglecting air buoyancy) is S = Wₐ/(Wₐ − Wᵥ), where Wₐ = 4.00 N and Wᵥ = 1.00 N. Therefore S = 4/3. Step 2 — Compute derivatives of S with respect to both independently measured readings. For small maximum errors: δS/S = (Wᵥ δWₐ)/[Wₐ(Wₐ − Wᵥ)] + δWᵥ/(Wₐ − Wᵥ). Step 3 — Insert δWₐ = δWᵥ = 0.02 N: δS/S = (1×0.02)/(4×3) + 0.02/3 = 1/600 + 1/150 = 1/120. Step 4 — Percentage error is 100/120% = 5/6%. The question denotes it by P/6, so P = 5. Answer: 5.