JEE Main 2026 Physics question
The density ρ of a uniform cylinder is determined by measuring its mass m, length l and diameter d. The measured values of m, l and d are 97.42±0.02 g, 8.35±0.05 mm and 20.20±0.02 mm, respectively. Calculated percentage fractional error in ρ is .
Answer: (B)
Solution
ρ=4m/(πd²l), so Δρ/ρ=Δm/m+2Δd/d+Δl/l=0.02/97.42+2(0.02/20.20)+0.05/8.35=0.000205+0.001980+0.005988≈0.0082, i.e. 0.82%.