A sliver wire has mass (0.6±0.006)g, radius (0.5±0.005)mm and length (4±0.04)cm. The maximum percentage error in the measurement of its density will be :

M=(0.6±0.006)g r=(0.5±0.005)mm l=(4±0.04)cm ρ=mV ⇒Δρρ=Δmm+2Δrr+Δll (Volume of cylinder =πr2l =0.0060.6+2×0.0050.5+0.044 100×Δρρ=4×10-2×100 Δρρ×100=4%

A sliver wire has mass (0.6±0.006)g, radius (0.5±0.005)mm and length (4±0.04)cm. The maximum percentage error in the measurement of its density will be :

M=(0.6±0.006)g r=(0.5±0.005)mm l=(4±0.04)cm ρ=mV ⇒Δρρ=Δmm+2Δrr+Δll (Volume of cylinder =πr2l =0.0060.6+2×0.0050.5+0.044 100×Δρρ=4×10-2×100 Δρρ×100=4%

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A sliver wire has mass $(0.6 \pm 0.006) g$ , radius $(0.5 \pm 0.005) mm$ and length $(4 \pm 0.04) cm$ . The maximum percentage error in the measurement of its density will be :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

4 %

3 %

6 %

7 %

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$M = (0.6 \pm 0.006) g r = (0.5 \pm 0.005) mm l = (4 \pm 0.04) cm ρ = \frac{m}{V} \Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ m}{m} + \frac{2 Δ r}{r} + \frac{Δ l}{l} (Volume of cylinder = π r^{2} l) = \frac{0.006}{0.6} + \frac{2 \times 0.005}{0.5} + \frac{0.04}{4} 100 \times \frac{Δ ρ}{ρ} = 4 \times 10^{- 2} \times 100 \frac{Δ ρ}{ρ} \times 100 = 4 %$