Two intervals of time are measured as Δt1=(2±0.02)S and Δt2=(4±0.02)S. The value of (Δt1)(Δt2) with correct significant figures and error is

T=(Δt1)(Δt2)=(2)(4)=8=2.828 = 2.83 ±ΔTT=12Δt1t1+12Δt2t2 ΔT=(120.022+120.024)2.83 ΔT=0.02121So after rounding off ΔT=0.02 T=(2.83±0.02)

Two intervals of time are measured as Δt1=(2±0.02)S and Δt2=(4±0.02)S. The value of (Δt1)(Δt2) with correct significant figures and error is

T=(Δt1)(Δt2)=(2)(4)=8=2.828 = 2.83 ±ΔTT=12Δt1t1+12Δt2t2 ΔT=(120.022+120.024)2.83 ΔT=0.02121So after rounding off ΔT=0.02 T=(2.83±0.02)

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

Two intervals of time are measured as $Δ t_{1} = (2 \pm 0.02) S$ and $Δ t_{2} = (4 \pm 0.02) S .$ The value of $\sqrt{(Δ t_{1}) (Δ t_{2})}$ with correct significant figures and error is

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

$(2.828 \pm 0.0075) S$

$(2.83 \pm 0.0075) S$

$(2.828 \pm 0.01) S$

$(2.83 \pm 0.02) S$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$T = \sqrt{(Δ t_{1}) (Δ t_{2})} = \sqrt{(2) (4)} = \sqrt{8} = 2.828 = 2.83$
$\pm \frac{Δ T}{T} = \frac{1}{2} \frac{Δ t_{1}}{t_{1}} + \frac{1}{2} \frac{Δ t_{2}}{t_{2}}$
$Δ T = (\frac{1}{2} \frac{0.02}{2} + \frac{1}{2} \frac{0.02}{4}) 2.83$
$Δ T = 0.02121$
So after rounding off $Δ T = 0.02$
$T = (2.83 \pm 0.02)$