A steel wire of cross sectional area 2 mm2 is just stretched horizontally between two fixed points at a temperature of 35∘C. The temperature of the wire changed to 25∘C. Then (Coefficient of linear expansion of steel =11×10−6/0C;Ysteel=2×1011Nm−2)

F=YA α Δt=11×10−6×2×10−6×2×1011×10=44NFA=2×1011×11×10−6×10=22×106=2.2×107 N/m2Δ ll=α Δt=11×10−6×10=11×10−5

A steel wire of cross sectional area 2 mm2 is just stretched horizontally between two fixed points at a temperature of 35∘C. The temperature of the wire changed to 25∘C. Then (Coefficient of linear expansion of steel =11×10−6/0C;Ysteel=2×1011Nm−2)

F=YA α Δt=11×10−6×2×10−6×2×1011×10=44NFA=2×1011×11×10−6×10=22×106=2.2×107 N/m2Δ ll=α Δt=11×10−6×10=11×10−5

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 steel wire of cross sectional area $2 {mm}^{2}$ is just stretched horizontally between two fixed points at a temperature of $35^{\circ} C$ . The temperature of the wire changed to $25^{\circ} C .$ Then (Coefficient of linear expansion of steel $= 11 \times 10^{- 6} /^{0} C; Y_{steel} = 2 \times 10^{11} {Nm}^{- 2}$ )

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

the tension in the wire at temperature $25^{\circ} C$ is 44N

the thermal stress is $2.2 \times 10^{7} N / m^{2}$

the thermal strain is $11 \times 10^{- 5}$

all the above

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$F = Y A α Δ t$

$= 11 \times 10^{- 6} \times 2 \times 10^{- 6} \times 2 \times 10^{11} \times 10 = 44 N$

$\frac{F}{A} = 2 \times 10^{11} \times 11 \times 10^{- 6} \times 10$

$= 22 \times 10^{6} = 2.2 \times 10^{7} N / m^{2}$

$\frac{Δ l}{l} = α Δ t = 11 \times 10^{- 6} \times 10 = 11 \times 10^{- 5}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

A steel wire of cross sectional area 2 mm2 is just stretched horizontally between two fixed points at a temperature of 35∘C. The temperature of the wire changed to 25∘C. Then (Coefficient of linear expansion of steel =11×10−6/0C;Ysteel=2×1011Nm−2)

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App