Physics, asked by shaikthouficahmed, 10 months ago

A bar with a crack at its centre buckles as a result of temperature rise of 32 °C. If the fixed
distance Lo is 3.77 m and the coefficient of linear expansion of the bar is 25x10-6 °C'. The
rise x of the centre is
a) 5.5x102 m
b) 6.5x102 m
c) 7.5*102 m
d) 8.5x10-2 m​

Answers

Answered by abhi178
9

answer : option (c) 7.5 × 10¯² m

It has given that a bar with a crack at its centre buckles as a result of temperature rise of 32°C. if the fixed distance L0 is 3.77 m and the coefficient of linear expansion of the bar is 25 × 10^-6/°C.

we have to find the rise x of the centre.

BC = Lo/2 = 3.77/2 = 1.885 m

using formula, ∆l = (Lo/2)α∆T

= 1.885 × 25 × 10^-6 × 32

= 1.508 × 10¯³ m

AB = Lo/2 + ∆l

= 1.885 + 1.508 × 10¯³

= 1.885 + 0.001508 m

= 1.886508 m

now applying Pythagoras theorem,

AB² = BC² + x²

⇒(1.886508)² = (1.885)² + x²

⇒x² = (1.886508)² - (1.885)²

= 0.00568743406

⇒x = 0.07541 ≈ 7.5 × 10¯² m

therefore option (c) 7.5 × 10¯² m is correct choice.

Attachments:
Answered by sankhsr3
1

Answer:

c)7.5*10^-2 m

Explanation:

It has given that a bar with a crack at its centre buckles as a result of temperature rise of 32°C. if the fixed distance L0 is 3.77 m and the coefficient of linear expansion of the bar is 25 × 10^-6/°C.

we have to find the rise x of the centre.

BC = Lo/2 = 3.77/2 = 1.885 m

using formula, ∆l = (Lo/2)α∆T

= 1.885 × 25 × 10^-6 × 32

= 1.508 × 10¯³ m

AB = Lo/2 + ∆l

= 1.885 + 1.508 × 10¯³

= 1.885 + 0.001508 m

= 1.886508 m

now applying Pythagoras theorem,

AB² = BC² + x²

⇒(1.886508)² = (1.885)² + x²

⇒x² = (1.886508)² - (1.885)²

= 0.00568743406

⇒x = 0.07541 ≈ 7.5 × 10¯² m

therefore option (c) 7.5 × 10¯² m is correct answer.

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