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
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.
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.