A sonometer wire having cross -sectional area 0.85 into 10power -6 m2 is streached between two rigid support 1.2m apart. a tension 20N applied at its free end If the temprature reduced by 12०C calculate final tension in the wire.take coffiecent of linear expansion( α ) and isothermal young 's modules( γ ) to be 1.5 into 10 power-5 per K and 2.0into 10power11Nm-2
Answers
Answered by
7
Y = (F / A) / (ΔL/L)
F = ΔL * YA/L
When the wire contracts due to lowering of temperature, it develops an additional tension in the wire that is proportional to ΔL = α L ΔT
So ΔF = α ΔT * Y A
= 1.5 * 10⁻⁵ * 12* 2.0 * 10¹¹ * 0.85 * 10⁻⁶ N
= 30.60 N
Tension total = 20 + 30.60 = 50.60 N
F = ΔL * YA/L
When the wire contracts due to lowering of temperature, it develops an additional tension in the wire that is proportional to ΔL = α L ΔT
So ΔF = α ΔT * Y A
= 1.5 * 10⁻⁵ * 12* 2.0 * 10¹¹ * 0.85 * 10⁻⁶ N
= 30.60 N
Tension total = 20 + 30.60 = 50.60 N
kvnmurty:
click on red heart thanks above pls
Answered by
4
young modulus (Y)= stress /strain
stress = F/A
strain =∆L/L
we know, according to concept of thermal expansion ,
∆L =La∆T
where a is coefficient of linear thermal expansion. ∆T is temperature change .
now,
Y =(∆F/A) ( L/∆L)
=(∆F/A)1/a∆T)
∆F =YA.a∆T
=2× 10¹¹ × 0.85×10^-6 × 1.5×10^-5×(12)
= -1.7 ×1.5 × 12 N
F -F° = 1.7×1.5×12N
F =(20 +1.7×1.5×12 ) N
=(30.6 + 20 ) N
=50.6 N
stress = F/A
strain =∆L/L
we know, according to concept of thermal expansion ,
∆L =La∆T
where a is coefficient of linear thermal expansion. ∆T is temperature change .
now,
Y =(∆F/A) ( L/∆L)
=(∆F/A)1/a∆T)
∆F =YA.a∆T
=2× 10¹¹ × 0.85×10^-6 × 1.5×10^-5×(12)
= -1.7 ×1.5 × 12 N
F -F° = 1.7×1.5×12N
F =(20 +1.7×1.5×12 ) N
=(30.6 + 20 ) N
=50.6 N
Similar questions