Properties of Matter 1 Question 21

21. A thin rod of negligible mass and area of cross-section 4×106m2, suspended vertically from one end, has a length of 0.5m at 100C. The rod is cooled to 0C, but prevented from contracting by attaching a mass at the lower end. Find

(a) this mass and

(b) the energy stored in the rod, given for the rod. Young’s modulus =1011N/m2, coefficient of linear expansion =105K1 and g=10m/s2.

(1997 C, 5M)

Show Answer

Solution:

  1. (a) The change in length due to decrease in temperature,

Δl1=LαΔθ=(0.5)(105)(0100)Δl1=0.5×103m

Negative sign implies that length is decreasing. Now, let M be the mass attached to the lower end. Then, change in length due to suspension of load is

Δl2=(Mg)LAY=(M)(10)(0.5)(4×106)(1011)Δl2=(1.25×105)MΔl1+Δl2=0 or (1.25×105)M=(0.5×103)M=(0.5×1031.25×105)kg or M=40kg

(b) Energy stored,

At 0C the natural length of the wire is less than its actual length; but since a mass is attached at its lower end, an elastic potential energy is stored in it. This is given by

U=12(AYL)(Δl)2

Here, Δl=|Δl1|=Δl2=0.5×103m

Substituting the values,

U=12(4×106×10110.5)(0.5×103)2=0.1J

NOTE

Comparing the equation

Y=F/AΔ//L or F=(AYL)Δl

with the spring equation F=KΔx, we find that equivalent spring constant of a wire is k=(AYL)

Therefore, potential energy stored in it should be

U=12k(ΔI)2=12(AYL)(ΔI)2



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक