Question

If coefficient of real expansion of a liquid is 1/5500°c.the temperature at which its density is 1%less than density at 0°c is?

Answer 1

Answer:

The  coefficient of real expansion of a liquid is 1/5500°c.

Explanation:

pt= p0/1+y Δt

=> p0= pt+ pt y Δt

=> (pt-p0)/pt= -yΔ t

given pt= 0.99p0    (p0 density at 0 )

(0.00p0-p0)/p0= -y (t-0)

=> -y t= - 0.01/0.99

t= 1/99y

Answer 2

Answer:

To get the density we need that density at origin is do and density at temperature t will be dt.

Thus, we know the formulae that (dt - do)/dt = -ΥΔt.

Since the temperature at which given density will be 1% less than the density at 0 degree will be.

Now, dt = 0.99do thus, (0.99do - do)/0.99do = -Υ(t - 0).

Thus, on solving we will get 0.1/0.99 = Υt.

So, the value of t will be 1/99Υ.