Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Bro sorry I don’t know these answer but keeep trying
Answer:
a. 05cm from x axis
b. 8cm from x axis
Explanation:
If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below
a. Given that i1= 5A and i2=3A
Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.
Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(4-x)
i1/x=i2/(4-x)
5/x=3/(4-x)
20-5x=3x
8x=20
8x=2.5cm
since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.
The null point is 0.5cm from the origin x axis
b. If both current are flowing in opposite direction, the null point lies outside of the current.
Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.
Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(x-4)
i1/x=i2/(x-4)
5/x=3/(x-4)
5x-20=3x
2x=20
x=10cm.
This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.
The null point is 8cm from the x axis.
Check the attachment to see the diagram of the current and the null points
