it is on the thermal expansion of solids

3. Q: What is Thermal Expansion
???????
Ans: Increase in size of any object like solid,
liquid and gas on heating is called thermal
expansion.

4. Q: How many types of Thermal expansion
are there???????
Ans: There are 3 types of Thermal expansion
1. Expansion of solids
2. Expansion of liquids
3. Expansion of gases

5. Q: There are again there types of Expansion in
solid??
Ans: 1 Linear Expansion
2. Superficial Expansion
3. Cubical Expansion

6. Q: What is Linear Expansion???
Ans: Increase in length of any solid on heating is
called linear expansion.
L= real length .
ΔL = change in length.
Δt= change in temperature.
α = coefficient of linear expansion.

7. Q: What is Superficial Expansion???
Ans: Increase in area on heating is called superficial
expansion.
A0-Is the original area .
∆A- Is the change in the area.
∆ t = change in temperature.
α = coefficient of superficial expansion.

8. Q: What is Cubical Expansion ???
Ans: Increase in volume on heating is called cubical
expansion.
V0 = real volume.
∆V=change in volume.
Δt = change in temperature.
α = coefficient of cubical expansion.

9. 1. There is sum space left between the two rail tracks so
as they can expand freely and fill the space

10. 2. The electric cable are loosely fixed to the
electric pole so as when during hot weather the
conductor will expand and the cable will not cut
down

11. 3. The iron railings which are present on the road of
the over bridge
the gap is present so as they can expand freely in
hot weather

12. 1.Calculate the increase in length of brass rod, which measures 220 cm at 10 °C,
such that it is heated to 950 °C. [α for brass = 0.000018 /°C]. Also calculate
the overall length at 950° C ?
sol: Given L0 =220 m, α = 0.000018 /°C and t = 950 – 10 = 940 °C
Increase in length (Lt – L0)= L0 × α × t
=220 × 0.000018 × 940
=3.72 cm
∴ Length at 950 °C =220 + 3.72
=223.72 cm

13. 2. A brass rod 2 m long, when heated through 80 °C,
increases in length by 3.2 mm. Calculate the
coefficient of linear expansion for brass?
Sol: Given, original length=2 m.
Rise in temperature=80 °C
Increase in length=3.2 mm = 3.2 × 10–3 m.
α =increase in length/original length × rise in
temperature
=(3.2 × 10–3)/(2 × 80)
=0.00002 per °C.