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 ?
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 |
An iron plate of dimensions 17 cm × 40 cm and at 18 °C is heated in a furnace, when the final area of plate is 703 cm2. If the coefficient of linear expansion of iron is 0.000012 /°C, find the temperature of furnace?
Initial area of plate A0 | = | 17 × 40 cm2 |
= | 680cm2 | |
Final area of plate At | = | 703 cm2. |
∴ Coefficient of linear expansion (α) | = | 0.000012 /°C |
∴ Coefficient of superficial expansion (β) | = | 2 × α |
= | 2 × 0.000012 /°C | |
= | 0.000024 /°C | |
We know At – A0 | = | A0 × β × t |
∴ t | = | (At – A0)/(A0× β) |
= | (703 – 680)/(680 × 0.000024) | |
= | 23/(68 × 0.00024) | |
= | 1409.31 °C | |
∴ Temp of furnace | = | Initial temperature + rise in temperature |
= | 18 °C + 1409.31 | |
= | 1427.31 °C. |
Calculate the length of brass rod, which will have same expansion as 14 m of iron rod. Coefficient of linear expansion for iron and brass are 12 × 10–6 /°C and 18 × 10–6 /°C respectively ?
Increase in length for brass | = | Increase in length of iron |
(Lt – L0) for brass | = | ((Lt – L0) for iron.) |
LBrass × αbrass* t | = | LIron × αiron *t |
LBrass × 18 × 10–6 | = | 14 × 12 × 10 – 6 |
∴ LBrass | = | (14 × 12)/18 |
= | 9.33 m |
The constant difference in lengths of iron rod and brass rod is 3.2 cm, when both the rods are heated through same range of temperature. Calculate the length of each rod. Coefficients of linear expansion for iron and brass are 12× 10–6 /°C and 18 × 10–6 /°C ?
Increase in length of brass | = | Increase in length of iron |
LBrass × αbrass × t | = | LIron × αiron × t |
LBrass × 18 × 10–6 | = | LIron × 12 × 10–6 |
LBrass / LIron | = | 12/18 |
= | 2/3 | |
LBrass | = | (2/3) × LIron |
Also by the condition of questions: | ||
LIron – LBrass | = | 3.2 cm |
∴ LIron – (2/3) × LIron | = | 3.2 cm |
∴ (1/3) × LIron | = | 3.2 cm |
LIron | = | 9.6 cm |
∴ LBrass | = | (2/3) × 9.6 cm |
= | 6.4 cm |
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?
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. |
Find the ratio of lengths of iron and brass rods, which will have the same linear expansion due to rise in temperature. Coefficient of linear expansion of brass and iron are 19 × 10–6 °C–1 and 12 × 10–6°C–1 respectively?
Let the original lengths of iron and brass rods be L and L' respectively.
Now. increase in length | = | α × original length × use in temperature. |
For the equal increase in length due to rise in temperature by t °C. | ||
Increase in length of iron rod | = | Increase in length of brass rod. |
(or) L α t | = | L ‘α’ t |
(or) L × 12 × 10–6 × t | = | L' × 19 × 10–6 × t |
(or) L/L' | = | 19/12 |
(or) length of iron rod : length of brass rod | = | 19 : 12 |
The coefficient of linear expansion of copper is 17 × 10–6 per °C. Calculate the increase in length (or change in length) of a copper wire 10 m long when heated from 30 °C to 60 °C ?
This problem is to be solved by using the formula :
α | = | L2 – L1/L1 × (T2 – T1) |
Here, Coefficient of linear expansion, α | = | 17 × 10–6 per °C. |
Increase in length, L2 – L1 | = | ? (To be calculated) |
Original length, L1 | = | 10 m |
And, Rise in temperature, (T2 – T1) | = | 60° – 30° |
= | 30°C | |
Now, putting these values in the above formula, we get: | ||
17 × 10–6 | = | (L2 – L1)/( 10 × 30) |
L2 – L1 | = | 17 × 10–6 × 10 × 30 |
= | 0.0051 m |
Thus, the increase in length of copper wire is 0.0051 metre. It can also be expressed as 0.51 centimetre or 5.1 millimetre.
A brass rod measures 50.2 cm at 16.6 °C and 50.279 cm at 99.5 °C. Calculate the coefficient of linear expansion of brass ?
The formula for calculating the coefficient of linear expansion is :
α | = | (L2 – L1)/(L1 × (T2 – T1)) |
Here, Original length of brass rod, L1 | = | 50.2 cm |
Final length of brass rod, L2 | = | 50.279 cm |
Initial temperature, T1 | = | 16.6 °C |
Final temperature, T2 | = | 99.5 °C |
Now, putting all these values in the above formula, we get: | ||
α | = | (50.279 – 50.2)/(50.2 × (99.5 – 16.6)) |
= | 0.079 /(50.2 × 82.9) | |
= | 18.9 × 10–6 per °C. |