Question

Two metal blocks A & B are made of the same material. The dimensions of blocks are shown in the figure below. If they both are heated to the same temperature T, what would the ratio of cubical expansion of A to that of B?​
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Answer 1

Explanation:

The heat temperature through A per second

Q1=K1A(100−t)/l

The heat transferred through B per second

Q2=K2A(t−0)/l

At steady state, K1A(100−t)/l=K2A(t−0)/l

⇒300(100−t)=200(t−0)

⇒300−3t=2t⇒t=60∘C.

Answer 2

Given:

Metal Block A:

Length = L

Height = L

Width = 2L

Metal Block B:

Length = 2L

Height = 2L

Width = 2L

To Find:

$\frac{del V_{A} }{del V_{B}}=\frac{V_{A} }{V_{B}}$

Step 1 of 3

It is found that the increase in volume due to heating the metals are directly proportional to the volume and change in temperature.

$delV=\beta$×$V$×$del T$

Step 2 of 3

In the present case, the change in temperature for both the blocks are same and the material with which the blocks are made are same.

So, the change in volume or cubical expansion depends on the original volume of the metal blocks.

$\frac{delV_{A} }{delV_{B}}=\frac{V_{A} }{V_{B}}$

$delV_{A}=L$×$L$×$2L$

$delV_{B}=2L$×$2L$×$2L$

Step 3 of 3

$delV_{A}=2L^{3}$

$delV_{B}=8L^{3}$

$\frac{delV_{A} }{delV_{B}}=\frac{2L^{3} }{8L^{3} }$

$\frac{delV_{A} }{delV_{B}}=\frac{2}{8}$

$\frac{delV_{A} }{delV_{B}}=\frac{1}{4}$

Thus, the ratio of cubical expansion of A to that of B is 1:4.

Answer:

The ratio of cubical expansion of A to that of B is 1:4.