To convert from Celsius to Fahrenheit, we use the formula:

\[
F = \frac{9}{5}C + 32
\]

Given \( C = 25^\circ \):

\[
F = \frac{9}{5} \times 25 + 32
\]
\[
F = 45 + 32 = 77
\]

So, the corresponding temperature is 77ºF.

Given the relation between temperature scales \( A \) and \( B \):

\[
\frac{A - 42}{100} = \frac{B - 7}{220}
\]

To find the temperature at which both scales show the same reading, set \( A = B = x \):

\[
\frac{x - 42}{100} = \frac{x - 7}{220}
\]

Cross-multiplying:

\[
220(x - 42) = 100(x - 7)
\]

Expanding and solving for \( x \):

\[
220x - 9240 = 100x - 700
\]
\[
120x = 8540
\]
\[
x = \frac{8540}{120} = 12
\]

Thus, the temperature at which both scales read the same is 12.

On heating an object its photographic expansion takes place. so, distance between any two point increases.

To find absolute zero in Fahrenheit, we use the relationship between Celsius and Fahrenheit:

\[
F = \frac{9}{5}C + 32
\]

Absolute zero in Celsius is \(-273.15^\circ C\). Substitute this into the formula:

\[
F = \frac{9}{5}(-273.15) + 32
\]
\[
F = -491.67 + 32 = -459.67
\]

Rounding to the nearest whole number, we get \(-460^\circ F\).

The change in length \( \Delta L \) due to temperature change is given by:

\[
\Delta L = L_0 \alpha \Delta T
\]

where:
- \( L_0 = 10 \, \text{cm} \)
- \( \alpha = 11 \times 10^{-6} / ^\circ \text{C} \)
- \( \Delta T = 20^\circ \text{C} - 19^\circ \text{C} = 1^\circ \text{C} \)

Substitute the values:

\[
\Delta L = 10 \times 11 \times 10^{-6} \times 1 = 11 \times 10^{-5} \, \text{cm}
\]

Thus, the bar will be \( 11 \times 10^{-5} \, \text{cm} \) shorter at 19ºC.

Using the relation \(\frac{C}{5} = \frac{F - 32}{9}\), let \(C = F = x\). This gives \(\frac{x}{5} = \frac{x - 32}{9} ⇒ 9x = 5x - 160 ⇒ x = -40\).