Thermo Dynamics

Step 1: Identify given values: initial length $l = 20$ cm, change in temperature $\Delta T = 40^\circ C - 20^\circ C = 20^\circ C$, and $\alpha = 1.2 \times 10^{-5} / ^\circ C$.
Step 2: Calculate the increase in length using the formula $\Delta l = l \alpha \Delta T$.
Step 3: Substitute the values to find the expansion: $\Delta l = 20 \times 1.2 \times 10^{-5} \times 20 = 0.48 \times 10^{-2}$ cm.
Step 4: Calculate the actual length of the ruler at $40^\circ C$: $l' = l + \Delta l = 20 + 0.0048 = 20.0048$ cm.

Step 1: Let the length of the brass rod be $l_b$ and the steel rod be $l_s$. Given $l_s - l_b = 5$ cm, so $l_s = l_b + 5$.
Step 2: For the difference in lengths to remain constant at all temperatures, the change in length of both rods must be equal: $\Delta l_s = \Delta l_b$.
Step 3: Apply the linear expansion formula: $l_s \alpha_s \Delta T = l_b \alpha_b \Delta T$.
Step 4: Substitute the given values: $(l_b + 5) \times 12 \times 10^{-6} = l_b \times 18 \times 10^{-6}$.
Step 5: Solve for $l_b$: $12 l_b + 60 = 18 l_b \implies 6 l_b = 60 \implies l_b = 10$ cm.

Step 1: Identify given values: diameter $d = 0.4$ m, so radius $r = 0.2$ m. Initial temperature $T_1 = 273$ K, final temperature $T_2 = 360$ K.
Step 2: Calculate the initial surface area of the ball: $A_0 = 4 \pi r^2 = 4 \times \pi \times (0.2)^2 = 0.5024 m^2$.
Step 3: Calculate the change in temperature: $\Delta T = 360 - 273 = 87$ K.
Step 4: Use the areal expansion formula to find the increase in surface area: $\Delta A = \beta A_0 \Delta T$.
Step 5: Substitute the values: $\Delta A = 0.000034 \times 0.5024 \times 87 = 1.486 \times 10^{-3} m^2$.

Step 1: Identify the given values: initial volume $V = 10000 cm^3$, change in temperature $\Delta T = 40^\circ C - 25^\circ C = 15^\circ C$, and increase in volume $\Delta V = 4 cm^3$.
Step 2: Calculate the coefficient of cubical (volume) expansion using the formula $\gamma = \frac{\Delta V}{V \cdot \Delta T}$.
Step 3: Substitute the values to find $\gamma$: $\gamma = \frac{4}{10000 \times 15} = 26.67 \times 10^{-6} / ^\circ C$.
Step 4: Use the relation between linear expansion coefficient $\alpha$ and volume expansion coefficient $\gamma$, which is $\alpha = \frac{\gamma}{3}$.
Step 5: Calculate the coefficient of linear expansion: $\alpha = \frac{26.67 \times 10^{-6}}{3} = 8.89 \times 10^{-6} / ^\circ C$.

Step 1: Identify the given values: initial volume $V = 10^{-6} m^3$, area of cross-section $A = 2 \times 10^{-7} m^2$, change in temperature $\Delta T = 100^\circ C$, and coefficient of volume expansion $\gamma = 18 \times 10^{-5} / ^\circ C$.
Step 2: Calculate the increase in volume of the mercury using the formula $\Delta V = V \times \gamma \Delta T$.
Step 3: Substitute the values to find $\Delta V$: $\Delta V = 10^{-6} \times 18 \times 10^{-5} \times 100 = 18 \times 10^{-9} m^3$.
Step 4: Relate the increase in volume to the increase in length of the mercury column in the capillary tube using the formula $\Delta V = A \times \Delta l$.
Step 5: Substitute the known values to solve for $\Delta l$: $18 \times 10^{-9} = 2 \times 10^{-7} \times \Delta l$.
Step 6: Calculate the final increase in length: $\Delta l = \frac{18 \times 10^{-9}}{2 \times 10^{-7}} = 9 \times 10^{-2} m = 9 cm$.

Given, coefficient of linear expansion, $\alpha = 12 \times 10^{-6} /^\circ C$
Change in temperature, $\Delta \theta = 40^\circ C - 20^\circ C = 20^\circ C$
Total time in a day, $t = 24 \times 60 \times 60 = 86400$ s
The time lost by the clock in a day is given by the formula:
$\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$
Substituting the given values into the formula:
$\Delta t = \frac{1}{2} \times 12 \times 10^{-6} \times 20 \times 86400$
$\Delta t = 12 \times 10^{-5} \times 86400$
$\Delta t = 10368 \times 10^{-3}$
$\Delta t = 10.368$ s
Rounding to one decimal place, the time lost is $10.4$ s.
Since the temperature increases, the length and time period of the pendulum increase, meaning the clock runs slowly and loses time.
Therefore, the loss in time per day is 10.4 s.

Given, time period of second's pendulum, $T = 2$ s
Calibration temperature, $\theta_1 = 20^\circ C$
Final temperature, $\theta_2 = 30^\circ C$
Change in temperature, $\Delta \theta = 30^\circ C - 20^\circ C = 10^\circ C$
Coefficient of linear expansion of steel, $\alpha = 1.2 \times 10^{-5} /^\circ C$
Total time in one week, $t = 7 \times 24 \times 3600 = 604800$ s
Change in time period is given by the formula:
$\Delta T = \frac{1}{2} T \alpha \Delta \theta$
Substituting the given values:
$\Delta T = \frac{1}{2} \times 2 \times 1.2 \times 10^{-5} \times 10$
$\Delta T = 1.2 \times 10^{-4}$ s
The new time period is:
$T' = T + \Delta T = 2 + 1.2 \times 10^{-4} = 2.00012$ s
Time lost in one week is given by the formula:
$\Delta t = \frac{\Delta T}{T'} \times t$
Substituting the values:
$\Delta t = \frac{1.2 \times 10^{-4}}{2.00012} \times 604800$
$\Delta t = 36.28$ s
Since the time period increases, the clock runs slowly and loses time.
Therefore, the clock loses 36.28 s in one week.