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m = 0.354g, T\(_1\) = 273^\circC = 273 + 273 = 546K P\(_1\) = 114cmHg, V\(_1\) = 2667cm\(^3\) at STP T\(^2\) = 273K, P\(_2\) = 76cmHg, V\(_2\) = ? \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_1}\) = 2000.25cm\(^3\)

Conduction takes places in gases when air is pumped out of a discharged tube under reduced pressure.

Depth of sea can be measured by echo, a reflected sound waves.

Newton's First law of motion describes the straight motion of a body.

Vacuum is an enclosure with no air or agent of heat transfer

Iris is similar to diaphragm.

Potential energy at the highest point = mgh PE = 4 \times 10 \times 40 = 1600J

\(\text{Circumference} \, (d) = 2 \pi r = 2 \pi (0.1) \approx 0.6283 \, \text{m}\) \(\text{Time for one revolution} \, (t) = 60 \, \text{s}\) \(\text{Speed} \, (v) = \frac{d}{t} = \frac{0.6283}{60} \approx 0.01047 \, \text{m/s} \approx 1.05 \times 10^{-2} \, \text{ms}^{-1}\)

For the parallel arrangement = 2 + 4 = 6\muf For the series arrangement = 1 C\(_T\) = 1 2 + 1 3 + 1 6 + 1 4 1 C\(_T\) = 15 12 C\(_T\) = 12 15 = 0.8\muf

C\(_{\theta}\) = 50, C\(_0\) = 2, C\(_{100}\) = 100 \theta = C\(_{\theta}\) - C\(_0\) C\(_{100}\) - C\(_0\) \times 100 = 51 - 2 100 - 2 \times 100 = 49 98 \times 100 \theta = 50^\circC.

m = 500kg, W = mg = 500 \times 10 = 5000N F = 2000N \mu = Force at constant velocity weight = 2000 5000 = 0.4

Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted. (i.e, P.E\(_g\) = \omegah, where \omega = mg)

For the first copper rod: \(L_0 = 5 \, \text{m} = 5000 \, \text{mm}, \quad \Delta T = 10 \, \text{^\circC}, \quad \Delta L = 1 \, \text{mm}\) Coefficient of linear expansion \( \alpha \): \(\alpha = \frac{1 \, \text{mm}}{5000 \, \text{mm} \cdot 10} = \frac{1}{50000} \, \text{mm}^{-1}\) For the second copper rod: \(L_0 = 2.5 \, \text{m} = 2500 \, \text{mm}, \quad \Delta T = 5 \, \text{^\circC}\) Expansion: \(\Delta L = L_0 \cdot \alpha \cdot \Delta T = 2500 \, \text{mm} \cdot \frac{1}{50000} \, \text{mm}^{-1} \cdot 5 = 0.25 \, \text{mm}\) The second copper rod will expand by: \(0.25 \, \text{mm}\)

- Topping is done with distilled water - Naked flame should be avoided when charging the battery - Direct connection of wires to the terminals should be avoided.

RH = mass of water vapour in volume of air mass of water vapour for saturation RH = m M = 0.05 0.15 \times 100 = 5 0.15 = 500 15 = 33.33%