CBSE Class 9 Physics Change of State and Relative Humidity MCQs Set 06

Answer: (b) joule

Explanation:
1. Heat is energy, so the SI unit is the joule; the kilojoule is a multiple and the joule per second is the watt.

Teacher's Note:
1. 1 kJ = 1000 J.
2. J/s is the unit of power.

Answer: (b) Ice cream

Explanation:
1. Ice, water and ice in contact with water are all at 0°C.
2. Ice cream is stored below 0°C, so it is colder.

Teacher's Note:
1. Ice in contact with water stays at 0°C until it melts.
2. Compare the temperatures, not the states.

Answer: (a) less than 27°C

Explanation:
1. SVP at the dew point equals the actual vapour pressure, which is 95% of the SVP at 27°C, so it is smaller than the SVP at 27°C.
2. SVP is smaller at lower temperatures, so the dew point is below 27°C (it would equal 27°C only at 100% humidity).

Teacher's Note:
1. The dew point is below the air temperature unless the air is saturated.
2. A humidity close to 100% gives a dew point close to the air temperature.

Answer: (c) 760 mm of Hg

Explanation:
1. 100°C is the boiling point of water at normal pressure, where the saturation vapour pressure equals the atmospheric pressure of 760 mm of mercury.

Teacher's Note:
1. Boiling occurs when SVP equals the external pressure.
2. SVP at 20°C is only about 17.5 mm.

Answer: (d) none of the above

Explanation:
1. R.H. = \( \dfrac{3}{9} \times 100 = 33.3\% \), which is not 10%, 20% or 30%.

Teacher's Note:
1. Always divide actual pressure by saturation pressure.
2. 33.3% matches none of the first three options.

Answer: (a) 85%

Explanation:
1. At high humidity less sweat evaporates from the body surface, so the body cools less and we feel hotter.

Teacher's Note:
1. Evaporation takes heat from the body.
2. Dry air makes evaporation fast.

Answer: (b) the air is completely dry

Explanation:
1. Dew can form only if there is water vapour in the air; completely dry air contains none, so cooling it never produces dew.

Teacher's Note:
1. Saturated air gives dew at the air temperature itself.
2. Any humidity above zero has a dew point.

Answer: (b) 25°C

Explanation:
1. No evaporation occurs at 100% relative humidity, so the wet bulb is not cooled and reads the same as the dry bulb, 25°C.

Teacher's Note:
1. The wet bulb reads lower only when evaporation takes place.
2. The bigger the difference of readings, the drier the air.

Answer: (b) 1008 kJ

Explanation:
1. \( Q = mL = 3 \times 336\,000 = 1\,008\,000 \) J = 1008 kJ.

Teacher's Note:
1. Latent heat is released on freezing.
2. 80 x 4.2 = 336.

Answer: (d) 4°C

Explanation:
1. Dew point = mean of the temperatures at which dew forms and disappears = \( \dfrac{3.2 + 4.8}{2} = 4 \)°C.

Teacher's Note:
1. The two readings bracket the true dew point.
2. Averaging reduces the observation error.

Answer: (a) 52.5%

Explanation:
1. The dew point is 10°C, so the actual vapour pressure = SVP at 10°C = 9.21 mm.
2. R.H. = \( \dfrac{9.21}{17.54} \times 100 = 52.5\% \).

Teacher's Note:
1. The dew point gives the actual vapour pressure.
2. Relative humidity cannot exceed 100%.

Answer: (a) 5.67 x 10⁵ J

Explanation:
1. \( Q = mL = 0.25 \times 22.7 \times 10^5 = 5.675 \times 10^5 \) J.

Teacher's Note:
1. Convert 250 g to 0.25 kg.
2. There is no rise of temperature, only a change of state.

Answer: (a) 5.98 x 10⁵ J

Explanation:
1. Melting: \( 1.5 \times 3.36 \times 10^5 = 5.04 \times 10^5 \) J.
2. Heating the water: \( 1.5 \times 4180 \times 15 = 0.94 \times 10^5 \) J. Total = 5.98 x 10⁵ J.

Teacher's Note:
1. Two stages: change of state, then change of temperature.
2. Add the two heats.

Answer: (a) \( \dfrac{Pt}{M} \)

Explanation:
1. The power P just balances the rate at which heat is lost, so in t seconds the substance gives out heat Pt while solidifying.
2. Pt = ML, so \( L = \dfrac{Pt}{M} \).

Teacher's Note:
1. Heat given out in solidifying = mass x latent heat.
2. Check units: W s / kg = J/kg.