Thermal Properties of Matter Numerical on Temperature Scales and Thermal Expansion Class-11 Nootan ISC Physics Solutions ch-17. Step by step solutions of Kumar and Mittal Physics as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Thermal Properties of Matter Numerical on Temperature Scales and Thermal Expansion Class-11 Nootan ISC Physics Solutions ch-17
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-17 | Thermal Properties of Matter |
| Topics | Numerical on Temperature Scales and Thermal Expansion |
| Academic Session | 2024-2025 |
Numerical on Temperature Scales and Thermal Expansion
(Thermal Properties of Matter Numerical-I Class-11 Nootan ISC Physics Solutions ch-17)
Que-1: The brass scale of a barometer gives correct reading at 0°C. The barometer reads 75 cm at 27°C. What is the correct atmospheric pressure at 27°C. The coefficient of linear expansion α of brass is 2.0 × 10^-5°C^-1.
Ans: Correct reading => L + Lo α Δt
=> 75 + 75 x 2 x 10^-5 x 27
=> 75.0405 cm
Que-2: An iron scale is calibrated at 0°C. The length of a zinc rod is measured to be 100 cm by the scale when the rod and the scale both are at 0°C. What will be the length of the rod as measured by the scale when both are at 100°C. Given: α iron = 1.2 × 10^-5°C^-1 and α zinc = 2.6 × 10^-5 °C^-1.
Ans: Extra length measures
=> (Lzinc x αzinc – Liron x αiron) x Δt
=> (100 x 2.6 x 10^-5 – 100 x 1.2 x 10^-5) x 100
=> 0.14 cm
Length measured = 100 + 0.14 = 100.14 cm
Que-3: A steel cylinder of diameter ‘exactly’ 1 cm at 30°C is to be fitted into a hole in a steel plate. The diameter of the hole is 0.99970 cm at 30°C. To what temperature should the plate be heated? Given : α steel = 1.1 × 10^-5 °C^-1.
Ans: L + Lo α Δt = 1 cm
=> L + ( 1 + α Δt) = 1 cm
=> Δt = (1 / Lo – 1) x 1/α
=> [1 / (0.9997 – 1)] x [1 / (1.1 x 10^-5)] = 27°
∴ final temperature = 30° + 27° = 57°
Que-4: A steel sphere is to be passed through a circular brass ring. At 20°C, the outer diameter of the sphere is 25 cm and the inner diameter of the ring is 24.9 cm. If both are heated together, find the temperature at which the sphere will just pass through the ring. Given : α steel = 1.2×10^-5 °C^-1. and α brass = 2.0 x 10^-5 °C^-1.
Ans: Ls + Ls αs Δt = Lb + Lb αb Δt
=> (Ls + Lb) / ( Lb αb – Ls αs) = Δt = (t – 20°C)
=> (25 – 24.9) / (24.9 x 2 x 10^-5 – 25 x 1.2 x 10^-5) = t – 20°
=> 0.1 x 10^5 / 49.8 – 30.0 = t – 20
=> 505 = t – 20
=> t = 525 °C
Que-5: The volume of a thin brass vessel and that of a solid brass cube are both equal to 1 litre exactly. What will be their new volumes when heated through 25°C ? α for brass is 1.9 × 10^-5 °C^-1.
Ans: Vt = Vo + Vo γ Δt
{γ = 3α , 1 litre = 1000 cm³}
=> Vt = 1000 + 1000 x 3α x Δt
=> Vt = 1000 (1 + 3 x 1.9 x 10^-5)
=> 1001.425 cm³
expansion is independent of shape
Que-6: A mercury thermometer has a bulb of volume 0.300 cm³ and a stem of diameter 0.0100 cm. Find the rise of mercury meniscus in the stem when the temperature rises through 15°C. Given : γ mercury = 1.82 × 10^-4 °C^-1. Ignore the expansion of the bulb.
Ans: Vt = Vo + Vo γ Δt
=> Vt = Vo + (1 + γ Δt )
=> Vt = 0.300 (1 + 1.82 x 10^-4 x 15)
=> Vt = 0.300819
∴ increase in volume = 0.000819
area of stem = πr² = 3.14 x (0.005)² = 7.85 x 10^-5
∴ height of mercury column = V / πr² = 0.000819 / 7.85 x 10^-5 = 10.4 cm
Que-7: A glass vessel of volume 256 cm³ is just filled with mercury at 20°C. How much mercury will overflow when the temperature is raised to 100°C? Given : α glass= 4 x 10^-6 °C^-1 and γ mercury=1 = 1.8 x 10^-4 °C^-1.
Ans: ΔV = VHg γHg Δt – Vglass + αglass Δt
=> ΔV = (256 x 1.8 x 10^-4 – 256 x 4 x 10^-6) x 100
=> 256 x 100 x 10^-4 x (1.8 – 4 x 10^-2)
=> 256 x 100 x 10^-4 x (1.8 – 0.04)
=> 2.56 x 1.76 = 3.36 cm³
Que-8: A glass tube of uniform bore and length 133 cm is to be filled with mercury so that the volume of the tube above the mercury level remains same at all temperatures. Calculate the length of the mercury column. Given: γ glass = 2.6 × 10^-5 °C^-1. and γ mercury = 18.2 x 10^-5 °C^-1.
Ans: The change in volume of glass and mercury must be the same at all temp i.e.
=> VHg γHg Δt – Vglass + γglass Δt
=> A x lHg x γHg = A x 133 x γglass
=> lHg = (133 x 2.6 x 10^-5) / (18.2 x 10^-5)
=> lHg = 19 cm
Que-9: A sphere of diameter 7.0 cm and mass 266.5 g floats in a liquid bath. On heating the bath, the sphere just begins to sink when the temperature reaches 35°C. The density of the liquid at 0°C is 1.527 g cm³. Find the coefficient of cubical expansion of the liquid. Neglect thermal expansion of the sphere.
Ans: density of sphere
=> M / V = 266.5 / (4/3 π x (3.5)³) = 1.484 g/cm³
=> now density of liquid at 35°C should be equal to density of sphere
=> 1.484 = 1.527 (1 -γ Δt )
=> γ = (1.527 – 1.484) / 1.527 x 35
=> 8 x 10^-4 °C^-1
Que-10: At what temperature, do the readings of Celsius and Fahrenheit scales coincide?
Ans: F – 32 / 9 = C / 5
=> According to the question F = C = x (let)
=> x – 32 / 9 = x / 5
=> 9x = 5x – 160
=> 4x = -160
=> x = -40°
Que-11: Normal temperature of the human body is 98.4°F. Find the temperature on Celsius and absolute scale.
Ans: F – 32 / 9 = C / 5
=> 98.4 – 32 / 9 = C / 5
=> C = 36.88 °C
again K = 273.15 + C
=> 273.15 + 36.88 °C
=> 310.03 K
— : end of Thermal Properties of Matter Numerical on Temperature Scales and Thermal Expansion :–
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