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Karun Deshpande

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  1. Asked: January 20, 2025In: Class-11-Physics

    Two spheres of same material have radii 1 m and 4 m and temperatures 4,000 K and 2,000 K respectively. The energy radiated per second by the first sphere is

    Karun Deshpande
    Added an answer on January 20, 2025 at 7:07 am

    According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is P = σ A T⁴ WhereRead more

    According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is

    P = σ A T⁴

    Where,
    – P is the power radiated,
    – σ is the Stefan-Boltzmann constant,
    – A is the surface area of the sphere
    – T is the temperature of the sphere.

    For a sphere, the surface area A is given by:

    A = 4 π r²

    Now, let’s calculate the energy radiated by both spheres:

    For the first sphere:
    – Radius r₁ = 1 m,
    – Temperature T₁ = 4000 K.

    Surface area:

    A₁ = 4 π (1)² = 4 π m²

    The power radiated:
    P₁ = σ A₁ T₁⁴ = σ · 4 π · (4000)⁴

    For the second sphere:
    – Radius r₂ = 4 m,
    – Temperature T₂ = 2000 K.

    Surface area:

    A₂ = 4 π (4)² = 64 π m²

    The power radiated:

    P₂ = σ A₂ T₂⁴ = σ · 64 π · (2000)⁴

    Now, comparing P₁ and P₂:

    – P₁ has a small surface area but a much greater temperature.
    – P₂ has a larger surface area but a lower temperature.

    However, the temperature factor dominates as the power varies as the fourth power of the temperature. So, 4000⁴ will be much larger than 2000⁴.

    Hence, even though the surface area of the first sphere is smaller, the energy radiated per second by the first sphere will be greater than that by the second.

    Conclusion:
    The energy radiated per second by the first sphere is greater than that by the second.

    Checkout for more:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/

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  2. Asked: January 20, 2025In: Class-11-Physics

    The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the North Star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the sun and the North Star is

    Karun Deshpande
    Added an answer on January 20, 2025 at 7:06 am

    We can use Wien's Law to solve this problem. Wien's Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is: λₘₐₓ T = b Where: - λₘₐₓ is the wavelength at which the maximum intensity occurs, - T is the temperature of the black body, - b is WiRead more

    We can use Wien’s Law to solve this problem. Wien’s Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is:

    λₘₐₓ T = b

    Where:
    – λₘₐₓ is the wavelength at which the maximum intensity occurs,
    – T is the temperature of the black body,
    – b is Wien’s constant, which is approximately 2.898 × 10⁶ nm·K.

    Let’s denote the temperatures of the Sun and the North Star as Tₛᵤₙ and Tₙₒᵣₜₕₛₜₐᵣ, and their corresponding maximum wavelengths as λₛᵤₙ and λₙₒᵣₜₕₛₜₐᵣ.

    For the Sun:
    λₛᵤₙ = 510 nm

    For the North Star:
    λₙₒᵣₜₕₛₜₐᵣ = 350 nm

    Applying Wien’s Law for both stars, we can write:

    λₛᵤₙ Tₛᵤₙ = λₙₒᵣₜₕₛₜₐᵣ Tₙₒᵣₜₕₛₜₐᵣ

    Now, solving for the ratio of their temperatures:

    Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = λₙₒᵣₜₕₛₜₐᵣ / λₛᵤₙ

    Substituting the values:

    Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = 350 / 510 ≈ 0.686

    Therefore, the ratio of surface temperatures of Sun and North Star is approximately 0.69.

    Answer: 0.69

    See more:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/

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  3. Asked: January 20, 2025In: Class-11-Physics

    A spherical black body with a radius 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

    Karun Deshpande
    Added an answer on January 20, 2025 at 7:06 am

    We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship: P = σ A T⁴ where, P = power radiated σ = Stefan-Boltzmann constant A = surface area of the sphere T = temperature of the sphere Now, the surface area of the sphere is gRead more

    We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship:
    P = σ A T⁴
    where,
    P = power radiated
    σ = Stefan-Boltzmann constant
    A = surface area of the sphere
    T = temperature of the sphere
    Now, the surface area of the sphere is given as
    A = 4 π r²
    Initially,
    – The radius of the sphere is r = 12 cm = 0.12 m,
    – The temperature is T = 500 K,
    – The power radiated is P = 450 W.

    Let us first calculate how much power is given out initially using the Stefan-Boltzmann law:

    P₁ = σ A₁ T₁⁴

    Now, when the radius is halved or become r₂ = 0.06 m and the temperature is doubled or become T₂ = 1000 K, then the new power will be as follows:

    P₂ = σ A₂ T₂⁴

    Since the area A is proportional to r², we can write the ratio of the new power to the initial power as:

    P₂ / P₁ = (A₂ / A₁) × (T₂⁴ / T₁⁴)

    Substitute the expressions for the areas and temperatures:

    P₂ / P₁ = (r₂² / r₁²) × (T₂⁴ / T₁⁴)

    Substitute the values:

    P₂ / P₁ = (0.06² / 0.12²) × (1000⁴ / 500⁴)

    Simplify:

    P₂ / P₁ = (1/4) × (16) = 4

    Therefore:

    P₂ = 4 × P₁ = 4 × 450 W = 1800 W

    Answer: 1800 W

    Click for more info:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/

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  4. Asked: January 20, 2025In: Class-11-Physics

    A block of steel heated to 100° C is left in a room to cool. Which of the curves shown in the figure, represents the correct behaviour?

    Karun Deshpande
    Added an answer on January 20, 2025 at 6:49 am

    The steel block cools according to Newton's Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay. In figure: Curve A It represents the steep and sharp drop, suggesting a rapid coolinRead more

    The steel block cools according to Newton’s Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay.

    In figure:
    Curve A
    It represents the steep and sharp drop, suggesting a rapid cooling.
    – Curve B: Indicates a linear cooling trend.
    – Curve C: This shows a gradual cooling curve, which is similar to exponential decay.

    The correct answer is Curve C, as it curves like one would expect according to Newton’s Law of Cooling.

    Click here for more solutions:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/

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  5. Asked: January 20, 2025In: Class-11-Physics

    A sphere, a cube and a thin circular plate, all made of the same material and having the same mass, are initially heated to a temperature of 3,000°C. Which of these will cool fastest?

    Karun Deshpande
    Added an answer on January 20, 2025 at 6:49 am

    The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that 1. Sphere: Has the minimum surface area for the same volume or mass. 2. Cube: Has a moderate surface area comparedRead more

    The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that

    1. Sphere: Has the minimum surface area for the same volume or mass.
    2. Cube: Has a moderate surface area compared to the sphere and plate.
    3. Thin Circular Plate: Has the largest surface area among the three.

    Since the thin circular plate has the largest surface area, it will lose heat fastest and cool the quickest.

    Click this for more:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/

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