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  1. 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

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  2. 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.

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  3. 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.

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

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    • 10
  4. To lift heavy pots safely, seek assistance from others to distribute the weight. Use proper posture by bending your knees and keeping your back straight to prevent strain. Hold the pot firmly and ensure your grip is secure. Avoid overexertion and assess the weight before lifting. If unsure, ask forRead more

    To lift heavy pots safely, seek assistance from others to distribute the weight. Use proper posture by bending your knees and keeping your back straight to prevent strain. Hold the pot firmly and ensure your grip is secure. Avoid overexertion and assess the weight before lifting. If unsure, ask for help or use equipment like trolleys to transport pots. These precautions prevent injuries and accidents during gardening activities.

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  5. Planting seeds in a kitchen garden involves preparing nutrient-rich soil and ensuring proper aeration. Make small holes or furrows for the seeds and plant them at the appropriate depth based on their type. Cover the seeds lightly with soil, water gently to moisten, and maintain consistent hydration.Read more

    Planting seeds in a kitchen garden involves preparing nutrient-rich soil and ensuring proper aeration. Make small holes or furrows for the seeds and plant them at the appropriate depth based on their type. Cover the seeds lightly with soil, water gently to moisten, and maintain consistent hydration. Place the garden in a location with adequate sunlight, as it is essential for seed germination and healthy growth. Regular care ensures optimal results.

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