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How can plants be protected from any kind of harm from animals and pests?
Plants are safeguarded from harm through various measures. Fences or netting prevent animals like rabbits or stray cattle from entering the garden. Organic pesticides or neem-based solutions effectively deter harmful pests without damaging plants. Companion planting, such as growing marigolds, naturRead more
Plants are safeguarded from harm through various measures. Fences or netting prevent animals like rabbits or stray cattle from entering the garden. Organic pesticides or neem-based solutions effectively deter harmful pests without damaging plants. Companion planting, such as growing marigolds, naturally repels certain insects. Additionally, regular monitoring of plants helps detect pests or diseases early. Proper care and timely actions ensure plants remain healthy and grow without disturbances.
See lessWhat were the two most interesting things you learnt during your visit?
Two key learnings during the visit were the concept of companion planting and the role of soil preparation. Companion planting, such as using marigolds to repel pests, showcased a natural pest-control method. Additionally, understanding soil preparation techniques like aerating and adding organic maRead more
Two key learnings during the visit were the concept of companion planting and the role of soil preparation. Companion planting, such as using marigolds to repel pests, showcased a natural pest-control method. Additionally, understanding soil preparation techniques like aerating and adding organic matter emphasized how proper groundwork impacts plant growth. Both practices highlight sustainable and efficient gardening strategies that promote a thriving kitchen garden with minimal environmental impact.
See lessA 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
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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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
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?
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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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
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?
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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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/