The x-axis in three-dimensional space consists of all points where the y and z coordinates are zero. Therefore, the equations of the x-axis are: y = 0 and z = 0 In parametric form, the x-axis can be represented as: x = t, y = 0, and z = 0, where t is a parameter. So, the correct answer is: α Click hRead more
The x-axis in three-dimensional space consists of all points where the y and z coordinates are zero. Therefore, the equations of the x-axis are:
y = 0 and z = 0
In parametric form, the x-axis can be represented as:
x = t, y = 0, and z = 0, where t is a parameter.
The reflection of a point (α, β, γ) in the xy-plane involves changing the sign of the z-coordinate while keeping the x and y coordinates unchanged. Therefore, the reflection of the point (α, β, γ) in the xy-plane is: (α, β, -γ) Thus, the correct answer is: (α, β, -γ) Click here for more: https://wwwRead more
The reflection of a point (α, β, γ) in the xy-plane involves changing the sign of the z-coordinate while keeping the x and y coordinates unchanged.
Therefore, the reflection of the point (α, β, γ) in the xy-plane is: (α, β, -γ)
The direction cosines of a line are given as k, k, and k. The sum of the squares of the direction cosines must equal 1: k² + k² + k² = 1 3k² = 1 k² = 1/3 k = ±1/√3 Thus, the correct answer is: k = 1/√3 or -1/√3 Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapteRead more
The direction cosines of a line are given as k, k, and k. The sum of the squares of the direction cosines must equal 1:
According to Wien's displacement law, the wavelength at which the black body emits maximum energy varies inversely with temperature: λ_max * T = constant Let the initial temperature be 20 K with a wavelength of 14 µm, and the final temperature be 1000 K. Given that using Wien's law, λ₁ * T₁ = λ₂ * TRead more
According to Wien’s displacement law, the wavelength at which the black body emits maximum energy varies inversely with temperature:
λ_max * T = constant
Let the initial temperature be 20 K with a wavelength of 14 µm, and the final temperature be 1000 K. Given that using Wien’s law,
According to Newton's law of cooling, the rate of cooling is proportional to the difference between the temperature of the object and the surrounding temperature: Rate of cooling = k(T_object - T_surrounding) Let the initial temperature of the object be 600 K and the surrounding temperature be 300 KRead more
According to Newton’s law of cooling, the rate of cooling is proportional to the difference between the temperature of the object and the surrounding temperature:
Rate of cooling = k(T_object – T_surrounding)
Let the initial temperature of the object be 600 K and the surrounding temperature be 300 K. If the rate of cooling at 600 K is R, the rate of cooling at a temperature T is proportional to (T – 300).
For the temperature 600 K, the rate of cooling is:
R = k(600 – 300) = 300k
Now, for the temperature 300 K, the rate of cooling will be zero because the temperature difference is zero.
The equations of x-axis in space are
The x-axis in three-dimensional space consists of all points where the y and z coordinates are zero. Therefore, the equations of the x-axis are: y = 0 and z = 0 In parametric form, the x-axis can be represented as: x = t, y = 0, and z = 0, where t is a parameter. So, the correct answer is: α Click hRead more
The x-axis in three-dimensional space consists of all points where the y and z coordinates are zero. Therefore, the equations of the x-axis are:
y = 0 and z = 0
In parametric form, the x-axis can be represented as:
x = t, y = 0, and z = 0, where t is a parameter.
So, the correct answer is: α
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-11
The reflection of the point (α, β, γ) in the xy-plane is
The reflection of a point (α, β, γ) in the xy-plane involves changing the sign of the z-coordinate while keeping the x and y coordinates unchanged. Therefore, the reflection of the point (α, β, γ) in the xy-plane is: (α, β, -γ) Thus, the correct answer is: (α, β, -γ) Click here for more: https://wwwRead more
The reflection of a point (α, β, γ) in the xy-plane involves changing the sign of the z-coordinate while keeping the x and y coordinates unchanged.
Therefore, the reflection of the point (α, β, γ) in the xy-plane is: (α, β, -γ)
Thus, the correct answer is: (α, β, -γ)
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-11
If the direction cosines of a line are k, k, k then
The direction cosines of a line are given as k, k, and k. The sum of the squares of the direction cosines must equal 1: k² + k² + k² = 1 3k² = 1 k² = 1/3 k = ±1/√3 Thus, the correct answer is: k = 1/√3 or -1/√3 Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapteRead more
The direction cosines of a line are given as k, k, and k. The sum of the squares of the direction cosines must equal 1:
k² + k² + k² = 1
3k² = 1
k² = 1/3
k = ±1/√3
Thus, the correct answer is: k = 1/√3 or -1/√3
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-11
A black body at 20 K is found to emit maximum energy at a wavelength 14 µm. When its temperature is raised to 1000K, then wavelength at which maximum energy emitted is
According to Wien's displacement law, the wavelength at which the black body emits maximum energy varies inversely with temperature: λ_max * T = constant Let the initial temperature be 20 K with a wavelength of 14 µm, and the final temperature be 1000 K. Given that using Wien's law, λ₁ * T₁ = λ₂ * TRead more
According to Wien’s displacement law, the wavelength at which the black body emits maximum energy varies inversely with temperature:
λ_max * T = constant
Let the initial temperature be 20 K with a wavelength of 14 µm, and the final temperature be 1000 K. Given that using Wien’s law,
λ₁ * T₁ = λ₂ * T₂
14 µm * 20 K = λ₂ * 1000 K
λ₂ = (14 µm * 20 K) / 1000 K
λ₂ = 280 µm / 1000
λ₂ = 0.28 µm
Thus, the correct answer is: 2.8 µm
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-11
Rate of cooling at 600 K, if surrounding temperature is 300 K is R.
According to Newton's law of cooling, the rate of cooling is proportional to the difference between the temperature of the object and the surrounding temperature: Rate of cooling = k(T_object - T_surrounding) Let the initial temperature of the object be 600 K and the surrounding temperature be 300 KRead more
According to Newton’s law of cooling, the rate of cooling is proportional to the difference between the temperature of the object and the surrounding temperature:
Rate of cooling = k(T_object – T_surrounding)
Let the initial temperature of the object be 600 K and the surrounding temperature be 300 K. If the rate of cooling at 600 K is R, the rate of cooling at a temperature T is proportional to (T – 300).
For the temperature 600 K, the rate of cooling is:
R = k(600 – 300) = 300k
Now, for the temperature 300 K, the rate of cooling will be zero because the temperature difference is zero.
Thus, the correct answer is: 2 R
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-11