Class 12th

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New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

I? = I? + I? ⇒ I? /I? = 1 + I? /I? ⇒ 1/α = 1 + 1/β = (β+1)/β ⇒ α = β/ (1+β)

⇒ 1/β = 1/α - 1 = (1-α)/α ⇒ β = α/ (1-α)

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Find the number of solutions for 2tan(x) = π/2 - x in [0, 2π].
This is equivalent to finding the number of intersection points of the graphs y = tan(x) and y = (π/4) - x/2.
Let's sketch the graphs:

y = tan(x) has vertical asymptotes at x = π/2, 3π/2.

y = (π/4) - x/2 is a straight line with a negative slope.
At x=0, y=π/4.
At x=π/2, y=0.
At x=π, y=-π/4.
At x=2π, y=-3π/4.
By observing the graphs, there will be one intersection in (0, π/2), one in (π/2, 3π/2), and one in (3π/2, 2π].
Total number of solutions is 3.

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Always possible as it is associated only with β? decay


New answer posted

7 months ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

θ? > I = θ ⇒ sinθ? > sinθ ⇒ 1/μ > sinθ

⇒ μ < 1/sin

Note: If we assume θ = 45° ⇒ μ < 1.414, then red colour light ray will come out from face PR of prism.

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Given that x, y, z are in A.P., so 2y = x + z.
The determinant is:
| 3 4√2 x |
| 4 5√2 y | = 0
| 5 k z |

Apply the operation R? → R? + R? - 2R?:
The first row becomes:
(3 + 5 - 24) (4√2 + k - 25√2) (x + z - 2y)
= 0 (k - 6√2) (0)
So the determinant becomes:
| 0 k-6√2 0 |
| 4 5√2 y | = 0
| 5 k z |

Expanding along the first row:
-(k - 6√2)(4z - 5y) = 0.
This implies k - 6√2 = 0 or 4z - 5y = 0.
k = 6√2 or y = 4z/5.
The condition y = 4z/5 is stated as not possible.
Therefore, k = 6√2, which means k² = (6√2)² = 36 * 2 = 72.

New answer posted

7 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

I = (V/R) (1 - e^ (-t/τ) where τ = L/R

E = (1/2)LI² = (1/2)L (V²/R²) (1 - e^ (-t/τ)² ⇒ Energy stored in inductor

According to Question, we can write
(1/4) * (1/2) (LV²/R²) = (1/2) (LV²/R²) (1 - e^ (-t/τ)²
⇒ 1/4 = (1 - e^ (-t/τ)² ⇒ 1/2 = 1 - e^ (-t/τ) ⇒ e^ (-t/τ) = 1/2

⇒ t/τ = ln (2) ⇒ t = τln (2) = (L/R)ln (2)

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

R = mv / (qB) = √ (2mK) / (qB) ⇒ K = (R²q²B²) / (2m)

⇒ K? /Kα = (R? ²q? ²B² / 2m? ) * (2mα / (Rα²qα²B²) = (mα/m? ) * (R? /Rα)² * (q? /qα)²

⇒ K? /Kα = (4) * (2)² * (1/2)² = 4:1

New answer posted

7 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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