Class 12th
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New answer posted
7 months agoContributor-Level 10
Given the refractive index μ = λ? / λ = 3/2 and v = 10m, the object distance is u = - (3/2)v = -15m.
Using the lens maker's formula:
μ/v - 1/u = (μ - 1)/R
(3/2)/10 - 1/ (-15) = (3/2 - 1)/R
This gives the radius of curvature R:
R = -30/13 m
New answer posted
7 months agoContributor-Level 9
Resonance frequency is independent of R.
Quality factor = ωL/R ⇒ Quality factor decreases with increase in R.
Bandwidth of resonance circuit = R/L ⇒ increases with increase in R.
New answer posted
7 months ago
Contributor-Level 10
Let the charge on C? be q µC. For the capacitor network:
q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16
New answer posted
7 months agoContributor-Level 9
λ? /λ? = (m? v? ) / (m? v? ) ⇒ m? = m? (v? /v? ) (λ? /λ? ) = m? (1/4) (1/2) = (1/8)m?
New answer posted
7 months agoContributor-Level 10
The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.
New answer posted
7 months ago
Contributor-Level 10
Let V? = 10V, V? = xV, V? = 0V, and V? = yV.
Applying Kirchhoff's current law at node B:
(x - 10)/100 + (x - y)/15 + (x - 0)/10 = 0 ⇒ 53x - 20y = 30 . (1)
Applying Kirchhoff's current law at node D:
(y - 10)/60 + (y - x)/15 + (y - 0)/5 = 0 ⇒ 17y - 4x = 10 . (2)
Solving equations (1) and (2), we get:
x = 0.865 and y = 0.792
The current i? is:
i? = (x - y) / 15 = 4.87 mA
New answer posted
7 months agoContributor-Level 10
dy/dx = xy - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)
⇒ ∫ dy / (1 + y) = ∫ (x - 1) dx ⇒ ln (1 + y) = x²/2 - x + C
For y (0) = 0, c = 0. ∴ ln (1 + y) = x²/2 - x
1 + y = e^ (x²/2 - x)
y (1) = e^ (1/2 - 1) - 1 = e^ (-1/2) - 1
New answer posted
7 months agoContributor-Level 9
In external magnetic field, a magnetic force acts on every small part of the loop in direction perpendicular to the wire. Thus, loop assumes a shape (circular) in which it covers maximum area
New answer posted
7 months agoContributor-Level 10
The Balmer series lies in the visible region of the electromagnetic spectrum.
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