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T = 2π√ (l/g) ⇒ g = 4π²l/T²
Percentage error: Δg/g = Δl/l + 2 (ΔT/T) = (0.1/10.0) + 2 (0.005/0.5) = 0.03
Percentage error = (Δg/g) × 100 = 3%
ω = 2πf = 100π rad/s
i_rms = i? /√2
While current changes from its maximum to its rms value, its phase changes by π/4 rad.
t = (π/4)/ω = π/ (4 × 100π) = 2.5 × 10? ³ s = 2.5ms.

Assuming the rope in the boy's hand is vertical and using a Free Body Diagram (FBD):
f? = T
R + T = 90 ⇒ R = 90 - T
For the piece of wood not to move, f? ≤ µR:
T ≤ 0.5 (90 - T) ⇒ T ≤ 30N

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l = 10.0 ± 0.1cm.
T = (100 ± 1) / 200 = 0.5 ± 0.005

Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0

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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

Kindly consider the following figure

The x and y coordinates of the center of mass are given by:
x? = y? = 4a / 3π

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.

The minimum force F? is calculated as:
F? = (μmg) / √* (1 + μ²)* = ( (1/√3) * 1 * 10 ) / √* (1 + (1/√3)²) = 5N

Harsha Institutions BA course fees on average is either at par or slightly higher than othersimilar institutes offering BA courses. However, it is important to note here that the BA courses offered by Harsha Institutions is a dual degree programme with both BA and BEd being covered as part of the same course. This makes the value of the Harsha Institutions BA course much higher than other similar institutes.

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

Kindly consider then following figure

A? √* (K? /m)* = A? √* (K? /m)* ⇒ A? √K? = A? √K?

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λ? /λ? = (m? v? ) / (m? v? ) ⇒ m? = m? (v? /v? ) (λ? /λ? ) = m? (1/4) (1/2) = (1/8)m?

According to the question,  
  (7 × 4) + (n × 6) = 52
6n = 24 ⇒ n = 4

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