Physics NCERT Exemplar Solutions Class 11th Chapter Five

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New answer posted

4 weeks ago

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R
Raj Pandey

Contributor-Level 9

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

New answer posted

4 weeks ago

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R
Raj Pandey

Contributor-Level 9

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 * 10 * (10 - 5) = 10m/s

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4 weeks ago

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R
Raj Pandey

Contributor-Level 9

Potential difference across resistor at time t = V = 30/3 = 10V
Current, I = 10 / (5 * 10? ) = 2? A

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4 weeks ago

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R
Raj Pandey

Contributor-Level 9

For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32

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4 weeks ago

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R
Raj Pandey

Contributor-Level 9

Angle from direction of flow = 90° + θ = 120°
⇒ θ = 30°
sin 30° = u/v ⇒ u/10 = 1/2 ⇒ u = 5 m/s

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a month ago

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R
Raj Pandey

Contributor-Level 9

Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Deceleration, a = u² / 2S = 10² / (2 * 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 * 100 = 10N

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a month ago

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R
Raj Pandey

Contributor-Level 9

R_eq = 10 + (50 * 20) / (50 + 20) = 170/7 Ω
⇒ I = 170 / (170/7) = 7A
⇒ x = 10 * 7 = 70V ⇒ Voltage across 10Ω resistor

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a month ago

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R
Raj Pandey

Contributor-Level 9

1/C = 5/ (ε? * 100) + 5/ (10ε? * 100)
⇒ C = (ε? * 1000) / 55 = (8.85 * 10? ¹² * 1000) / 55 = 1.61 * 10? ¹? F = 161 pF

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a month ago

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R
Raj Pandey

Contributor-Level 9

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

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