Class 12th
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New answer posted
4 months agoContributor-Level 10
Voltage across secondary source
P = V? i? = V? i?
V? = P/i? = 60/0.11 ≈ 545 V
Since voltage across secondary source is more than primary source (220V)
⇒ Step- up transformer.
New answer posted
4 months agoContributor-Level 10
According to Question, we can write
E = λ/ (2πε? r) ⇒ σ = ε? E = λ/ (2πr)
⇒ σ = 8*10? / (2*3.14*3) = 0.424 nCm? ²
New answer posted
4 months agoContributor-Level 10
Using lens maker formula, we can write
1/f = (µ/µ? - 1) (1/R? - 1/R? )
µ = µ? ⇒ 1/f = 0 ⇒ f → Infinite
New answer posted
4 months agoContributor-Level 10
Using de-Broglie equation:
λ = h/p = h/√ (2mK) = h/√ (2meV)
λ ∝ 1/√m
λ? /λ? = √ (m? /m? ) = √ (1.00727/0.00055) = √1831 ≈ 43:1
New answer posted
4 months agoContributor-Level 10
A = Activity = λN = (ln (2)/t? /? ) N
N = (1.5*10? ³ / 198) * 6*10²³
A = (0.693 / (2.7*24*3600) * (1.5*10? ³ / 198) * 6*10²³ disintegration/s
A ≈ 1.32 * 10¹³ Bq = (1.32*10¹³ / 3.7*10¹? ) Ci ≈ 357 Ci
New answer posted
4 months agoContributor-Level 10
d = √ (2hT R) + √ (2hR R) = 2√ (2hR) (assuming towers are identical)
h = d²/ (8R) = (45*10³)²/ (8*6400*10³) ≈ 39.55 m
New answer posted
4 months agoContributor-Level 10
As we know that magnetic force acting on a charge particle will be
F = q (v * B)
W = F.dl
Since force and displacement will be always perpendicular so work done is always zero.
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