Class 12th
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New answer posted
2 months agoContributor-Level 10
CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 * CD * AB = 1/2 * 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10
New answer posted
2 months agoContributor-Level 10
lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6
New answer posted
2 months agoContributor-Level 10
x + 2y + z = 2
αx + 3y – z = α
–αx + y + 2z = –α
Δ = | (1, 2, 1), (α, 3, -1), (-α, 1, 2) | = 1 (6+1) – 2 (2α–α) + 1 (α+3α) = 7+2α
α = –7/2
New answer posted
2 months agoContributor-Level 10
hc/λ = K? + φ [given φ is negligible]
So, hc/λ = K?
λ? = h/√ (2mK? ²) ⇒ K? = h²/ (2mλ? ²)
(hc)/λ = h²/ (2mλ? ²) ⇒ λ = (2mc/h)λ? ²
New answer posted
2 months agoContributor-Level 10
Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm
New answer posted
2 months agoContributor-Level 9
Theory based
Capacitor in parallel removes the AC ripple from the rectified output.
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