Class 12th
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New answer posted
2 months agoContributor-Level 10
Power delivered = 1000 W
Voltage = 220 V
Transmission 'I' = 1000/220
Power loss = I²R
Efficiency = (1000x100)/ (1000 + I²R)
New answer posted
2 months agoContributor-Level 10
ln |R| = ln|R? | - λt
and T? /? ∝ 1/λ
where 'N' is slope
V² = 2gh
h = √2gh
h = -√2gh + gt²/2
gt² - 2√2ght - 2h = 0
t = (2√2gh ± √ (8gh + 8gh)/2g
t = (2√2gh + 4√gh)/2g as cannot be negative.
∴ t = (2 + √2)√ (h/g) → 3.4√ (h/g)
T? /? (A):T? /? (B):T? /? (C) = 10/6 : 5/6 : 5/2 = 2:1:3
New answer posted
2 months agoContributor-Level 10
V = I? (G + R? )
1 = I? (G + R? )
2 = I? (G + R? + R? )
2 = 1 + I? R?
I? R? = 1
R? = G + R?
New answer posted
2 months agoContributor-Level 10
Photodiode operate in reverse bias. The photocurrent increases initially and saturates fi
New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
There are multiple factors that make the carbonyl group a strong ligand. Check the list below for the reasons.
- Unlike other alkyl ligands, it is an unsaturated compound.
- Due to its unsaturated nature, it has difficulty donating? electron density.
- It has a tendency to accept? (Pie) antibonding electrons.
- CO ligand acts as Lewis acid and donates a lone pair of electrons to form a metal-carbon bond.
- The? -acidic nature of CO gives a strong field and greater d-orbital splitting.
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