Class 11th
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New answer posted
2 months agoContributor-Level 10
I = 0.8 kgM², |μ? | = 20Am²
U? = -μ? ⋅ B? = 0
Uf = -μBcos (30°) = -20 x 4 x √3/2
U? – Uf = 40√3 = ½ I ω² = 0.4ω²
ω² = 100√3; ω = 10 (3¹/? )
New answer posted
2 months agoContributor-Level 10
In steady state rate of flow of heat in all three rods are same.
dQ/dt = (k? A (100 – 70)/? = (k? A (70 – 20)/? = (k? A (20 – 0)/?
30k? = 50k? = 20k?
∴ k? : k? = 2:3 & k? : k? = 2:5
New answer posted
2 months agoContributor-Level 9
2 Pb (NO? )? → 2PbO + 2NO? + O?
2NO? → N? O? colourless
N? O? + 2NO → 2 N? O?
Oxidation state in N? O? is +3
New answer posted
2 months agoContributor-Level 10
Given (2x² + 3x + 4)¹? = Σ (r=0 to 20) a? x?
replace x by 2/x in above identity :-
2¹? (2x²+3x+4)¹? / x²? = Σ (r=0 to 20) a?2? /x?
⇒ 2¹? Σ (r=0 to 20) a? x? = Σ (r=0 to 20) a?2? x²? (from (i)
now, comparing coefficient of x? from both sides
(take r = 7 in L.H.S. and r = 13 in R.H.S.)
2¹? a? = a?2¹³ ⇒ a? /a? = 2³ = 8
New answer posted
2 months agoContributor-Level 10
max {n (A), n (B)} ≤ n (A U B) ≤ n (U)
⇒ 76 ≤ 76 + 63 - x ≤ 100
⇒ -63 ≤ -x ≤ -39
⇒ 63 ≥ x ≥ 39
New answer posted
2 months agoContributor-Level 10
Σ (r=30 to 50) 50-rC? =? C? +? C? +? C? + . + ³? C?
=? C? +? C? +? C? + . + (³? C? + ³? C? ) - ³? C?
=? C? +? C? +? C? + . + (³¹C? + ³¹C? ) - ³? C?
=? C? +? C? - ³? C?
=? ¹C? - ³? C?
New answer posted
2 months agoContributor-Level 10
Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),
which passes through (9,0) ⇒ b² = 2a²
So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)
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