Class 12th
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New answer posted
10 months agoContributor-Level 10
Reaction (1) is SN1 (rate independent of [OH? ]). Reaction (2) is E2 (rate depends on [OH? ]).
Statement (B) is correct. Changing concentration of base will have no effect on reaction (1).
New answer posted
10 months agoContributor-Level 9
Cesium has lowest ionisation enthalpy and hence it can show photoelectric effect to the maximum extent hence it is used in photo electric cell.
New question posted
10 months agoNew answer posted
10 months agoContributor-Level 10
P = E/t
= 2/235 * (6.023*10²? *200*1.6*10? ¹? )/ (30*24*60*60) = 60W
New answer posted
10 months agoContributor-Level 10
Pitch = (Vcosθ)T = (Vcosθ)2πm/eB
= (4*10? cos60°) (2π (1.67*10? ²? )/ (0.3*10? ¹? *1.69*10¹? )
= 4 cm
New answer posted
10 months agoContributor-Level 10
From Given equation
µ = 0.6
A? = µA?
(A? - A? )/2 = A? = 5
(A? + A? )/2 =?
A? - A? = 3
From equation (1)+ (2)
A? = 8
From equation (1)- (2)
A? = 2
New answer posted
10 months agoContributor-Level 10
Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36
New answer posted
10 months agoContributor-Level 10
hc/λ = Φ + eV
hc/3λ = Φ + eV/4
from (1) and (2)
hc/λ (1-1/3) = 3/4 eV
eV = 8/9 hc/λ
eV = 8/9 (hc/λ - Φ)
Φ = hc/9λ
λ? = 9λ ∴k=9
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