Class 12th

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New answer posted

10 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-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 ago

0 Follower 1 View

New answer posted

10 months ago

0 Follower 17 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 4 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 35 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-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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