Given that a photon of light of wavelength
10000 A has energy 1.23 eV. Now when light of
intensity lo and wavelength 5000 Å falls on a photo
cell the saturation current and stopping potential
are 0.40 uA and 1.36 V respectively. The work
function is
Answers
Explanation:
\huge \sf\underline\color{blue}{Given}\dag
Given†
if the matries
A = ((3 ,2, -1 )(2 ,-2 ,0)(1 ,3 ,0)) B = ((-3 ,-1 ,0)(2 ,1 ,3 )(4 ,-1 ,2)) and X = A+ B then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25\huge \sf\underline\color{blue}{Given}\dag
Given†
if the matries
A = ((3 ,2, -1 )(2 ,-2 ,0)(1 ,3 ,0)) B = ((-3 ,-1 ,0)(2 ,1 ,3 )(4 ,-1 ,2)) and X = A+ B then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d
Then c×10 -(b-a)=d
so now going by this logic we have
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25
3,4,9=89
9×10-(4–3)=89
4,5,7=69
7×10-(5–4)=69
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
3,4,9=89q
7×10-(5–4)=69
\huge \sf\underline\color{blue}{Given}\dag
Given†
if the matries
A = ((3 ,2, -1 )(2 ,-2 ,0)(1 ,3 ,0)) B = ((-3 ,-1 ,0)(2 ,1 ,3 )(4 ,-1 ,2)) and X = A+ B then find x
\huge \sf\underline\color{blue}{Solution}\dag
Solution†
Let’s represent the series in a different perspective,
Below are the positions,
Name Positions 1st 2nd 3rd
So, representation would be,
Positions
1 2 3 (Position's Alias/Naming)
1 5 8 = 76
2 7 3 = 25
3 4 9 = 89
4 5 7 = 69
5 3 8 = ?
Now, (by observation)
Take 1st and 3rd positions of each expression to make a number like this,
Number = 3rd position number (concatenate) 1st position number
e.g., if expression is 2 5 8 Then Number would be 82.
Then subtract 2nd position number from the Number would be 82 - 5 = 77.
Therefore,
By Observation,
5 3 8 = 85(Number) - 3 = 82
Answer = 82.
LETS TRY TO ANSWER THIS
FIRST OF ALL FIND THE LOGIC BEHIND IT
IF THE NUMBER ARE
Let a,b,c = d
Then c×10 -(b-a)=d
so now going by this logic we have
1,5,8=76
8×10-(5–1) =76
2,7,3=25
3×10 -(7–2)=25
3,4,9=89
9×10-(4–3)=89
4,5,7=69
7×10-(5–4)=69
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82
So, 5,3,8=????
8×10-(3–5)=82
The answer is 82