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Case 1 : Exactly one digit repeated twice --

eg AABCD     Words= 10C1 X 9C3 X {5!/2!} =50400

Case 2 : Exactly one digit repeated thrice --

eg AAABC     Words= 10C1 X 9C2 X {5!/3!}= 7200

Case 3 : Exactly one digit repeated four times --

eg AAAAB     Words= 10C1 X 9C1 X {5!/4!}= 450

Case 4 : Exactly one digit repeated five times --

eg AAAAA     Words= 10C1 X {5!/5!}= 10

Case 5 : Exactly two digits repeated two times each --

eg AABBC     Words= 10C2 X 8C1 X {5!/(2! 2!)}= 10800

Case 6 : Exactly two digits repeated one of them twice and the other thrice --

eg AAABB     Words= 10C1 X 9C1 X {5!/(2! 3!)}= 900

FINAL ANSWER= Sum of all cases= 69760

density of water = 1g/ml

mass of 18ml of water = densityx volume =( 1g/ml)*(18ml) = 18g

no. of mol of water = 18g/18g/mol = 1 mol

1molecule of water contain 8 electrons

1mol of water contain 8*(6.02x1023)

= 48.16x1023

In order to find the locus of the centres of the circles intersecting the given circles we subtract the equations of the two circles.On subtracting them we get 8x-12y+5=0.This is a linear equation in x and y.Therefore locus is a straight line.

Option C as Mg and O will exchange their valency which is 2 and 3 respectively

Hard work , persevearence, mock tests, good guide.

150 m/s

The compound has 49.3% carbon, 6.84% hydrogen, and remaining is oxygen, which is 43.86%, As these elements have relative atomic weight (roughly) of 12, 1 and 16 respectively. Their ratios in the compound is C:H:O = (49.3/12) : (6.84/1) : (43.86/16) = 4.108: 6.84: 2.74, and dividing by lowest number 2.74, we get C:H:O = 1.5: 2.5 :1 = 3:5:2, Therefore the empirical formula is C3H5O2 and molecular formula is (C3H5O2)*n , So, molecular weight (12*3 + 1* 5 + 2* 16) *n = 73n, As, vapour density is 73, Molecular weight is 2* 73 =146 and Hence, 73n = 146, implies n =2, Therefore, Molecular formula is C6H10O4. And Option D is the correct answer.

Answer C

The correct answer is given by option (D).

Explanation: The potential drop across R2 and 30 Ohm will be same as they are in parallel. Now current across R2 is clearly 2.25-1.5 = 0.75 A. Thus,

0.75 x R2 = 30 x 1.5

Therefore, R2 = 60 Ohm = R1.

The equivalent resistance of R2 and 30 Ohm will be

R3 = (60 x 30)/(60 + 30) = 20 Ohm.

Therefore, the total resistance of the circuit is

R4 = R1 + R3 = 80 Ohm

Therefore,

V = 80 x 2.25 = 180 V

According to Balmer's Formula,

Wave no. = RH [1/n1^2 – 1/n2^2]

where RH Rydberg constant for hydrogen

Here n1 = 2, n2 = 4, RH = 109678

Putting these values in the equation we get

Wave no. = 109678 (1/2^2 - 1/4^2) = 109678  (1/4-1/16) = 109678 * 12/64 = 109678 * 3/16

Also λ = 1/ wave number

Therefore λ = 16 / 109678 x 3 = 486 nm