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Answered 3 days ago Learn Sequence and Series

Deepika Agrawal

Let us assume that the man saved Rs a in the first year. In each succeeding year, an increment of Rs. 200 is made. So it forms an A.P. whose first term is a Common difference =200 n=20 Sn=n⁄2 We are given, S20=66000 S20=20⁄2=66000 2a+19×200=6600 ⇒a=1400 read more
Let us assume that the man saved Rs a in the first year.
In each succeeding year, an increment of Rs. 200 is made.
So it forms an A.P. whose first term is a
Common difference =200
n=20
Sn=n⁄2[2a+(n1)d]
We are given, S20=66000
S20=20⁄2[2a+19d]=66000
2a+19×200=6600
a=1400
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Answered 4 days ago Learn Probability

Deepika Agrawal

Given that, Since there52 cards n(S)=Total number of cards=52 There are 13 diamond cards Let A be event that diamond card is withdrawn So,n(A)=13 probability of A=P(A) =NumberofdiamondcardsTotalNumberofcards =n(A)n(S) =1352=14 read more

Given that,

Since there52 cards
n(S)=Total number of cards=52
There are 13 diamond cards
Let A be event that diamond card is withdrawn
So,n(A)=13
probability of A=P(A)
=NumberofdiamondcardsTotalNumberofcards
=n(A)n(S)
=1352=14
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Answered 3 days ago Learn Permutations and Combinations

Deepika Agrawal

Hence, 40320 words with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once.
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Answered 3 days ago Learn Permutations and Combinations

Deepika Agrawal

(10 – 6)! = 4! = 4 × 3 × 2 × 1 = 24
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Answered 3 days ago Learn Chapter 9-Mechanical Properties of Solids

Deepika Agrawal

Mass of the big structure, M = 50,000 kgInner radius of the column, r = 30 cm = 0.3 mOuter radius of the column, R = 60 cm = 0.6 mYoung’s modulus of steel, Y=2×1011Pa Total force exerted, F=Mg=50000×9.8N Stress = Force exerted on a single column =50000×9.84=122500NYoung’s... read more

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y=2×1011Pa

Total force exerted, F=Mg=50000×9.8N

Stress = Force exerted on a single column =50000×9.84=122500N
Young’s modulus, Y=StressStrain

Strain =(F/A)Y
Where,
Area, A=π(R2r2)=π((0.6)2(0.3)2)

Strain =122500/[π((0.6)2(0.3)2)×2×1011]=7.22×107

Hence, the compressional strain of each column is 7.22×107.

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Answered 2 days ago Learn Unit 10-s -Block Elements (Alkali and Alkaline Earth Metals) +1 Chemistry

Deepika Agrawal

The ionization potentials of potassium and cesium are lower than that of lithium. On exposure to light, potassium and caesium easily emit electrons however lithium does not. Hence, potassium and caesium are used in the photoelectric cell.
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Answered 2 days ago Learn Unit 10-s -Block Elements (Alkali and Alkaline Earth Metals) +1 Chemistry

Deepika Agrawal

Different concentrations of alkali metals in liquid ammonia results in different colours. The dilute solutions are blue in colour due to presence of ammoniated electrons. The concentrated solutions have copper bronze colour as ammoniated metal ions are bound by free electrons. read more

Different concentrations of alkali metals in liquid ammonia results in different colours. The dilute solutions are blue in colour due to presence of ammoniated electrons. The concentrated solutions have copper bronze colour as ammoniated metal ions are bound by free electrons.

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Answered 2 days ago Learn Unit 10-s -Block Elements (Alkali and Alkaline Earth Metals) +1 Chemistry

Deepika Agrawal

Be and Mg atoms due to their small size, bind their electrons more strongly (because of higher effective nuclear charge). Hence, require high excitation energy and are not excited the energy of the flame with the result that no flame colour is shown by them.
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Answered 2 days ago Learn Unit 10-s -Block Elements (Alkali and Alkaline Earth Metals) +1 Chemistry

Deepika Agrawal

This process involves several intermediate reactions. Carbon dioxide is passed through a brine solution which is saturated with ammonia gas and ammonium bicarbonate is formed. This ammonium bicarbonate reacts with brine solution and forms sodium bicarbonate.
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