ef : jk :: no : ?

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Solution

The correct option is D tu
(D) In the question pair 'ef. jk'. The letters of the first term 'ef' are in natural alphabetic sequence. So is the second term 'jk'. The letter 'no' are also in the natural alphabetic sequence.

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Similar questions

Q. If AB = CD and CD = 2 EF then

Q. If

Δ P Q R ≅ Δ E D F,

then is it true to say that

P R = E F ?

If yes enter 1. else if No enter 0.

Q. Assume

△ H^{o} a n d △ S^{o}

to be independent of temperature, at what temperature will the reaction given below becomes spontaneous?

N_{2} (g) + O_{2} (g) \to 2 N O (g)

S_{N_{2} (g)}^{o} = 191.4 J K^{- 1} m o l^{- 1}

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△ H^{o} = 180.8 k J m o l^{- 1}

PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF is (No gravity is there)

Q. Assume

△ H^{o} a n d △ S^{o}

to be independent of temperature, at what temperature will the reaction given below becomes spontaneous?

N_{2} (g) + O_{2} (g) \to 2 N O (g)

S_{N_{2} (g)}^{o} = 191.4 J K^{- 1} m o l^{- 1}

S_{O_{2} (g)}^{o} = 204.9 J K^{- 1} m o l^{- 1}

S_{N O (g)}^{o} = 210.5 J K^{- 1} m o l^{- 1}

△ H^{o} = 180.8 k J m o l^{- 1}

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ef : jk :: no : ?

The correct option is D tu(D) In the question pair 'ef. jk'. The letters of the first term 'ef' are in natural alphabetic sequence. So is the second term 'jk'. The letter 'no' are also in the natural alphabetic sequence.

The correct option is D tu
(D) In the question pair 'ef. jk'. The letters of the first term 'ef' are in natural alphabetic sequence. So is the second term 'jk'. The letter 'no' are also in the natural alphabetic sequence.