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7 months ago

Class 12th NEET

According to Bohr's theory, why do electrons not fall into the nucleus?

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Pallavi Pathak

Contributor-Level 10

According to Bohr's postulates, in specific stable orbits, electrons revolve without emitting energy. Energy is only absorbed or emitted when electrons jump between these orbits, preventing them from spiraling into the nucleus.

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7 months ago

Class 12th Probability

In Exercise 64 above, P(B | $A^{'}$ ) is equal to

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o E x e r c i s e 64, w e h a v e \\ P (B) = \frac{3}{5}, P (A / B) = \frac{1}{2}, P (A \cup B) = \frac{4}{5} \\ P (B / A') = \frac{P (B \cap A')}{P (A')} = \frac{P (B) - P (A \cap B)}{1 - P (A)} = \frac{\frac{3}{5} - \frac{3}{10}}{1 - \frac{1}{2}} = \frac{\frac{3}{10}}{\frac{1}{2}} = \frac{3}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

7 months ago

Class 12th Probability

You are given that $A$ and $B$ are two events such that $P (B) = \frac{3}{5}$ , $P (A / B) = \frac{1}{2}$ , and $P (A \cup B) = \frac{4}{5}$ , then $P (A)$ equals

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{3}{5}, P (A / B) = \frac{1}{2}, P (A \cup B) = \frac{4}{5} \\ W e k n o w t h a t P (A / B) = \frac{P (A \cap B)}{P (B)} \Rightarrow \frac{1}{2} = \frac{P (A \cap B)}{3 / 5} \\ ∴ P (A \cap B) = \frac{3}{10} \\ N o w P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{4}{5} = P (A) + \frac{3}{5} - \frac{3}{10} \\ \Rightarrow P (A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10} = \frac{1}{5} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

7 months ago

Class 12th NCERT

In understanding atomic structure, what is the significance of Bohr's model?

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Pallavi Pathak

Contributor-Level 10

The concept of quantized energy levels for electrons in atoms is introduced by Bohr's model. Although, Bohr's model could not explain the spectra of multi-electron atoms. It helped lay the foundation for quantum mechanics and successfully explained the spectral lines of hydrogen.

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7 months ago

Class 12th Probability

A and B are events such that $P (A) = 0.4$ , $P (B) = 0.3$ , and $P (A \cup B) = 0.5$ . Then $P (B' \cap A)$ equals

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = 0.4, P (B) = 0.3, P (A \cup B) = 0.5 \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ 0.5 = 0.4 + 0.3 - P (A \cap B) \\ \Rightarrow P (A \cap B) = 0.4 + 0.3 - 0.5 = 0.2 \\ ∴ P (B' \cap A) = P (A) - P (A \cap B) \\ = 0.4 - 0.2 = 0.2 = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

7 months ago

Class 12th Probability

If $A$ and $B$ are two events and $A \neq ϕ$ , $B \neq ϕ$ , then

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : A = ? a n d B = ?, \\ t h e n P (A / B) = \frac{P (A \cap B)}{P (B)} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

7 months ago

Class 12th Probability

If $P (A) = 0.4$ , $P (B) = 0.8$ , and $P (B / A) = 0.6$ , then $P (A \cup B)$ is equal to:

(a) 0.24

(b) 0.3

(c) 0.48

(d) 0.96

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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New answer posted

7 months ago

Class 12th Probability

If A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A ? B) = \frac{1}{4}$ , then $P (A^{'} \cap B^{'})$ equals:

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{1}{2}, P (B) = \frac{1}{3} a n d P (A / B) = \frac{1}{4} \\ P (A / B) = \frac{P (A \cap B)}{P (B)} \Rightarrow \frac{1}{4} = \frac{P (A \cap B)}{\frac{1}{3}} \\ ∴ P (A \cap B) = \frac{1}{4} * \frac{1}{3} = \frac{1}{12} \\ N o w P (A' \cap B') = 1 - P (A \cup B) \\ = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{1}{2} + \frac{1}{3} - \frac{1}{12}] = 1 - [\frac{5}{6} - \frac{1}{12}] \\ = 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

7 months ago

Class 12th Probability

If $P (A) = \frac{2}{5}$ , $P (B) = \frac{3}{10}$ , and $P (A \cap B) = \frac{1}{5}$ , then $P (A ∣ B) \cdot P (B^{'} ∣ A^{'})$ is equal to:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = \frac{2}{5}, P (B) = \frac{3}{10} a n d P (A \cap B) = \frac{1}{5} \\ P (A') = 1 - \frac{2}{5} = \frac{3}{5}, P (B') = 1 - \frac{3}{10} = \frac{7}{10} \\ a n d P (A' \cap B') = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{2}{5} + \frac{3}{10} - \frac{1}{5}] = 1 - [\frac{1}{5} + \frac{3}{10}] = 1 - \frac{5}{10} = \frac{1}{2} \\ ∴ P (A' / B') = \frac{P (A' \cap B')}{P (B')} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{5}{7} \\ a n d P (B' / A') = \frac{P (A' \cap B')}{P (A')} = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{5}{6} \\ ∴ P (A' / B') . P (B' / A') = \frac{5}{7} * \frac{5}{6} = \frac{25}{42} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

7 months ago

Class 12th Probability

If $P (A) = \frac{3}{10}$ , $P (B) = \frac{2}{5}$ , and $P (A \cup B) = \frac{3}{5}$ , then $P (B ? A) + P (A ? B)$ equals:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (B) = \frac{2}{5}, P (A) = \frac{3}{10} a n d P (A \cup B) = \frac{3}{5} \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{3}{5} = \frac{3}{10} + \frac{2}{5} - P (A \cap B) \\ P (A \cap B) = \frac{3}{10} + \frac{2}{5} - \frac{3}{5} = \frac{3 + 4 - 6}{10} = \frac{1}{10} \\ N o w P (A / B) + P (B / A) = \frac{P (A \cap B)}{P (B)} + \frac{P (A \cap B)}{P (A)} = \frac{\frac{1}{10}}{\frac{2}{5}} + \frac{\frac{1}{10}}{\frac{3}{10}} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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