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Class 12th NEET

Why regardless of the intensity of light, the photoelectric emission not occur below the threshold frequency?

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Pallavi Pathak

Contributor-Level 10

Photons do not have enough energy ( $E = h ?$ )to overcome the work function (? ) of the material below the threshold frequency.The energy per photon remains too low to liberate electrons even if the light intensity is increased, so photoelectric emission cannot occur.

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Class 12th NEET

In modern physics, what is the significance of de Broglie's hypothesis?

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Pallavi Pathak

Contributor-Level 10

According to de Broglie's hypothesis, all matter exhibits wave-like behavior. It laid the foundation for quantum mechanics by introducing the concept of matter waves. The dual nature of particles (both wave and particle) is essential in designing technologies like electron microscopes and helps explain phenomena like electron diffraction.

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Class 12th Probability

Two dice are thrown. If it is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum of 3, is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t s h o w i n g t h e s u m o f t h e n u m b e r s o n t h e t w o d i c e w a s l e s s t h a n 6 \\ ∴ E_{1} = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (2, 3), (3, 2)} \\ n (E_{1}) = 10 \\ . E_{2} b e t h e e v e n t t h a t t h e s u m o f t h e n u m b e r s i s 3 . \\ ∴ E_{2} = {(1, 2), (2, 1)} \\ n (E_{2}) = 2 a n d n (E_{1} \cap E_{2}) = 2 \\ ∴ P (E_{2} / E_{1}) = \frac{n (E_{1} \cap E_{2})}{n (E_{1})} = \frac{2}{10} = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Eight coins are tossed together. The probability of getting exactly 3 heads is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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New answer posted

a year ago

Class 12th Probability

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} p r o b a b i l i t y = P (d e a d) . P (d e a d) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{3}{8} . \frac{2}{7} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

A box contains 3 orange balls, 3 green balls, and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P robability o f d r a w i n g 2 g r e e n a n d 1 b l u e b a l l s \\ = P (G) . P (G) . P (B) + P (G) . P (B) . P (G) + P (B) . P (G) . P (G) \\ = \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{2}{8} . \frac{3}{7} . \frac{2}{6} = \frac{12}{336} + \frac{12}{336} + \frac{12}{336} = \frac{36}{336} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f g e t t i n g e v e n n u m b e r o n t h e d i e . E_{2} b e t h e e v e n t o f s e l e c t i n g a s p a d e c a r d . \\ ∴ P (E_{1}) = \frac{3}{6} = \frac{1}{2} a n d P (E_{2}) = \frac{13}{52} = \frac{1}{4} \\ S o, P (E_{1} \cap E_{2}) = P (E_{1}) . P (E_{2}) = \frac{1}{2} . \frac{1}{4} = \frac{1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl, given that the family has at least one girl is

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t G d e n o t e s t h e g i r l a n d B d e n o t e s t h e b o y o f t h e g i v e n f a m i l y . \\ S o, n (S) = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (B B B), (G G G)} \\ L e t E_{1} b e t h e e v e n t t h a t t h e f a m i l y h a s a t l e a s t o n e g i r l \\ ∴ E_{1} = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (G G G)} \\ . E_{2} b e t h e e v e n t t h a t t h e e l d e s t c h i l d i s a g i r l . \\ ∴ E_{2} = {(G B G), (G G B), (G B B), (G G G)} \\ (E_{1} \cap E_{2}) = {(G B B), (G G B), (G B G), (G G G)} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{4 / 8}{7 / 8} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Three persons, A, B, and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3, and 0.2 respectively. The probability of exactly two hits is:

(a) 0.024

(b) 0.188

(c) 0.336

(d) 0.452

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alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = 0.4, P (B) = 0.3, P (C) = 0.2 \\ A l s o, P (\bar{A}) = 1 - 0.4 = 0.6, P (\bar{B}) = 1 - 0.3 = 0.7, P (\bar{C}) = 1 - 0.2 = 0.8 \\ ∴ Probabilities o f t w o h i t s \\ = P (A) . P (B) . P (\bar{C}) + P (A) . P (\bar{B}) . P (C) + P (\bar{A}) . P (B) . P (C) \\ = 0.4 * 0.3 * 0.8 + 0.4 * 0.7 * 0.2 + 0.6 * 0.3 * 0.2 \\ = 0.096 + 0.056 + 0.036 = 0.188 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Refer to Question 74 above. The probability that exactly two of the three balls were red, with the first ball being red, is

0 Follower 17 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o Q u e s t i o n 74, \\ L e t E_{1} b e e v e n t t h a t f i r s t b a l l i s r e d . \\ E_{2} b e e v e n t t h a t e x a c t l y t w o o f t h e t h r e e b a l l s a r e r e d . \\ ∴ P (E_{1}) = P (R) . P (R) . P (B) + P (R) . P (R) . P (R) + P (R) . P (B) . P (R) + P (R) . P (B) . P (B) \\ = \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} + \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{3}{7} . \frac{2}{6} \\ = \frac{60}{336} + \frac{60}{336} + \frac{60}{336} + \frac{30}{336} = \frac{210}{336} \\ P (E_{1} \cap E_{2}) = P (R) . P (B) . P (R) + P (R) . P (R) . P (B) \\ = \frac{5}{8} . \frac{3}{7} . \frac{4}{6} + \frac{5}{8} . \frac{4}{7} . \frac{3}{6} = \frac{60}{336} + \frac{60}{336} = \frac{120}{336} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{120 / 336}{210 / 336} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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