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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A ray of light incident at an angle d on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is mad? of a material of refractive index 1.5, the angle of incidence is

(a) 7.5°

(b) 5°

(c) 15°

(d) 2.5°

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) since deviation $δ = (μ - 1) A$ = (1.5-1)5⁰= 2.5⁰

But $δ = θ - r$ so $θ = 7.5 °$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} H e r e, A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ | A | = 0 | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | - 1 | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | \\ = 0 - 1 (0 - 1) + 1 (1 - 0) \\ = 1 + 1 = 2 \neq 0 (n o n - singular m a t r i x) \\ N o w, c o - f a c t o r s, \\ a_{11} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{12} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{13} = + | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1 \\ a_{21} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{22} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{23} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1 \\ a_{31} = + | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1, a_{32} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1, a_{33} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1 \\ A d j (A) = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'} = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} A d j (A) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ H e n c e, A^{2} = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ N o w, w e h a v e t o p r o v e t h a t A^{- 1} = \frac{A^{2} - 3 I}{2} \\ R . H . S . = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}{2} \\ = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]}{2} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ = A^{- 1} = L . H . S . \\ H e n c e, p r o v e d . \end{array}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} | \begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A + c o s A \\ - c o s^{2} B - c o s B \end{array} & \begin{array}{l} c o s^{2} B + c o s B \\ - c o s^{2} C - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A - c o s^{2} B \\ + c o s A - c o s B \end{array} & \begin{array}{l} c o s^{2} B - c o s^{2} C \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} (c o s A + c o s B) * \\ (c o s A - c o s B) \\ + (c o s A - c o s B) \end{array} & \begin{array}{l} (c o s B + c o s C) * \\ (c o s B - c o s C) \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ T a k i n g (c o s A - c o s B) a n d (c o s B - c o s C) c o m m o n f r o m C_{1} a n d C_{2} r e s p e c t i v e l y . \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & 1 + c o s C \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 & c o s^{2} C + c o s C \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [1 | \begin{matrix} 1 & 1 \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 \end{matrix} |] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [(c o s B + c o s C + 1) - (c o s A + c o s B + 1)] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [c o s B + c o s C + 1 - c o s A - c o s B - 1] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) (c o s C - c o s A) = 0 \\ \Rightarrow c o s A - c o s B = 0 o r c o s B - c o s C = 0 o r c o s C - c o s A = 0 \\ \Rightarrow c o s A = c o s B o r c o s B = c o s C o r c o s C = c o s A \\ \Rightarrow ∠ A = ∠ C o r ∠ B = ∠ C \Rightarrow ∠ A = ∠ B \\ H e n c e, Δ A B C i s a n i s o s c e l e s t r i a n g l e . \end{array}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f t h e g i v e n points l i e o n a s t r a i g h t l i n e, t h e n t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e \\ points p a i r w i s e i s z e r o . \\ S 0, | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 7 & - 7 & 0 \\ - 2 & 3 & 0 \\ a & a & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ \Rightarrow 1 . | \begin{matrix} 7 & - 7 \\ - 2 & 3 \end{matrix} | = 21 - 14 = 7 u n i t s \\ A s 7 \neq 0 . H e n c e, t h e t h r e e points d o n o t l i e o n a s t r a i g h t l i n e f o r a n y v a l u e o f a . \end{array}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f a_{1}, a_{2}, a_{3}, \dots a_{r} b e t h e t e r m s o f G . P ., t h e n \\ a_{n} = A R^{n - 1} (w h e r e A i s t h e f i r s t t e r m a n d R i s t h e c o m m o n r a t i o o f t h e G . P .) \\ ∴ a_{r + 1} = A R^{r + 1 - 1} = A R^{r}; a_{r + 5} = A R^{r + 5 - 1} = A R^{r + 4} \\ a_{r + 9} = A R^{r + 9 - 1} = A R^{r + 8}; a_{r + 7} = A R^{r + 7 - 1} = A R^{r + 6} \\ a_{r + 11} = A R^{r + 11 - 1} = A R^{r + 10}; a_{r + 15} = A R^{r + 15 - 1} = A R^{r + 14} \\ a_{r + 17} = A R^{r + 17 - 1} = A R^{r + 16}; a_{r + 21} = A R^{r + 21 - 1} = A R^{r + 20} \\ ∴ T h e determinant b e c o m e s \\ | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} | \\ T a k i n g A R^{r}, A R^{r + 6} a n d A R^{r + 10} c o m m o n f r o m R_{1,} R_{2,} a n d R_{3} r e s p e c t i v e l y . \\ A R^{r} . A R^{r + 6} . A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} | \\ = A R^{r} . A R^{r + 6} . A R^{r + 10} | 0 | [? R_{1} a n d R_{2} a r e i d e n t i c a l r o w s] \\ = 0 \\ H e n c e, t h e g i v e n determinant i s i n d e p e n d e n t o f r . \end{array}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] = 0 \\ | A | = | \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow | \begin{matrix} 12 + x & 12 + x & 12 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ T a k i n g (12 + x) c o m m o n f r o m R_{1} \\ \Rightarrow (12 + x) | \begin{matrix} 1 & 1 & 1 \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow (12 + x) | \begin{matrix} 0 & 0 & 1 \\ 2 x & - 2 x & 4 + x \\ 0 & 2 x & 4 - x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (12 + x) [1 . | \begin{matrix} 2 x & - 2 x \\ 0 & 2 x \end{matrix} |] = 0 \\ \Rightarrow (12 + x) (4 x^{2} - 0) = 0 \Rightarrow 12 + x = 0 o r 4 x^{2} = 0 \\ \Rightarrow x = - 12 o r x = 0 \end{array}$

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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In many experimental set-ups, the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

1/f=1/v-1/u

-u+V=D

U=-(D-v)

Putting $\frac{1}{D - v} + \frac{1}{v} = \frac{1}{f}$

On solving v²-Dv+Dt=0

So v= $\frac{D}{2} ? \frac{\sqrt[]{D^{2} - 4 D f}}{2}$

When the objects distance is $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The image forms at $\frac{D}{2} ? \frac{\sqrt[]{D^{2} - 4 D f}}{2}$

Similarly when the objects distance is $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The image forms at $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The distance between the poles for these two objects distance is

\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}

\frac{D}{2} - \frac{\sqrt[]{D^{2} - 4 D t}}{2}

\sqrt[]{D^{2} - 4 D t}

Let d =

\sqrt[]{D^{2} - 4 D t}

If u = D/2+d/2 then the image is at v=D/2-d/2

The magnification m₁= $\frac{D + d}{D - d}$

If u =D-d/2then v= D+d/2

The magnification m₂= $\frac{D + d}{D - d}$

$\frac{m_{2}}{m_{1}} = \frac{{(D + d)}^{2}}{{(D - d)}^{2}}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0 \\ | A | = | \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0 \\ T a k i n g 7 c o m m o n f r o m C_{1} \\ \Rightarrow = 7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} 1 & s i n 3 θ \\ - 7 & - 2 \end{matrix} | + 2 | \begin{matrix} 1 & s i n 3 θ \\ 3 & c o s 2 θ \end{matrix} | = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 (c o s 2 θ - 3 s i n 3 θ) = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 c o s 2 θ - 6 s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 c o s 2 θ + s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 4 s i n^{3} θ - 4 s i n^{2} θ + 3 s i n θ = 0 \\ \Rightarrow - s i n θ (4 s i n^{2} θ + 4 s i n θ - 3) = 0 \\ - s i n θ = 0 o r 4 s i n^{2} θ + 4 s i n θ - 3 = 0 \\ ∴ θ = n π o r 4 s i n^{2} θ + 6 s i n θ - 2 s i n θ - 3 = 0 w h e n n \in I \\ \Rightarrow 2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3) = 0 \\ \Rightarrow (2 s i n θ + 3) (2 s i n θ - 1) = 0 \\ \Rightarrow &thinsp \end{array}$

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New answer posted

7 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If the coordinates of the vertices of an equilateral triangle with sides of length 'a' are $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ , then

${[\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}]}^{2} = \frac{3 a^{2}}{4}$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0, 0) and an object is placed at (-50 cm, 0). Find the coordinates of the image.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'.

Applying lens formula, we have

1/v-1/u=1/f $\frac{1}{D - v} + \frac{1}{v} = \frac{1}{f}$ Z

= $\frac{1}{- 50} + \frac{1}{25}$

V= 50cm

Magnification m = v/u= 50/-50=-1

So coordinates of image are (50cm, -1cm)

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