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

Give an example of matrices $A$ , $B$ , and $C$ such that $A B = A C$ , where $A$ is a non-zero matrix, but $B \neq C$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], B = [\begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix}] a n d C = [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}] \\ A B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 + 0 & 2 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}] \\ A C = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}] \Rightarrow A C = [\begin{matrix} 1 + 0 & 2 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}] \\ H e n c e, A B = A C f o r m a t r i x A i s n o n - z e r o a n d B \neq C . \end{array}$

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

If $A = [3 5], B = [7 3]$ , then find a non-zero matrix $C$ such that $A C = B C$ .

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Payal Gupta

Contributor-Level 10

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

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

If X and Y are $2 * 2$ matrices, then solve the following matrix equations for X and Y:

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}], 3 X + 2 Y = [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] .$

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : \\ 2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \dots (1) \\ 3 X + 2 Y = [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \dots (2) \\ M u l t i p l y i n g e q . (1) b y 3 a n d e q . (2) b y 2, w e g e t, \\ 3 [2 X + 3 Y] = 3 [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \Rightarrow 6 X + 9 Y = [\begin{matrix} 6 & 9 \\ 12 & 0 \end{matrix}] \dots (3) \\ 2 [3 X + 2 Y] = 2 [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \Rightarrow 6 X + 4 Y = [\begin{matrix} - 4 & 4 \\ 2 & - 10 \end{matrix}] \dots (4) \\ O n s u b t r a c t i n g e q . (4) f r o m e q . (3) w e g e t \\ 5 Y = [\begin{matrix} 6 + 4 & 9 - 4 \\ 12 - 2 & 0 + 10 \end{matrix}] \\ 5 Y = [\begin{matrix} 10 & 5 \\ 10 & 10 \end{matrix}] \Rightarrow Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] \\ N o w, p u t t i n g t h e v a l u e o f Y i n e q u a t i o n (1) w e g e t, \\ 2 X + 3 [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \\ \Rightarrow 2 X + [\begin{matrix} 6 & 3 \\ 6 & 6 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] - [\begin{matrix} 6 & 3 \\ 6 & 6 \end{matrix}] \\ \Rightarrow 2 X = [\begin{matrix} 2 - 6 & 3 - 3 \\ 4 - 6 & 0 - 6 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} - 4 & 0 \\ - 2 & - 6 \end{matrix}] \\ X = \frac{1}{2} [\begin{matrix} - 4 & 0 \\ - 2 & - 6 \end{matrix}] \Rightarrow X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] \\ H e n c e, X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] a n d Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] . \end{array}$

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New answer posted

6 months ago

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

Solve for x and y :

Kindly Consider the following

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Payal Gupta

Contributor-Level 10

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

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New answer posted

6 months ago

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

Given Is ${(A B)}^{'} = B^{'} A^{'}$ ?

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Payal Gupta

Contributor-Level 10

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

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New answer posted

6 months ago

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

Show by an example that for $A \neq O, B \neq O$ , $A B = O$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] \\ A B = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] \\ A B = [\begin{matrix} 1 - 1 & 1 - 1 \\ - 1 + 1 & - 1 + 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 \\ H e n c e, A = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] . \end{array}$

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New answer posted

6 months ago

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

If possible, find $B A$ and $A B$ , where

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Payal Gupta

Contributor-Level 10

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

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New answer posted

6 months ago

Class 12th Electrostatic Potential and Capacitance

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\hat{n}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $σ \hat{n}$ / $ε_{0}$ .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed lo

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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

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alok kumar singh

Contributor-Level 10

2.16 Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body is $E_{2} .$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

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$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

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New answer posted

6 months ago

Class 12th Electrostatic Potential and Capacitance

2.31 (a) Two large conducting spheres carrying charges $Q_{1}$ and $Q_{2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_{1} Q_{2}$ /4 $π ε_{0} r^{2}$ , where r is the distance between their centres?

(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if t

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(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

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alok kumar singh

Contributor-Level 10

2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elli

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2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

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New answer posted

6 months ago

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

If then verify ${(B A)}^{2} \neq B^{2} A^{2}$

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Payal Gupta

Contributor-Level 10

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

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