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6 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Ten

BeO reacts with HF in presence of ammonia to give [A] which on thermal decomposition produces [B] and ammonium fluoride. Oxidation state of Be in [A] is____________.

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Payal Gupta

Contributor-Level 10

$B e O + 2 N H_{3} + 4 H F \to {(N H_{4})}_{2} [B e F_{4}]$

${(N H_{4})}_{2} [B e F_{4}] \to_{}^{Δ} B e F_{2} + 2 N H_{4} F$

here, oxidation state of be in A is +2.

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6 months ago

Class 12th Structure of Atom

Consider the following set of quantum numbers.

	n	$l$	$m_{l}$
A.	3	3	-3
B.	3	2	-2
C.	2	1	+1
D.	2	2	+2

The number of correct sets of quantum numbers is____________.

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Payal Gupta

Contributor-Level 10

n = 1, 2, 3 …….; $l = 0$ ……. to n – 1, $m_{l} = l . . . . . . . 0 . . . . . . + l$

A. n = 3 $l = 3$ $m_{l} = - 3$ is incorrect as $l$ can not be equal to n.

B. n = 3 $l = 2$ $m_{l}$ = -2 is correct set.

C. n = 2 $l = 1$ $m_{l}$ = +2 is incorrect set as $l$ can not be equal to n.

So; correct set of quantum numbers is B and C.

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6 months ago

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

A paramagnetic sample shows a net magnetisation of $6 A / m$ when it is placed in an external magnetic field of $0.4 T$ at a temperature of $4 K$ . When the sample is placed in an external magnetic field of $0.3 T$ at a temperature of $24 K$ , then the magnetisation will be :

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alok kumar singh

Contributor-Level 10

Sol. $ρ v g - m g = m a$

$\Rightarrow \frac{ρ v g}{m} = g + a$

$\Rightarrow m = \frac{ρ v g}{g + a}$

$\Rightarrow \frac{10^{3} (\frac{4}{3} π * 10^{- 6}) (9.8)}{9.898}$

$\Rightarrow 4.15 g m$

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6 months ago

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

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 $N$ over a distance of $15 m$ . Thereafter, he gets progressively tired and his applied force reduces linearly with distance to $100 N$ . The total distance through which the box has been moved is $30 m$ . What is the work done by the person during the total movement of the box?

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alok kumar singh

Contributor-Level 10

Sol. (n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

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6 months ago

Class 12th Some Basic Concepts of Chemistry

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ &n

...more

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Payal Gupta

Contributor-Level 10

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$ &n

...more

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} * 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} * 100 % = 6.5 %$

$% o f O = \frac{60}{116} * 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$

H 6.5 g $\frac{6.5}{1} = 6.5$ $\frac{6.5}{3.23} \approx 2$

O 51.7 g $\frac{51.7}{16} = 3.23$ $\frac{3.23}{3.23} = 1$

Ratio of moles of C, H and O is 1 : 2 : 1

A. CH₃COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1

B. HCHO ⇒ C : H : O = 1 : 2 : 1

C. CH₃OOCH₃ ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1

D. CH₃CHO ⇒ C : H : O = 2 : 4 : 1

So, CH₃COOH and HCHO formulae agree with the given data

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

The electric field of a plane electromagnetic wave is given by $\overset{?}{E} = E_{0} (\hat{x} + \hat{y}) s i n ? (k z - ω t)$ Its magnetic field will be given by:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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6 months ago

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

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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6 months ago

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

A series $L - R$ circuit is connected to a battery of emf $V$ . If the circuit is switched on at $t =$ 0 , then the time at which the energy stored in the inductor reaches $(1 / n)$ times of its maximum value, is :

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alok kumar singh

Contributor-Level 10

Sol. Mono atomic $? C_{v} = \frac{3 R}{2} C_{p} = \frac{5 R}{2}$

Di- atomic $? C_{V} = \frac{5 R}{2} C_{p} = \frac{7 R}{2}$

(Rigid)

Di-atomic $? C_{v} = \frac{7 R}{2} C_{p} = \frac{9 R}{2}$

(Non-Rigid)

Tri-atomic $? C_{V} = 3 R C_{P} = 4 R$

(Rigid)

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6 months ago

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

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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Vishal Baghel

Contributor-Level 10

$h = \frac{u^{2}}{2 g} \Rightarrow u = \sqrt[]{2 g h}$

$\frac{h}{3} = \sqrt[]{2 g h} t - 5 t^{2}$

$5 t^{2} - \sqrt[]{20 h} t + \frac{h}{3} = 0$

$t = \frac{\sqrt[]{20 h} \pm \sqrt[]{20 h - 4 * 5 * \frac{h}{3}}}{10}$

$∴ \frac{t_{1}}{t_{2}} = \frac{\sqrt[]{20 h} - \sqrt[]{\frac{40}{3} h}}{\sqrt[]{20 h} + \sqrt[]{\frac{40}{3} h}} = \frac{1 - \sqrt[]{\frac{2}{3}}}{1 + \sqrt[]{\frac{2}{3}}} = \frac{\sqrt[]{3} - \sqrt[]{2}}{\sqrt[]{3} + \sqrt[]{2}}$

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6 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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