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Class 12th Maths

Let X = $[\begin{matrix} 1 & 1 & 1 \end{matrix}] a n d A = [\begin{matrix} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{matrix}] . F o r k \in N, i f X' A^{k} X = 33,$ then k is equal to

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Payal Gupta

Contributor-Level 10

$A = [\begin{matrix} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{matrix}]$

$A^{2} = [\begin{matrix} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{matrix}] [\begin{matrix} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{matrix}]$

$= [\begin{matrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I + B, B = [\begin{matrix} 0 & 0 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$A^{4} = [\begin{matrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] B^{2} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0$

$A^{2 n} = {(I + B)}^{n}$

I + nB + 0 + 0 +……….

$A^{2 n} = I + [\begin{matrix} 0 & 0 & 6 n \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 6 n \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$X' A^{k} x = [33]$

$= [111] A^{k} [\begin{matrix} 1 & 1 & 1 \end{matrix}]$

$[111] A^{2 n} [\begin{matrix} 1 & 1 & 1 \end{matrix}] (k = 2 n)$

= $[1 + 1 + 6 n + 1] = [6 n + 3] = [33] n = 5$

k = 10

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Class 12th Chemistry

The number of chlorine atoms in bithionol is_____________.

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Payal Gupta

Contributor-Level 10

Structure of Bithinol is;

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Class 12th Chemistry

A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344mL of H₂ gas at STP. The number of alcoholic hydrogens present in compound 'X' is_________.

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Payal Gupta

Contributor-Level 10

Mole of polyhydric alcohol = $\frac{1.84 * 1 0^{- 3}}{92} = 2.0 * 1 0^{- 5} m o l e .$

Mole of H₂ gas produced = $\frac{1.344}{22400} = 6.0 * 1 0^{- 5} m o l e .$

No of OH gp present = $\frac{6.0 * 1 0^{- 5}}{2.0 * 1 0^{- 5}} = 3$

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Class 12th Chemistry

Consider the following sulphur based oxoacids.

H₂SO₃, H₂SO₄, H₂S₂O₈ and H₂S₂O₇.

Amongst these oxoacids, the number of those with peroxo (O-O) bond is______________.

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Payal Gupta

Contributor-Level 10

Only H₂S₂O₈ has per-oxo bond.

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Class 12th Chemistry Coordination Compounds

Sum of oxidation state (magnitude) and coordination number of cobalt in $N a [C o (b p y) C l_{4}]$ is___________.

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Payal Gupta

Contributor-Level 10

Oxidation state of Co = 3

And co-ordination No = 6

Sum = 3 + 6 = 9

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8 months ago

Class 12th Chemistry

Assuming 1 $μ g$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is __________* 10^-1 $μ g$

[Given: In 10 = 2.303; log 2 = 0.30]

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Payal Gupta

Contributor-Level 10

$λ t = l n \frac{N_{o}}{N_{t}}$

$\frac{0.693}{t_{1 / 2}} * t = l n \frac{1}{N_{t}}$

Or $\frac{0.693 * 100}{30} = l n \frac{1}{N_{t}}$

$\frac{1}{N_{t}} = 10$

$∴ N_{t} = \frac{1}{10} = 0.1$

Or N_t = 1 * 101 $μ g$

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Class 12th Chemistry

1.80g of solute A was dissolved in 62.5cm³ of ethanol and freezing point of the solution was found to be 155.1 K . The molar mass of solute A is __________g mol^-1.

[Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm^-3. Freezing point depression constant of ethanol is 2.00 K kg mol^-1]

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Payal Gupta

Contributor-Level 10

$Δ T_{f} = K_{f} * m$

$(156 - 155.1) = 2 * \frac{1.8}{M W} * \frac{1000}{(62.5 * 0.8)}$

0.9 = $\frac{2 * 1.8 * 1000}{M W * 50}$

$∴ M W = 80 g m / m o l$

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8 months ago

Class 12th Physics

A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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Payal Gupta

Contributor-Level 10

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 * 1 0^{- 3} * 5 * 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 * 1 0^{6} * 100 * 200$

= 10A

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Class 12th Physics Nuclei

Two radioactive materials A and B have decay constants 25 $λ$ and 16 $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be “e” after a time $\frac{1}{a λ} .$ The value of a is__________.

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Payal Gupta

Contributor-Level 10

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 N_A = N_B = N₀

after t = $\frac{1}{a λ} : - \frac{N_{B}}{N_{A}} = \frac{N_{0} e^{- 16 λ t}}{N_{0} e^{- 25 λ t}} = e^{(25 λ - 16 λ) t}$

e = $e^{(9 λ \frac{1}{a λ})}$

$\Rightarrow \frac{9}{a} = 1 \Rightarrow a = 9$

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8 months ago

Class 12th Physics

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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Payal Gupta

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $(μ A) : -$

$20 - i_{m i n} R - i_{m i n} R_{L} = 0$

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At maxm zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 103 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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