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Class 12th Chemistry

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is___________.

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Vishal Baghel

Contributor-Level 10

$2 K M n O_{4} + 3 H_{2} O_{2} \to_{}^{B a s i c m e d i u m} 2 M n O_{2} + 3 O_{2} + 2 H_{2} O + 2 K O H$

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Class 12th Chemistry

Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if iron is ___________.

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Vishal Baghel

Contributor-Level 10

Bauxite = AlOx (OH)_3-2x (where O < x < 1)

Siderite = FeCO₃

Cuprite = Cu₂O

Calamine = ZnCO₃

Haemetite = Fe₂O₃

Kaolinite = Al₂ (OH)₄Si₂O₅

Malachite = CuCO₃ Cu (OH)₂

Magneite = Fe₃O₄

Sphalerite = ZnS

Limonite = Fe₂O₃.3H₂O

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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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alok kumar singh

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 Chemical Kinetics

[A] -> [B]

Reactant Product

If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x * 10^-⁶ s^-¹. The value of x is___________. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{70 * 60} = \frac{6930}{7 * 6} * 1 0^{- 6}$

= 165 * 10^-6 s^-1

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Class 12th Solutions

When a certain amount of solid A is dissolved in 100g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is_________.

(Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$\frac{P^{0} - P_{S}}{P^{0}} = \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

$\frac{P^{0} - \frac{P^{0}}{2}}{P^{0}} = \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

$n_{s o l u t e} = \frac{n_{s o l v e n t}}{2} = \frac{100}{18 * 2} = 2.78 m o l$

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Class 12th 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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alok kumar singh

Contributor-Level 10

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 Þ 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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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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alok kumar singh

Contributor-Level 10

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

20 – i_maxR – 8 = 0

i_max R = 12At 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 max^m 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 * 10^-3 R = 12

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

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Class 12th Physics Electrostatic Potential and Capacitance

A parallel plate capacitor with width 4cm, length 8cm and separation between the plates of 4mm is connected to a battery of 20V. A dielectric slab of dielectric constant 5 having length 1cm, width 4cm and thickness 4mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ___________ $\in_{0} J .$

(Where $\in_{0}$ is the permittivity of free space)

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Vishal Baghel

Contributor-Level 10

$C_{e f f} = [\frac{ε_{0} (7 * 4)}{4 / 10} + \frac{5 ε_{0} (1 * 4)}{4 / 10}] * 1 0^{- 2}$

$C_{e f f} = 1.2 ε_{0}$

Energy = $\frac{1}{2} C_{e f f} V^{2}$ $\frac{}{}_{}^{}$

$\frac{1}{2} (1.2 ε_{0}) (20) (20) = 240 ε_{0}$

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Class 12th Physics

A modulating signal 2sin (6.28 * 10⁶)t is added to the carrier signal 4 sin(12.56 *10⁹)t for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passes through a band pass filter. The bandwidth of the output signal of band pass filter will be_________ MHz.

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