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Class 12th Physics

In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be 6V and 2V respectively. The modulation index is

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Payal Gupta

Contributor-Level 10

Modulation index = $\frac{A_{m a x} - A_{m i n}}{A_{m a x} + A_{m i n}} = \frac{6 - 2}{6 + 2} = 0.5 \approx 50 %$

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Class 12th Chemistry

Given below are the critical temperature of some of the gases:

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Class 12th Chemistry

Match List-I with List-II

List – I		List – II
(A)	4NH₃(g) + 5O₂(g) -> 4NO(g) + 6H₂O(g)	(I)	NO(g)
(B)	N₂(g) + 3H₂(g) -> 2NH₃(g)	(II)	H₂SO₄(I)
(C)	C₁₂H₂₂O₁₁(aq) + H₂O(I) -> C₆H₁₂O₆ + C₆H₁₂O₆ Glucose Fructose	(III)	Pt(s)
(D)	2SO₂(g) + O₂(g) -> 2SO₃(g)	(IV)	Fe(s)

Choose the correct answer from the options given below

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Vishal Baghel

Contributor-Level 10

(a) Nitric oxide formation -> Pt is used as catalyst

(b) Haber's process -> Fe is used as catalyst

(d) SO₃ formation -> NO is used as catalyst

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Class 12th Electrochemistry

Match List-I with List-II

List – I		List – II
(A)	Cd(s) + 2Ni(OH)₃(s) -> CdO(s) + 2Ni(OH)₂(s) + H₂O(I)	(I)	Primary battery
(B)	Zn(Hg) + HgO(s) -> ZnO(s) + Hg(I)	(II)	Discharging of secondary battery
(C)	2PbSO₄(s) + 2H₂O(I) -> Pb(s) + PbO₂(s) + 2H₂SO₄(aq	(III)	Fuel cell
(D)	2H₂(g) + O₂(g) -> 2H₂O(I)	(IV)	Charging of secondary battery

Choose the correct answer from the options given below

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Vishal Baghel

Contributor-Level 10

(a) $C d (s) + 2 N i {(O H)}_{3} (s) \to C d O (s) + 2 N i {(O H)}_{2} (s) + H_{2} O (l)$ $_{}_{}_{}$

During discharging of secondary battery this reaction takes place.

(b) $Z n (H g) + H g O (s) \to Z n O (s) + H g (l)$

Primary battery mercury cell reaction

During charging of secondary battery PbSO4 reacts and $H_{2} S O_{4}$ $_{}_{}$ generated

(d) $H_{2} & O_{2}$ $_{}_{}$ reacts in fuel cell to form $H_{2} O (l)$ $_{}$

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Class 12th Physics

Light enters form air into a given medium at angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is:

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alok kumar singh

Contributor-Level 10

Using snell's law –

sin I = $? s i n ? r$

->sin45° = $? s i n 30 °$

$? = \frac{s i n 45}{s i n 30} = \frac{1}{\sqrt[]{2}} * 2 = \sqrt[]{2}$

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Class 12th Physics

The vertical component of the earth's magnetic field is 6 * 10^-5 T at any place where the angle of dip is 37°. The earth's resultant magnetic field at that place will be (Given tan37° = $\frac{3}{4}$ )

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alok kumar singh

Contributor-Level 10

B_v = 6 * 10^-5 T

B_v = B_N sin37°

$∴ B_{N} = \frac{B_{v}}{s i n 37 °} = \frac{6 * 1 0^{- 5}}{3 / 5} = 1 0^{- 4} T$

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Class 12th Physics

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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alok kumar singh

Contributor-Level 10

W_total= W_D_®E + W_E->_F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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Class 12th Physics

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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alok kumar singh

Contributor-Level 10

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

...more

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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Class 12th Physics

An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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alok kumar singh

Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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Class 12th Physics Nuclei

Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.

Choose the correct option from the following options

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alok kumar singh

Contributor-Level 10

Radius, R = R₀A^1/3

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