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Class 12th Physics Electrostatic Potential and Capacitance

Capacitance of an isolated conduction sphere of radius R₁ becomes n times when it is enclosed by a concentric conducting sphere of radius R₂ connected to earth. The ration of their radii $(\frac{R_{2}}{R_{1}})$ is

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Payal Gupta

Contributor-Level 10

$C_{1} = 4 π R_{1} a n d$

$C_{2} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2} C_{1}}{R_{2} - R_{1}}$

$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2}}{R_{2} - R_{1}} = n$

$\Rightarrow \frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

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11 months ago

Class 12th Physics Moving Charges and Magnetism

The electric current in a circular coil of 2 turns produces a magnetic induction B₁ at its centre. The coil is unwound and is rewound into circular coil of 5 turns and at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction B₂ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is

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Payal Gupta

Contributor-Level 10

$B_{1} = \frac{N_{1} μ_{0} l}{2 R}$ given N1 = 2

When new loop is made the length of wire remains same, so

$N_{2} * 2 π r = N_{1} * 2 π R \Rightarrow r = \frac{N_{1} R}{N_{2}}$

$\Rightarrow \frac{B_{2}}{B_{1}} = {(\frac{N_{2}}{N_{1}})}^{2} = \frac{25}{4}$

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11 months ago

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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