Tags Class 12th

Class 12th

Get insights from 12k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

12k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Class 12th Physics

For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

0 Follower 25 Views

Payal Gupta

Contributor-Level 10

When glass slab is not put in between lens and screen

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{12} - \frac{1}{- 240} = \frac{1}{f} . . . . . . (1)$

When glass slab is put in between lens and screen

Shift = $t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} c m$

$\Rightarrow \frac{3}{35} - \frac{1}{- x} = \frac{1}{f} . . . . . . . . . (2)$

With the help of equations (1) and (2), we can write

$\frac{3}{35} - \frac{1}{- x} = \frac{1}{12} - \frac{1}{- 240}$

$\Rightarrow | Δ u | = 560 - 240 = 320 c m = 3.2 m$

0 0

New answer posted

a year ago

Class 12th Dual Nature of Radiation and Matter

The ratio of wavelengths of proton and deuteron accelerated by potential V_p and V_d is $1 : \sqrt[]{2}$ . Then, the ratio of V_p to V_d will be

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

According to de-Broglie's hypothesis, we can write

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m e V}}$

$\Rightarrow \frac{λ_{P}}{λ_{D}} = \frac{\sqrt[]{2 m_{D} e V_{D}}}{\sqrt[]{2 m_{P} e V_{P}}} = \frac{1}{\sqrt[]{2}} \Rightarrow \frac{2 m_{P} V_{D}}{m_{P} V_{P}} = \frac{1}{2} \Rightarrow \frac{V_{P}}{V_{D}} = 4 : 1$

0 0

New answer posted

a year ago

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

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

$C_{1} = 4 π R_{1} a n d$

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

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

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

0 0

New answer posted

a year 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

0 Follower 12 Views

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

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

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

0 0

New answer posted

a year 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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

0 0

New question posted

a year ago

Class 12th Chemistry

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

0 Follower 1 View

New answer posted

a year ago

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

0 Follower 6 Views

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

0 0

New answer posted

a year ago

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

0 Follower 33 Views

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 $H_{2} O (l)$ $_{}$

0 0

New answer posted

a year ago

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:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Using snell's law –

sin I = $? s i n ? r$

->sin45° = $? s i n 30 °$

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

0 0

New answer posted

a year ago

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° = $\frac{3}{4}$ )

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

B_v = 6 * 10^-5 T

B_v = B_N sin37°

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

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience