Physics

In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, ε₁/ε₂,, is:

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Physics Physics Ray Optics and Optical Instruments

Contributor-Level 10

ε? = 250 * Potential Gradient
ε? + ε? = 400 * Potential Gradient
(ε? )/ (ε? + ε? ) = 250/400 = 5/8
By solving above
ε? /ε? = 5/3

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5 months ago

A ray of light entering from air into a denser medium of refractive index 4/3, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle θ should be equal to :

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Physics Dual Nature of Radiation and Matter

Contributor-Level 10

By shell's law
(sin θ)/ (sin θ') = 4/3 . (i)
For TIR on second surface
sin θ' > sin θc
(θ' + θ' = 90)
sin (90 - θ') > sin θc
cos θ' > sin θc
1 - sin²θ' > sin²θc (By equation (i) )
1 - (3/4 sin θ)² > sin²θc
[sin θc = 3/4]
1 - (9/16)sin²θ > (9/16)
1 - 9/16 > (9/16)sin²θ
7/16 > (9/16)sin²θ
sinθ < 7/3

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5 months ago

What should be the order of arrangement of de-Broglie wavelength of electron (λ?), an α – particle (λα) and proton (λ?) given that all have the same kinetic energy ?

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Contributor-Level 9

λ = h/mv = h/√2mK
For same K:
λ ∝ 1/√m
λ? : λ? : λ? = 1/√m? : 1/√m? : 1/√4m?
As m? > m? ,
λ? > λ? > λ?

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5 months ago

A ray of laser of a wavelength 630 nm is incident at an angle of 30° at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

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Contributor-Level 9

μ? sin 30° = μ? sin θ

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Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : (where λ is the decay constant)

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Contributor-Level 9

3N? /4 = N? e?
⇒ e? = 4/3
t? = (1/λ) ln (4/3)
t? = ln2/λ
t? - t? = (1/λ)ln2 - (1/λ)ln (4/3)
= (1/λ)ln (2/ (4/3) = (1/λ)ln (3/2)

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5 months ago

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d? + a? sin(ωt); where d?, ω and a? are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

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Contributor-Level 9

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)

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Physics Class 11th

Match List I with List II.

List - I	List - II
(a) C - A - B = 0	(i) Diagram showing A + B = C
(b) A - C - B = 0	(ii) Diagram showing C + B = A
(c) B - A - C = 0	(iii) Diagram showing A + C = B
(d) A + B = -C	(iv) Diagram showing A + B + C = 0

Choose the correct answer from the options given below :

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Contributor-Level 9

From triangle law of addition, we can clearly write answer.

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5 months ago

Physics Physics Wave Optics

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

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Physics Physics Mechanical Properties of Solids

Contributor-Level 10

Amplitude is proportional to the slit – width, thus,

$\frac{A_{1}}{A_{2}} = 3$

$\frac{l_{m a x}}{l_{m i n}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}} = \frac{{(\frac{A_{1}}{A_{2}} + 1)}^{2}}{{(\frac{A_{1}}{A_{2}} - 1)}^{2}} = \frac{{(3 + 1)}^{2}}{{(3 - 1)}^{2}} = 4 .$

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5 months ago

If y, K and $η$ are the value of Young's modulus, bulk modulus of rigidity of any material respectively. Choose

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Contributor-Level 10

If $μ$ is Poisson's ratio,

Y = 3K (1 - 2 $μ$ ) ……… (1)

and Y = 2 $n (1 + μ)$ ……… (2)

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

$\frac{3}{Y} = \frac{1}{η} + \frac{1}{3 k} \Rightarrow K = \frac{η Y}{9 η - 3 Y}$

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Physics Thermodynamics

n mole of perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A->B : Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and Pressure changes from P₁ to P₂.

B->C : Isobaric compression at pressure P₂ to initial volume V₁.

C->A : Isobaric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is:

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