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10 months ago

Physics Class 12th

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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Raj Pandey

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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10 months ago

Physics Class 12th

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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Raj Pandey

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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10 months ago

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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Raj Pandey

Contributor-Level 9

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

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10 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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alok kumar singh

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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10 months ago

Physics Physics Mechanical Properties of Solids

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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alok kumar singh

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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10 months ago

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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alok kumar singh

Contributor-Level 10

$W_{A B} = n R T l n \frac{2 V_{1}}{V_{1}} = n R T l n 2 .$

$W_{B C} = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{1}{2} n R T$

[At B, 2P₂ V₁ = nRT]

$W_{C A} = 0$ [CA is isochoric process].

$W_{A B C A} = W_{A B} + W_{B C} + W_{C A} = n R T (l n (2) - \frac{1}{2})$

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10 months ago

Physics Physics System of Particles and Rotational Motion

Moment of inertia (M.I) of four bodies, having same mass and radius, and reported as:

I₁ = M.I. of thin circular ring about its diameter

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

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alok kumar singh

Contributor-Level 10

$l_{1} = \frac{1}{2} m r^{2}$

$\begin{array}{l} l_{2} = \frac{1}{2} m r^{2} \\ l_{3} = \frac{1}{2} m r^{2} \\ l_{4} = \frac{2}{5} m r^{2} \end{array}$

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10 months ago

Physics Class 11th

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The

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alok kumar singh

Contributor-Level 10

$m ω^{2} . \frac{2 d}{3} = \frac{G . m . 2 m}{d^{2}} \Rightarrow ω = \sqrt[]{\frac{3 G m}{d^{3}}}$

$T = \frac{2 π}{ω} = 2 π \sqrt[]{\frac{d^{3}}{3 G m}}$

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10 months ago

Physics Class 12th

If an emitter current is changed by 4mA, the collector current changes by 3.5mA. The value of b will be:

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alok kumar singh

Contributor-Level 10

$α = \frac{Δ l_{C}}{Δ l_{E}} = \frac{3.5}{4} = \frac{7}{8}$

$β = \frac{α}{1 - α} = \frac{7 / 8}{1 - 7 / 8} = 7$

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10 months ago

Physics Physics Ray Optics and Optical Instruments

The focal length f is related to the radius of curvature r of the spherical convex mirror by:

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alok kumar singh

Contributor-Level 10

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2} r .$

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