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Class 12th Physics

An audio signal V_m = 20 sin 2p(1500t) amplitude modulates a carrier

V_C = 80 sin 2p (100,000t)( The value of percent modulation is______________.

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alok kumar singh

Contributor-Level 10

Percentage Modulation = $\frac{A_{m}}{A_{c}} * 100 = \frac{20}{80} * 100 = 25 %$

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9 months ago

Class 12th Physics

In connection with the circuit drawn below, the value of current flowing through 2 $k Ω$ resistor is________ * 10^-4A.

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alok kumar singh

Contributor-Level 10

Potential difference across 2k $Ω$ is 5V, thus current through it,

$i = \frac{5}{2 * 1 0^{3}} = 25 * 1 0^{- 4} A .$

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9 months ago

Class 12th 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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9 months ago

Class 12th Electromagnetic Induction

The current in a coil of self inductance 2.0 H is increasing according to I = 2sin(t2)A. The amount of energy spent during the period when current changes from 0 to 2 A is…………J.

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Raj Pandey

Contributor-Level 9

Given, L = 2H

l = 2 sin (t²)

$u = \int_{0}^{2} L i d i = \frac{L}{2} {[i^{2}]}_{0}^{2} = \frac{L}{2} [4 - 0]$

$u = \frac{2}{2} * 4 = 4 J$

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9 months ago

Class 12th Maths

Let g(x) = x + 2| - 3|. If a denotes the number of relative minima, b denotes the number of relative maxima and c denotes the product of the zeros. Then the value of (a + 2b - c) is __________

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Vishal Baghel

Contributor-Level 10

c = -5, a = 2, b = 1

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9 months ago

Class 12th Maths

In a geometric progression the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49 and sum of the first and the third term is 35. The fifth term of the G.P. is p, the value of 4p is __________

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Vishal Baghel

Contributor-Level 10

Let five terms in G.P. be a/r², a/r, a, ar, ar²
Then, a (r? ² + r? ¹ + 1 + r + r²) / (1/a) (r? ² + r? ¹ + 1 + r? ¹ + r? ²) = 49
⇒ a² = 49 ⇒ a = ±7
Also, a/r² + a = 35
Therefore, a = -7 is not possible
Now, fifth term = ar² = a (7/28) = p ⇒ 4p = 7

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9 months ago

Class 12th Physics

The total current supplied to the circuit is shown in figure by the 5 V battery is…………….A.

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Raj Pandey

Contributor-Level 9

$\equiv R e q = 2.5 Ω$

$l = \frac{v}{R_{e q}} = \frac{5}{2.5} = 2 A$

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9 months ago

Class 12th Physics

A resistor develops 300 J of thermal energy in 15s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10s is…………….J.

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Raj Pandey

Contributor-Level 9

$H = I^{2} R t = 2^{2} * R * 15$

$\Rightarrow 300 = 60 R$

$R = 5 Ω$

Now for 3A, time = 10 sec

$H' = l^{2} R t = 3^{2} * 5 * 10 = 450 J$

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9 months ago

Class 12th Inverse Trigonometric Functions

The value of cot^-1(2² + 1/2) + cot^-1(2³ + 1/2²) + ……∞ is equal to tan^-1 λ, then the value of λ is equal to __________

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Vishal Baghel

Contributor-Level 10

T? = cot? ¹ (2²? ¹ + 1/2? ) = cot? ¹ (1 + 2²? ¹)/2? ) = tan? ¹ (2? / (1 + 2²? ¹)
T? = tan? ¹ (2? ¹ - 2? ) / (1 + 2? ¹ . 2? ) = tan? ¹ (2? ¹) – tan? ¹ (2? )
S∞ = π/2 - tan? ¹ (2) = cot? ¹ (2) = tan? ¹ (1/2)

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9 months ago

Class 12th Physics Electrostatic Potential and Capacitance

The equivalent capacitance between points A and B in below shown figure will be…………. $μ F .$