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Class 12th Physics Alternating Current

In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be--------.

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Payal Gupta

Contributor-Level 10

As we know that Q = $\frac{ω L}{R}$

$\Rightarrow \frac{Q'}{Q} = \frac{L'}{R'} * \frac{R}{L} = (\frac{L'}{L}) * (\frac{R}{R'}) = 2 * 2 \Rightarrow Q' = 4 Q = 4 * 100 = 400$

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7 months ago

Class 12th Physics Electrostatic Potential and Capacitance

If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $λ$ where C/V = $λ$

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Vishal Baghel

Contributor-Level 10

$λ = \frac{c}{v} = \frac{Q^{3}}{W^{2}} = \frac{I^{3} T^{3}}{M^{2} L^{4} T^{- 4}} \Rightarrow [λ] = [M^{- 2} L^{- 4} T^{7} l^{3}]$

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7 months ago

Class 12th Maths

$\underset{x \to 0}{l i m} \frac{\int_{0}^{x^{2}} (s i n \sqrt[]{t}) d t}{x^{3}}$ is equal to:

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alok kumar singh

Contributor-Level 10

$\underset{x \to 0}{l i m} \frac{\int_{0}^{x^{2}} (s i n \sqrt[]{t}) d t}{x^{3}}$ $(\frac{0}{0})$ applying L' Hospital rule

$= \underset{x \to 0}{l i m} \frac{s i n (\sqrt[]{x^{2}}) . 2 x}{3 x^{2}}$

$= \frac{2}{3}, f o r x > 0$

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7 months ago

Class 12th Physics

The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude wave x * 10^-2. The value of x is----------.

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Payal Gupta

Contributor-Level 10

Modulation Index $(μ) = \frac{A_{m a x} - A_{m i n}}{A_{m a x} + A_{m i n}} = \frac{16 - 8}{16 + 8} = \frac{1}{3} = 0.33 = 33 * 1 0^{- 2}$

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7 months ago

Class 12th Physics

In an electrical circuit, a battery is connected to pass 20 C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15V. The work done by the battery is--------- J.

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Payal Gupta

Contributor-Level 10

$W = q Δ V = 20 * 15 = 300 J$

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7 months ago

Class 12th Physics

The circuit contains two diodes each with a forward resistance of 50 $Ω$ and with infinite reverse resistance. If the battery voltage is 6 V, the current through the 120 $Ω$ resistance is---------mA.

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Payal Gupta

Contributor-Level 10

D₁ is forward biased & D₂is reverse biased.

$l = \frac{6}{120 + 130 + 50} = \frac{6}{300} = 2 * 1 0^{- 2} A = 20 * 1 0^{- 3} A = 20 m A$

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7 months ago

Class 12th Physics

The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $\vec{a}, \vec{b} a n d \vec{c}$ respectively. Then choose the correct relation for these vectors.

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Vishal Baghel

Contributor-Level 10

$\hat{r} - \hat{i} = 2 c o s θ \hat{n} . . . . . . (1)$

$\hat{i} . \hat{n} = - c o s θ . . . . . . (2)$

$\Rightarrow \hat{r} = \hat{i} - 2 (\hat{i} . \hat{n}) \hat{n}$

$\Rightarrow \vec{b} = \vec{a} - 2 (\vec{a} . \vec{c}) \vec{c}$

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7 months ago

Class 12th Differential Equations

If $\int \frac{c o s x - s i n x}{\sqrt[]{8 - s i n 2 x}} d x = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c,$ where c is a constant of integration, then the ordered pair (a, b) is equal to:

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alok kumar singh

Contributor-Level 10

$\int \frac{(c o s x - s i n x) d x}{\sqrt[]{8 - s i n 2 x}} = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c$

$\int \frac{(c o s x - s i n x) d x}{\sqrt[]{9 - {(s i n x + c o s x)}^{2}}} =$

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$= \int \frac{d t}{\sqrt[]{3^{2} - t^{2}}} = s i n^{- 1} \frac{t}{3} + c = s i n^{- 1} (\frac{s i n x + c o s x}{3}) + c$

= a = 1, b = 3

$∴ (a, b) \equiv (1, 3)$

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7 months ago

Class 12th Physics Wave Optics

In a Young's double slit experiment two slits are separated by 2mm and the screen is placed one meter away. When a light of wavelength 500nm used, the fringe separation will be

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Payal Gupta

Contributor-Level 10

$β = \frac{λ D}{d} = \frac{500 * 1 0^{- 9} * 1}{2 * 1 0^{- 3}} = 250 * 1 0^{- 6} m = 250 * 1 0^{- 3} m m = 0.25 m m$

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7 months ago

Class 12th Electric Charges and Fields

An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.

$[g = 9.8 m / s^{2}; s i n 30 ° = \frac{1}{2}; c o s 30 ° = \frac{\sqrt[]{3}}{2}]$

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Vishal Baghel

Contributor-Level 10

$N = m g c o s 30 ° + q E s i n 30 °$

$a = \frac{m g s i n θ - q E c o s θ - μ N}{m} = 2.30 m / s^{2}$

$S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow t = \sqrt[]{\frac{2 l}{a}} = 1.31 s e c$

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