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11 months ago

Class 12th Semiconductor Electronics

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

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alok kumar singh

Contributor-Level 10

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let the solution curve of the differential equation

$x \frac{d y}{d x} - y = \sqrt[]{y^{2} + 16 x^{2}}, y (1) = 3 b e y = y (x) .$ Then y(2) is equal to:

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} = \frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x}$

Put y = $V \cdot x$

differentiable worst = x

$\frac{d y}{d x} = V + x \cdot \frac{d V}{d x}$

V + $x \frac{d V}{d x} = V + \sqrt[]{V^{2} + 16}$

$x \frac{d V}{d x} = \sqrt[]{V^{2} + 16}$

apply variable separable method

$\int \frac{d V}{\sqrt[]{V^{2} + 16}} = \int \frac{d x}{x} + l n C$

$\Rightarrow l n | V + \sqrt[]{V^{2} + 16} | = l n C x$

$\frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x} = C x$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2} + \frac{\sqrt[]{y^{2} + 16.4}}{2} = 8.2$

y2 + 64 = (32 – y)2

y = 15

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

The probability that a randomly chosen 2 * 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to:

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alok kumar singh

Contributor-Level 10

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$= \frac{10 + 4 *^{10} ? C_{2}}{1 0^{4}} = \frac{10 + \frac{4 * 10.9}{2}}{1 0^{4}} = \frac{19}{1000}$

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11 months ago

Class 12th Physics Nuclei

$\sqrt[]{d_{1}} a n d \sqrt[]{d_{2}}$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be________.

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alok kumar singh

Contributor-Level 10

$\sqrt[]{d}$ $c o t \frac{θ}{2}$

cot² 30° = x cot² 45°

x = 3

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11 months ago

Class 12th Thermodynamics

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

The variation of applied potential and flowing through a given wire is shown in figure. The length of wire is 31.4cm. The diameter of wire is measured as 2.4cm. The resistivity of the given wire is measured as x * 10^-3 $Ω c m .$ The value of x is_________.

[Take p = 3.14]

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alok kumar singh

Contributor-Level 10

R = tanq = tan 45° = $\frac{ρ L}{A}$

$? 1 = ρ * \frac{31.4}{π {(1.2)}^{2}}$

$ρ = \frac{3.14 * 1.2 * 1.2}{31.4}$

$ρ = 12 * 12 * 1 0^{- 3}$

$ρ = 144 * 1 0^{- 3} = x * 1 0^{- 3} Ω - c m$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let Q and R be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $\sqrt[]{26}$ from the point P(4, 2, 7). Then the surface of the area of the triangle PQR is……………

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Vishal Baghel

Contributor-Level 10

Let PT perpendicular to QR

$\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2} = λ \Rightarrow T (2 λ - 1, 3 λ - 2, 2 λ + 1)$ therefore

$2 (2 λ - 5) + 3 (3 λ - 4) + 2 (2 λ - 6) = 0 \Rightarrow λ = 2$

$T (3, 4, 5) ∴ P T = \sqrt[]{1 + 4 + 4} = 3 ∴ Q T = \sqrt[]{26 - 9} = \sqrt[]{17}$

$∴ Δ P Q R = \frac{1}{2} * 2 \sqrt[]{17} * 3 = 3 \sqrt[]{17}$

Therefore square of $a r (Δ P Q R)$ = 153.

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let the function f(x) = 2x² – log_e x, x > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y² = 4ax at a point P on it passes through the point (8a, 8a – 1) but does not pass through the point $(- \frac{1}{a}, 0) .$ If the equation of the normal at P is $\frac{x}{α} + \frac{γ}{β} = 1,$ then a + b is equal to…………….

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Vishal Baghel

Contributor-Level 10

$f' (x) = \frac{4 x^{2} ? 1}{x}$ so f (x) is decreasing in $(0, \frac{1}{2}) a n d ? ? (\frac{1}{2}, ?)$ $? a = \frac{1}{2}$

Tangent at y² = 2x is y = mx + $\frac{1}{2 m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2} o r ? ? \frac{1}{4}$

So tangent may be $y = \frac{1}{2} x + 1 ? ? o r ? ? y = \frac{1}{4} x + 2 ? ? ? b u t ? ? y = \frac{1}{2} x + 1$ passes through (-2, 0) so rejected.

Equation of normal $\frac{x}{9} + \frac{y}{36} = 1$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), $- \frac{1}{2} < a < 2,$ then p is equal to…………

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Vishal Baghel

Contributor-Level 10

Slope of AH = $\frac{a + 2}{1}$ slope of BC = $- \frac{1}{p} ∴ p = a + 2$ $∴ C (\frac{18 p - 30}{p + 1}, \frac{15 p - 33}{p + 1})$

slope of HC = $\frac{16 p - p^{2} - 31}{16 p - 32}$

slope of BC * slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then is equal to………….

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Vishal Baghel

Contributor-Level 10

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

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