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Class 12th Physics

A long cylindrical volume contains a uniformly distributed charge of density $ρ C m^{- 3} .$ The electric field inside the cylindrical volume at a distance $x = \frac{2 ε_{0}}{ρ} m$ from its axis is_________Vm^-1.

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Payal Gupta

Contributor-Level 10

$x = \frac{2 ε_{0}}{ρ}$

$E 2 π x l = \frac{ρ π x^{2} l}{ε_{0}}$

$E_{x} = \frac{ρ x^{2}}{2 x ε_{0}} = \frac{ρ x}{2 ε_{0}}$

$= \frac{ρ}{2 ε_{0}} * \frac{2 ε_{0}}{ρ} = 1 v / m$

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11 months ago

Class 12th Physics

A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0mm² and the resistivity of copper is 1.7 * 10^-8 $Ω m,$ the force experienced by moving electron in the wire is _______* 10^-23N.

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Payal Gupta

Contributor-Level 10

i = 1 A $j = σ E$

A = 2mm²

$ρ = 1.7 * 1 0^{- 8} Ω m$ $F = e E = \frac{e j}{σ} = \frac{e i}{A σ} = \frac{e i ρ}{A}$

= $\frac{1.6 * 1 0^{- 19} * 1.7 * 1 0^{- 8}}{2 * 1 0^{- 6}}$

= 136 * 10^-23 N

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11 months ago

Class 12th Physics Alternating Current

To light, a 50 W, 100 V lamp is connected, in series with a capacitor of a capacitance $\frac{50}{π \sqrt[]{x}} μ F,$ with 200V, 50Hz AC source. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$R = \frac{v_{2}}{P} = \frac{10000}{50} \Rightarrow R = 200 Ω$

$V = \sqrt[]{V_{C}^{2} + V_{R}^{2}}$

$\Rightarrow 200 = \sqrt[]{V_{C}^{2} + {(100)}^{2}}$

$x_{c} = \frac{1}{ω_{c}} = \frac{π \sqrt[]{x}}{100 π * 50} = \frac{\sqrt[]{x}}{5000 * 1 0^{- 6}} = \frac{\sqrt[]{x} * 1 0^{3}}{5}$

$= 200 \sqrt[]{x}$

$V_{C} = i x_{C}$

$\Rightarrow \sqrt[]{3} . \sqrt[]{1 + x} = 2 \sqrt[]{x}$

$\Rightarrow 3 (1 + x) = 4 x \Rightarrow x = 3$

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11 months ago

Class 12th Physics Wave Optics

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are p/2 and p/3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$?_{A} = \frac{π}{2}, ?_{B} = \frac{π}{3}$

$I_{A} = I_{1} + I_{2} + 2 \sqrt[]{l_{1} l_{2}} c o s ? = l + 4 l + 2 \sqrt[]{l * 4 l} c o s \frac{π}{2} = 5 l$

$I_{B} = I + 4 I + 2 \sqrt[]{l . 4 l} c o s \frac{π}{3} = 5 l + 2 l = 7 l$

$Δ l = 7 l - 5 l = 2 l \Rightarrow x = 2$

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11 months ago

Class 12th Physics

In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance 30cm from the left side as shown at point D. If R is 5.6 $k Ω$ , then the value of unknown resistance 'S' will be ___________ $Ω .$

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Payal Gupta

Contributor-Level 10

$\frac{30}{70} = \frac{S}{5.6 k Ω} \Rightarrow S = \frac{3}{7} * 5.6 k Ω = 3 * 0.8 k Ω$

= 2.4k $Ω$ = 2400 $Ω$

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11 months ago

Class 12th Exam Compartment

When will the class 12th compartment exams be held?

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Saket Kumar

Contributor-Level 8

The CBSE Class 12 compartment exams for 2025 are scheduled to be held in July. Specifically, the exams are expected to take place from July 15, 2025. The practical exams are scheduled from July 10 to July 15, 2025. The admit cards for these exams are likely to be released in late June. The results for the compartment exams are expected to be announced in August.

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11 months ago

Class 12th Maths

Let x = $s i n (2 t a n^{- 1} α)$ and y = $s i n (\frac{1}{2} t a n^{- 1} \frac{4}{3}) .$ If S = ${α \in R : y^{2} = 1 - x},$ then $\sum_{a \in S}^{} 16 α^{3}$ is equal to

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Payal Gupta

Contributor-Level 10

$? x = s i n (2 t a n^{- 1} α) = \frac{2 α}{1 + α^{2}}$ ……. (i)

and

$y = s i n (\frac{1}{2} t a n^{- 1} \frac{4}{3}) = s i n (s i n^{- 1} \frac{1}{\sqrt[]{5}}) = \frac{1}{\sqrt[]{5}}$

Now,

$y^{2} = 1 - x$

$∴ α = 2, \frac{1}{2} ∴ \sum_{a \in S}^{} 16 α^{3} = 16 * 2^{3} + 16 * \frac{1}{2^{3}} = 130$

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11 months ago

Class 12th Maths

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ -3xy² + 6x² -5xy -8y² + 9x + 14 = 0 at the point (-2, 3) be A. Then 8A is equal to

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Payal Gupta

Contributor-Level 10

$4 x^{3} - 3 x y^{2} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$ differentiating both sides we get

$12 x^{2} - 3 y^{2} - 6 x y y' + 12 x - 5 y - 5 x y = 16 y y' + 9 = 0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$∴ A r e a = \frac{1}{2} * B a s e * H e i g h t$

$A = \frac{1}{2} * (\frac{- 4}{3} + \frac{31}{2}) (3) = \frac{1}{2} (\frac{85}{0}) . 3 = \frac{85}{4} = 8 A = 170$

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11 months ago

Class 12th Maths

Let f be a twice differentiable function on R. If $f' (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f' (t) d t = (e^{2 x} + e^{- 2 x}) c o s 2 x + \frac{2}{a} x,$ then (2a + 1)⁵a² is equal to

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Payal Gupta

Contributor-Level 10

$f (x) + \int_{0}^{x} (x - t) f' (t) d t = (e^{2 x} + e^{- 2 x}) c o s 2 x + \frac{2 x}{a} . . . . . . . (i)$

Here f (0) = 2 ………. (ii)

On differentiating equation (i) w.r.t. x we get :

$f' (x) + f \int_{0}^{x} f' (t) d t + x f' (x) - x f' (x)$

$4 = \frac{2}{a} \Rightarrow a = \frac{1}{2}$

$∴ {(2 a + 1)}^{5} . a^{2} = 2^{5} . \frac{1}{2^{2}} = 2^{3} = 8$

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11 months ago

Class 12th Maths

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y^{3} + 2 y x^{2}}{3 x y^{2} + x^{3}}, y (1) = 1 .$ .If for some $n \in N, y (2) \in [n - 1, n],$ then n is equal to

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Payal Gupta

Contributor-Level 10

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

Put y = vx

$\Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v (4 v^{2} + 2)}{(3 v^{2} + 1)}$

$l n (\frac{y^{2}}{8} + \frac{y}{2}) = 2 l n 2 \Rightarrow \frac{y^{3}}{8} + \frac{y}{2} = 4$

$\Rightarrow [y (2)] = 2$

$\Rightarrow n = 3$

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