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6 months ago

Physics Photoelectric Effect

In photoelectric effect experiment, the graph of stopping potential $\lor$ versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

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alok kumar singh

Contributor-Level 10

Sol. $e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 * 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 * 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 * 10^{- 19}]} \end{array}$

$= 2.27$

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6 months ago

Physics Physics Electrostatic Potential and Capacitance

A capacitor $C$ is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance. $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitor is:

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alok kumar singh

Contributor-Level 10

Sol. $g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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6 months ago

Physics Alternating Current

If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

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Payal Gupta

Contributor-Level 10

Dimension

RC → [T]

$\frac{L}{R} \to [T]$

$\sqrt[]{L c} \to [T]$

$\frac{L}{C} \to$ dimensionless

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6 months ago

Physics Moving Charges and Magnetism

Two long parallel conductors S₁ and S₂ are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).

A charge particle of 3p coulomb is passing through the point P with velocity $\vec{v} = (2 \hat{i} + 3 \hat{j})$ m/s; where $\hat{i} a n d \hat{j}$ represents unit vector along x & y axis respectively. The force acting on the charge particle is 4p * 10^-5 $(- x \hat{i} + 2 \hat{j}) N .$ The value of x is

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Payal Gupta

Contributor-Level 10

${\vec{B}}_{N e t} = \frac{μ_{0}}{4 π} * 2 [\frac{i_{1}}{d_{1}} - \frac{i_{2}}{d_{2}}] (- \hat{k})$

$= 1 0^{- 7} * 2 [\frac{4}{4} * 100 - \frac{2}{6} * 100] (- \hat{k})$

$= 2 * 1 0^{- 5} [1 - \frac{1}{3}] - \hat{k}$

${\vec{B}}_{N e t} = \frac{4}{3} * 1 0^{- 5} (- \hat{k})$

q = 3πc

$\vec{v} = 2 \hat{i} + 3 \hat{j}$

$\vec{F} = q (\vec{v} * \vec{B})$

= $3 π [(2 \hat{i} * 3 \hat{j}) * \frac{4}{3} * 1 0^{- 5} (- \hat{k})]$

$\vec{F} = 4 π * 1 0^{- 5} (- 3 \hat{i} + 2 \hat{j}) N$

x = 3

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6 months ago

Physics Electric Charges and Fields

Three identical charged balls each of charge 2 C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1 m. The ratio of net force on a charged ball to the force between any two charged balls will be

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Payal Gupta

Contributor-Level 10

$F_{23} = \frac{k q_{2} q_{3}}{d^{2}} = \frac{k * 2 * 2}{12} = 4 k$

$F_{21} = 4 k$

$F_{13} = 4 k$

Net force on q₂ = F_R = $\sqrt[]{F_{21}^{2} + F_{13}^{2} + 2 F_{21} F_{23} c o s θ}$

= $\sqrt[]{{(4 k)}^{2} + {(4 k)}^{2} + 2 * 4 k c o s 60 °}$

= $k \sqrt[]{48} - (1)$

Force b/w (1) & (2) F₂₁ = 4k – (2)p

$\frac{F_{R}}{F_{21}} = \frac{k \sqrt[]{48}}{4 k} = \frac{4 \sqrt[]{3}}{4} = \sqrt[]{3} : 1$

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6 months ago

Physics Electric Charges and Fields

If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be:

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Payal Gupta

Contributor-Level 10

$?_{f l o t} = \frac{q}{2 \in_{0}} (1 - c o s θ)$

When, 'q' is at the centre of the flat surface then, $θ = 50 .$

$?_{f l a t} = \frac{q}{2 \in_{0}} (1 - c o s 90 °)$

$= \frac{q}{2 \in_{0}}$

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6 months ago

Physics Oscillations

The equation of a particle executing simple harmonic motion is given by x = sin $(t + \frac{1}{3}) m$ . At t = 1s, the speed of particle will be

(Given : p = 3.14)

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Payal Gupta

Contributor-Level 10

$x = s i n π (t + \frac{1}{3})$

$v = \frac{d x}{d t} = c o s [π (t + \frac{1}{3})] * π$

$= c o s (π + \frac{π}{3}) * π$

$= - c o s \frac{π}{3} * (π)$

$= - \frac{π}{2} = - 1.57 m / s e c$

Speed = 157 cm/sec

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6 months ago

Physics Kinetic Theory

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecules per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).

Choose the most appropriate answer form the options given below

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Payal Gupta

Contributor-Level 10

$\bar{x}$ (mean free path) = $\frac{1}{\sqrt[]{2} π d^{2} n_{v}}$

$n_{v} = \frac{n N_{A}}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v} = \frac{P}{R_{T}}$

$\bar{x} \propto v \propto \frac{1}{ρ (d e n s i t y} & \bar{x} α T a t c o n s t a n t P)$

^{$ρ ↑ \Leftrightarrow \bar{x} ↓ | \bar{x} ↑ \Leftrightarrow T ↑$}

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is $= \frac{1}{2} k_{B} T$ (for Mono atomic gases)

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:

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Payal Gupta

Contributor-Level 10

At constant Temp

$P α \frac{1}{v}$

v₂ > v₁

⇒ P₁ > P₂

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6 months ago

Physics Gravitation

Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is