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11 months ago

Physics Class 12th

When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on

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Payal Gupta

Contributor-Level 10

This phenomenon works on resonance in AC circuit.

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New answer posted

11 months ago

Physics Physics Electromagnetic Waves

Light wave travelling in air along x-direction is given by Ey = $540 s i n π * 1 0^{4} (x - c t) V m^{- 1} .$ The, the peak value of magnetic field of wave will be (Given c = 3 * 10⁸ms^-1)

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Payal Gupta

Contributor-Level 10

$c = \frac{E_{0}}{B_{0}} \Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{540}{3 * 1 0^{8}} = 18 * 1 0^{- 8} T$

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11 months ago

Physics Class 12th

For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

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Payal Gupta

Contributor-Level 10

When glass slab is not put in between lens and screen

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{12} - \frac{1}{- 240} = \frac{1}{f} . . . . . . (1)$

When glass slab is put in between lens and screen

Shift = $t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} c m$

$\Rightarrow \frac{3}{35} - \frac{1}{- x} = \frac{1}{f} . . . . . . . . . (2)$

With the help of equations (1) and (2), we can write

$\frac{3}{35} - \frac{1}{- x} = \frac{1}{12} - \frac{1}{- 240}$

$\Rightarrow | Δ u | = 560 - 240 = 320 c m = 3.2 m$

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11 months ago

Physics Dual Nature of Radiation and Matter

The ratio of wavelengths of proton and deuteron accelerated by potential V_p and V_d is $1 : \sqrt[]{2}$ . Then, the ratio of V_p to V_d will be

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Payal Gupta

Contributor-Level 10

According to de-Broglie's hypothesis, we can write

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m e V}}$

$\Rightarrow \frac{λ_{P}}{λ_{D}} = \frac{\sqrt[]{2 m_{D} e V_{D}}}{\sqrt[]{2 m_{P} e V_{P}}} = \frac{1}{\sqrt[]{2}} \Rightarrow \frac{2 m_{P} V_{D}}{m_{P} V_{P}} = \frac{1}{2} \Rightarrow \frac{V_{P}}{V_{D}} = 4 : 1$

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11 months ago

Physics Physics Electrostatic Potential and Capacitance

Capacitance of an isolated conduction sphere of radius R₁ becomes n times when it is enclosed by a concentric conducting sphere of radius R₂ connected to earth. The ration of their radii $(\frac{R_{2}}{R_{1}})$ is

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Payal Gupta

Contributor-Level 10

$C_{1} = 4 π R_{1} a n d$

$C_{2} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2} C_{1}}{R_{2} - R_{1}}$

$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2}}{R_{2} - R_{1}} = n$

$\Rightarrow \frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

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New answer posted

11 months ago

Physics Class 11th

For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional forces must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?

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Payal Gupta

Contributor-Level 10

Let additional force is $\vec{F}$ , so

$\vec{F} + 5 (+ \hat{i}) + 6 (- \hat{i}) + 7 (+ \hat{j}) + 8 (- \hat{j}) = \vec{0}$

$\Rightarrow \vec{F} - (\hat{i} + \hat{j}) = \vec{0} \Rightarrow \vec{F} = \hat{i} + \hat{j} \Rightarrow F = \sqrt[]{2} N, 45 °$

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11 months ago

Physics Class 11th

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms^-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?

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Payal Gupta

Contributor-Level 10

$Δ p$ = Change in momentum of each ball

$Δ p = 2 m v = 2 * 0.05 * 10 = 1 N s$

$\Rightarrow F = \frac{Δ p}{Δ t} = \frac{2 * 0.05 * 10}{0.005} = 200 N$

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New answer posted

11 months ago

Physics Class 11th

A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension 7.5 * 10^-4Ncm^-1. The radius of drop in cm will be: (g = 10ms^-2)

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Payal Gupta

Contributor-Level 10

Since liquid drop is in equilibrium, so

mg = F_B + 2πRT

$\Rightarrow \frac{4}{3} π R^{3} ρ g = \frac{2}{3} π R^{3} σ g + 2 π R T$

$\Rightarrow R = \frac{3 T}{(2 ρ - σ) g} = \frac{15}{\sqrt[]{(2 ρ - σ)}} * 1 0^{- 2} m = \frac{15}{\sqrt[]{(2 ρ - σ)}} c m$

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11 months ago

Physics Physics Moving Charges and Magnetism

The electric current in a circular coil of 2 turns produces a magnetic induction B₁ at its centre. The coil is unwound and is rewound into circular coil of 5 turns and at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction B₂ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is

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Payal Gupta

Contributor-Level 10

$B_{1} = \frac{N_{1} μ_{0} l}{2 R}$ given N1 = 2

When new loop is made the length of wire remains same, so

$N_{2} * 2 π r = N_{1} * 2 π R \Rightarrow r = \frac{N_{1} R}{N_{2}}$

$\Rightarrow \frac{B_{2}}{B_{1}} = {(\frac{N_{2}}{N_{1}})}^{2} = \frac{25}{4}$

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New answer posted

11 months ago

Physics Class 12th

In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be 6V and 2V respectively. The modulation index is

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Payal Gupta

Contributor-Level 10

Modulation index = $\frac{A_{m a x} - A_{m i n}}{A_{m a x} + A_{m i n}} = \frac{6 - 2}{6 + 2} = 0.5 \approx 50 %$

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