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6 months ago

Physics Dual Nature of Radiation and Matter

Let K₁and K₂ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $λ_{1} a n d λ_{2},$ respectively are incident on a metallic surface. If $λ_{1} = 3 λ_{2}$ then:

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alok kumar singh

Contributor-Level 10

E Energy with which photons are incident on a metallic surface.

$λ_{1} = 3 λ_{2} \Rightarrow λ_{1} > λ_{2}$

E $α \frac{1}{λ},$ E₁ < E₂

We know E₁ = $?_{0} + k_{1}$

$\frac{h c}{λ_{1}} = ? + k_{1}$

k₁ = $\frac{h c}{λ_{1}} - ?_{0} - - - - (i)$

& E₂ = $?_{0} + k_{2}$

$\frac{h c}{λ_{2}} = ?_{0} + k_{2}$

$\frac{3 h c}{λ_{1}} = ? + k_{2}$

k₂ = $\frac{3 h c}{λ_{1}} - ?_{0}$ (i)

from (1) & (2)

$k_{z} = 3 [k_{1} + ?_{0}] - ?_{0} = 3 k_{1} + 2 ?_{0}$

k₂ > $3 k_{1} \Rightarrow \frac{k_{2}}{3} > k_{1}$

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6 months ago

Physics Physics Wave Optics

In young's double slit experiment performed using a monochromatic light of wavelength $λ,$ when a glass plate $(μ = 1.5)$ of thickness $x λ$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:

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alok kumar singh

Contributor-Level 10

Þ AC = d + t $(μ - 1) \to$ path length

BC = d

path different = AC – BC = $n λ$

$t (μ - 1) = μ λ$

for contal maxima

$x λ (1.5 - 1) = n λ$

0.5x = x

n = 0, 1,2- -

n = 0 Not possible

Now, n = 1

0.5x = 1

x = 2

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6 months ago

Physics Class 12th

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

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alok kumar singh

Contributor-Level 10

Given, $λ = 8 m m, v = 8 m m, v = 3 * 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 * 1 0^{- 3}} = \frac{π}{4} * 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{j} v / m$

Now $\frac{E_{0}}{B_{0}} = C$

B0 = $\frac{60}{3 * 1 0^{8}} = \frac{20}{1 0^{8}} = 2 * 1 0^{- 7}$ $\vec{B} = 2 * 1 0^{- 7} [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{k} T$

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6 months ago

Physics Physics Moving Charges and Magnetism

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 * 10^-6N, then the value of x is approximately:

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alok kumar singh

Contributor-Level 10

$\frac{F}{l} = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{d}$

$2 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π} * \frac{x * x}{d}$

x² = 2x = $\sqrt[]{2}$ = 1.4x = 1.4

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6 months ago

Physics Class 12th

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circuited.)

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alok kumar singh

Contributor-Level 10

$P = \frac{E^{2}}{R} = \frac{{(N A \frac{d B}{d t})}^{2} * 4 A C}{ρ l}$

$p' = {(\frac{N A}{2} \frac{d B}{d t})}^{2} * 4 A C$

$p' = 2 P$

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New answer posted

6 months ago

Physics Physics Moving Charges and Magnetism

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 * 10^-5. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

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alok kumar singh

Contributor-Level 10

$χ (S u s c e p t i b i l i t y) = 1.2 * 1 0^{- 5}$

fractional change = $= \frac{Δ B}{B} = \frac{B i n s i d e - B o u t}{B o u t}$

$= \frac{μ_{m} H - μ_{0} H}{μ_{0} H} = \frac{(μ_{0} μ_{r} - μ_{0}) H}{μ_{0} H}$ $[? μ_{m} = b_{0} μ_{0}]$

$= μ_{r} - 1$

= $= (1 + χ) - 1 = χ = 1.2 * 1 0^{- 5}$

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6 months ago

Physics Class 12th

Resistance of the wire is measured as at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is:

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alok kumar singh

Contributor-Level 10

R_T = R₀ $(1 + α Δ T)$

at t = 10°C, R_T = 2 $Ω$

2 = R₀ (1 + [10 – T0])- (1)

at T = 30°C, R_T= 3 $Ω$

3 = $R_{0} [1 + α (30 - T_{0})]-$ (2)

from (1)

$2 = R_{0} (1 + α 10)-$ (3)

from (2) 3 = R₀ (1 + 30) - (4)

$(4) \div (3)$

$\frac{3}{2} = \frac{1 + 30 α}{1 + 10 α}$

3 + 30 = 2T 60α = $\frac{1}{30} = 0.033$

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6 months ago

Physics Class 12th

Two point charges A and B of magnitude + 8 * 10^-6 and -8 * 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 * 10⁴ NC^-1. The distance 'd' between the point charges A and B is:

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

Enet = E₁ + E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$\Rightarrow 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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6 months ago

Physics Class 12th

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k ? q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

Enet = E₁+ E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$? 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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6 months ago

Physics Class 11th

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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