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11 months ago

Physics Electric Charges and Fields

In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $\frac{λ}{100},$ where $λ$ is the wavelength of the wave in free space. The phenomenon, which happens there will be

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Payal Gupta

Contributor-Level 10

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A bar magnet having a magnetic moment of 2.0 * 10⁵ JT^-1, is placed along the direction of uniform magnetic field of magnitude B = 14 * 10^-5 T. The work done in rotating the magnet slowly through 60° from the direction of field is:

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Vishal Baghel

Contributor-Level 10

$u = - M B c o s θ_{2} - (- M B c o s θ_{1})$

$\begin{array}{l} = - M B (c o s θ_{2} - c o s θ_{1}) \\ = - 2 * 1 0^{5} * 14 * 1 0^{- 5} (c o s 60 ° - c o s 0 °) \end{array}$

$= - 28 (\frac{1}{2} - 1)$

$= \frac{28}{2} = 14 J$

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11 months ago

Physics Electrostatic Potential and Capacitance

If Electric field intensity of a uniform plane electro magnetic wave is given as $E = - 301.6 s i n (k z - ω t) {\hat{a}}_{x} + 452.4 s i n (k z - ω t) \hat{a} t \frac{V}{m} .$

Then, magnetic intensity 'H' of this wave in Am^-¹ will be:

[Given : Speed of light in vacuum c = 3 * 10⁸ ms^-¹, Permeability of vacuum $μ_{0} = 4 π * 1 0^{- 7} N A^{- 2}$ ]

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Payal Gupta

Contributor-Level 10

Since

$\vec{E} . \vec{B} = 0 (\vec{E} ⊥ \vec{B})$

$& \frac{E_{0}}{B_{0}} = C = 3 * 1 0^{8} m / s e c$

for option D

$\frac{E_{0}}{B_{0}} = \frac{\sqrt[]{301 . 6^{2} + 452 . 4^{2}}}{1 0^{- 6} \sqrt[]{1 . 5^{2} + 1^{2}}} \begin{matrix} = 301.6 * 1 0^{6} & = 3.01 * 1 0^{8} m / s e c \end{matrix}$

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New answer posted

11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

The equivalent resistance between points A and B in the given network is:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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11 months ago

Physics Thermodynamics

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B}$

$r_{α} = \frac{m_{α} v}{q_{α} B}$

$r_{P} = \frac{m_{P} v}{q_{P} B}$

$\frac{r_{α}}{r_{P}} = \frac{m_{α} q_{P}}{m_{P} q_{α}} = \frac{m_{α}}{m_{P}} (\frac{q_{P}}{q_{α}})$

$= \frac{4 m}{m} (\frac{q}{2 q}) = 2 : 1$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Sixty four conducting drops each of radius 0.02m and each carrying a charge of $5 μ C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

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Vishal Baghel

Contributor-Level 10

Since, volume remain constant

$64 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$

R³ = r³ * 64

R = 4r

$\frac{σ_{b i g g e r}}{σ_{s m a l l d r o p}} = \frac{5 * 64 * 4 π^{2}}{4 π R^{2} * 5} = 64 {(\frac{r}{R})}^{2} = 64 {(\frac{r}{4 r})}^{2} = \frac{64}{16} = 4 : 1$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A parallel plate capacitor with plate area A and plate separation d = 2m has a capacitance of $4 μ F .$ The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be:

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Vishal Baghel

Contributor-Level 10

$G i v e n \frac{\in_{0} A}{d} = 4 μ F$

$C_{1} = \frac{\in_{0} A * 2}{d} \begin{matrix} = 2 * 4 & = 8 μ F \end{matrix}$

$C_{2} = \frac{k \in_{0} A}{d / 2} = 3 * 2 * 4 = 24 μ F$

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{8 * 24}{8 + 24} = \frac{8 * 24}{32} = 6 μ F$

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

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Payal Gupta

Contributor-Level 10

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} * 100 = (\frac{d l}{l} - \frac{d A}{A}) * 100$

= 0.4 (0.4)

= 0.8 %

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

The change on capacitor of capacitance $15 μ F$ in the figure given below is:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{10} + \frac{1}{15} + \frac{1}{20}$

$\frac{1}{C_{e q}} = \frac{6 + 4 + 3}{60} = \frac{13}{60}$

$L_{e q} = \frac{60}{30} μ F$

$Q = C_{e q} v = \frac{60}{13} * 13 μ C$

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11 months ago

Physics Nuclei

The magnetic flux through a coil perpendicular to its plane is varying according to the relation $ϕ = (5 t^{3} + 4 t^{2} + 2 t - 5)$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2s will be

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Payal Gupta

Contributor-Level 10

$V = \frac{- d ?}{d t} = - (15 t^{2} + 8 t + 2)$

at t = 2 sec, V = (15 * 4 + 8 * 2 + 2)

= 78 V

$I = \frac{V}{R} = \frac{78}{5} = 15.6 A$

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