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9 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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New answer posted

9 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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9 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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9 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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9 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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9 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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9 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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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively will be:

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Vishal Baghel

Contributor-Level 10

The average kinetic energy (per molecule)of any (ideal) gas (be it monoatomic like Argon, diatomic like oxygen or polyatomic) is always equal to $\frac{3}{2} K_{B} T .$ If depends only on temperature, and is independent of the nature of the gas. So, Answer is 1 : 1.

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9 months ago

Physics Electromagnetic Induction

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

In the light of above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

If p is the density and $η$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

$R e = \frac{ρ v d}{η}$

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