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11 months ago

Physics Work, Energy and Power

A particle is projected with velocity v₀ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = - αx². The distance at which the particle stops:

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Vishal Baghel

Contributor-Level 10

$F = - α x^{2} \Rightarrow \frac{m v d v}{d x} = - α x^{2}$

$\Rightarrow \int_{v_{0}}^{0} v d v = - \frac{α}{m} \int_{0}^{x} x^{2} d x \Rightarrow \frac{0^{2}}{2} - \frac{v_{0}^{2}}{2} = \frac{- α}{3 m} x^{3}$

$\Rightarrow \frac{- v_{0}^{2}}{2} = - \frac{α}{3 m} x^{3} \Rightarrow x = {(\frac{3 v_{0}^{2} m}{2 α})}^{1 / 3}$

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11 months ago

Physics Class 12th

Given below are two statements:

Statement I : PN junction diodes can be used to function as transistor simply by connecting two diodes, back to back, which acts as the base terminal.

Statement II : In the study of transistor, the amplification factor b indicates ratio of the collector current to the base current.

In the light of the above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

A tristor is formed by doping a semi conductor water from one side by N-type dopant (high concentration while comparatively lower concentration on other side thus forming NPN tristor & vice-versa.

So, statement (I) is false & statement (II) is true.

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11 months ago

Physics Magnetism and Matter

A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:

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Vishal Baghel

Contributor-Level 10

When a soft ferromagnetic substance is placed in external magnetic field, the size of domain lying in the opposite direction of external magnetic field increases while size of domain lying in the opposite direction of field decreases if field is weak. However, if field is strong then the domain rotate in the direction of external magnetic field due to strong torque.

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11 months ago

Physics Class 11th

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[use g = $\frac{G m}{R^{2}} = 9.8 m s^{- 2}$ and radius of earth, R = 6400km].

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Vishal Baghel

Contributor-Level 10

As we go pole to equator, acceleration due to gravity decreases. So, weight of the body will also reduces. Thus, weight on equator will be slightly smaller than 49 N.

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11 months ago

Physics Class 12th

Zener breakdown occurs in a P- n junction having P and n both:

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Vishal Baghel

Contributor-Level 10

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

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11 months ago

Physics Units and Measurement

The period of oscillation of a simple pendulum is $T = 2 π \sqrt[]{\frac{L}{g}} .$ Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{L}{g}} o r T^{2} = 4 π^{2} (\frac{L}{g})$

$\Rightarrow g = 4 π^{2} (\frac{L}{T^{2}})$

$\Rightarrow \frac{Δ g}{g} % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) % = [\frac{1 m m}{1 m} + \frac{2 * 0.01}{1.95}] * 100 % = (0.001 + 0.0102) * 100 = 1.13 %$

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11 months ago

Physics Class 12th

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current 'i' just after switch 'S' is closed will be:

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Vishal Baghel

Contributor-Level 10

R = $2 Ω$

L = 2 mH

E = 9V

$i = \frac{ε}{2 R} = \frac{9 v}{4 Ω} = 2.25 A$

Just after the switch 'S' is closed, the inductor acts as open circuit.

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11 months ago

Physics Class 11th

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

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Vishal Baghel

Contributor-Level 10

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

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11 months ago

Physics Electric Flux

The electric field in a region is given by $\vec{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \frac{N}{C} .$ The ratio of flux of reported field through the rectangular surface of are 0.2m² (parallel to y – z plane) to the of the surface of are 0.3m² (parallel to x – z plane) is a : b, where a = ……………

[Here $\hat{i}, \hat{j} a n d \hat{k}$ are the unit vectors along x, y and z – axes respectively]

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Vishal Baghel

Contributor-Level 10

$?_{1} = \frac{3}{5} E_{0} (0.2) N m^{2} C^{- 1}, a n d ?_{2} = \frac{4}{5} E_{0} (0.3) N m^{2} C^{- 1}$

$\Rightarrow \frac{?_{1}}{?_{2}} = \frac{3 * 0.2 * 5}{5 * 0.3 * 4} = \frac{1}{2}$

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11 months ago

Physics Motion in a Plane

A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……………… m/s. (Take g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

Let v is velocity at the highest position.

$T_{m a x} = 5 T_{m i n} \Rightarrow m g + \frac{m (v^{2} + 4 g l)}{l} = 5 (\frac{m v^{2}}{l} - m g) \Rightarrow 4 . \frac{v^{2}}{l} = 10 g$

$\Rightarrow v = \sqrt[]{\frac{5}{2} g l} = \sqrt[]{\frac{5}{2} * 10 * 1} = 5 m / s$

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