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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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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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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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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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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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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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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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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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Physics Class 11th

A mono-atomic gas of mass 4.0u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $\frac{x}{3 R} .$ Value of x is…………….

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Vishal Baghel

Contributor-Level 10

$Δ U + Δ K E = 0$

$\Rightarrow \frac{f}{2} n R Δ T = \frac{1}{2} m v^{2} \Rightarrow Δ T = \frac{m}{n} \frac{v^{2}}{f R} = 4 * 1 0^{- 3} * \frac{3 0^{2}}{3 R} = \frac{3.6}{3 R} K .$

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Physics Oscillation

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $U = \frac{α}{r^{10}} - \frac{β}{r^{5}} - 3$

Where, and are positive constants. The equilibrium between two atoms will be ${(\frac{2 α}{β})}^{\frac{a}{b}},$ ${\frac{}{}}^{\frac{}{}}$ where a =…………….

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Vishal Baghel

Contributor-Level 10

For equilibrium,

$\frac{d U}{d r} = 0 \Rightarrow \frac{- 10 α}{r^{11}} + \frac{5 β}{r^{6}} = 0 \Rightarrow r = {(\frac{2 α}{β})}^{\frac{1}{5}}$

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