Physics

Contributor-Level 10

Force on q₁ :

$F_{q_{1}} = q (V * B) = 4 π (0.5 c \hat{i}) * \frac{B_{0}}{\sqrt[]{2}} (c o s (k z - ω t) \hat{i} + c o s (k z - w t) \hat{j})$

$= 4 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{π}{k}) \hat{k}$

For on q₂:

= $2 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{3 π}{k}) \hat{k}$

So ratio will be 2 : 1

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Four particle each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

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Contributor-Level 10

For circular motion F_{net

$= \frac{M V^{2}}{R}$}

_{$\sqrt[]{2 F} + F' = \frac{M V^{2}}{R}$}

$\sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} R)}^{2}} + \frac{G M M}{{(2 R)}^{2}} = \frac{M V^{2}}{R}$

$\frac{G M}{R} [\frac{1}{\sqrt[]{2}} + \frac{1}{4}] = V^{2}$

$V = \frac{1}{2} \sqrt[]{\frac{G M}{R} (2 \sqrt[]{2} + 1)}$

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If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:

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Physics Physics Magnetism and Matter

Contributor-Level 10

Using F = MA = $m \frac{V}{T}$

=> m = FTV^-1

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Following plots show Magnetization (M) vs Magnetising field (H) and Magnetic susceptibility $(χ)$ vs temperature (T) graph:

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Contributor-Level 10

Magnetization (M) is directly proportional to magnetizing field and magnetic susceptibility does not depend on temperature so option 3 is correct.

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Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 * 10³ kg. The inner and outer radii of each column are 50cm and 100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.

[Use Y = 2.0 * 10¹¹ Pa, g = 9.8 m/s²]

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Contributor-Level 10

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³N.

Cross section area of the column = $π [{(1)}^{2} - {(0.5)}^{2}] = 2.355 m^{2}$

Using young's modulus : $ε = \frac{σ}{Y} = \frac{F}{A Y} = \frac{125 * 1 0^{3}}{2.355 * 2 * 1 0^{11}} = 2.65 * 1 0^{- 7}$

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Physics Class 12th

If VA and VB are the input voltages (either 5V or 0V) and Vo is the output voltage then the two gates represented in the following circuit (A) and (B) are:-

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Contributor-Level 10

Using truth table

A B V₀

0 0

0 1 1

1 0 1

1 1

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Physics Class 12th

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

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Contributor-Level 10

V = 10 $(1 - e^{- t / R C})$

2 = 20 $(1 - e^{- t / R C})$

$\frac{1}{10} = 1 - e^{- t / R C}$

$e^{t / R C} = \frac{10}{9}$

$\frac{t}{R C} = l n (\frac{10}{9}) = 0.105$

$C = \frac{t}{R * 0.105} = \frac{1 0^{- 6}}{10 * 0.105} = 0.95 μ F$

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Physics Class 12th

The equivalent resistance of the given circuit between the terminals A and B is:

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Physics Physics Ray Optics and Optical Instruments

Contributor-Level 10

Since $5 Ω$ is connected across conductor so we can remove it.

$R_{e q} = 1 Ω$

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5 months ago

A glass tumbler having inner depth of 17.5cm is kept on a table. A student starts pouring water $(μ = 4 / 3)$ into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

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Contributor-Level 10

As observer is at O So height of water observed by observer

$\Rightarrow \frac{H}{μ_{ω}} = \frac{H}{(4 / 3)} = \frac{3 H}{4}$

given diagram (17.5 - H) is height of observer

$S o \frac{3 H}{4} = 17.5 - H$

$\frac{7 H}{4} = 17.5$

7H = 70

H = 10

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For a body executing S.H.M. :

(a) Potential energy is always equal to its K.E.

(b) Average potential and kinetic energy over any given time interval are always equal.

(c) Sum of the kinetic and potential energy at any point of time is constant.

(d) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below:

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