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Physics Class 11th

Statement-I: If three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ are represented by three sides of triangle and, ${\vec{F}}_{1} + {\vec{F}}_{2} = - {\vec{F}}_{3},$ then these three forces are concurrent forces and satisfy the condition for equilibrium.

Statement-II: A triangle made up of three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ as its sides taken in the same order, satisfy the condition for translator equilibrium.

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

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Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

0 0

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a year ago

Physics Thermodynamics

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300K, pressure 400 kPa and mass 0.76g. The ratio of masses of oxygen to hydrogen will be:-

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Vishal Baghel

Contributor-Level 10

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{400 * 1 0^{3} * 500 * 1 0^{- 6}}{8.3 * 100} = 0.008$

=> $m_{1} = 0.12, m_{2} = 0.64$

=> $\frac{m_{2}}{m_{1}} = \frac{16}{3}$

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New answer posted

a year ago

Physics Electromagnetic Waves

The magnetic field vector of an electromagnetic wave is given by B = B₀ = $\frac{\hat{i} + \hat{j}}{\sqrt[]{2}} c o s (k z - ω t);$ where $\hat{i}, \hat{j}$ represents unit vector along x and y-axis respectively. At t = 0 s, two electric charges q1 of 4π coulomb and q2 and 2p coulomb located at $(0, 0, \frac{π}{k}) a n d (0, 0, \frac{3 π}{k}),$ respectively, have the same velocity of 0.5 c $\hat{i}$ (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is:

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Vishal Baghel

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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New answer posted

a year ago

Physics Class 11th

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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alok kumar singh

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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Physics Class 11th

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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Vishal Baghel

Contributor-Level 10

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

=> m = FTV^-1

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New answer posted

a year ago

Physics Physics Magnetism and Matter

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

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alok kumar singh

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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New answer posted

a year ago

Physics Class 11th

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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Vishal Baghel

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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a year ago

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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Vishal Baghel

Contributor-Level 10

Using truth table

A B V₀

0 0

0 1 1

1 0 1

1 1

0 0

New answer posted

a year ago

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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alok kumar singh

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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New answer posted

a year ago

Physics Class 12th

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

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Vishal Baghel

Contributor-Level 10

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

$R_{e q} = 1 Ω$

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