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Class 12th Physics Moving Charges and Magnetism

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. It the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be:

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Vishal Baghel

Contributor-Level 10

Radius of circular path R = $\sqrt[]{\frac{2 m k}{q B}}$ $\frac{}{}$

$q = \frac{\sqrt[]{2 m k}}{R B}$

$\frac{q_{1}}{q_{2}} = \sqrt[]{\frac{m_{1}}{m_{2}}} * \frac{R_{2}}{R_{1}} = \sqrt[]{\frac{9}{4}} * \frac{5}{6} = \frac{5}{4}$

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Class 12th Physics

A condenser of $2 μ F$ capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across it's plates with respect to the charge (Q) on the condenser?

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Vishal Baghel

Contributor-Level 10

Q = CV

$V = \frac{Q}{C}$

Straight line with slope = $\frac{1}{C ?}$ $\frac{}{}$

$S l o p e = \frac{1}{C} = \frac{1}{2 * 1 0^{- 6}} = 5 * 1 0^{5}$

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Class 12th Physics

An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} i & = \frac{V}{r} (1 - e^{- R t / L}) \\ = 4 (1 - e^{- 500 t}) \end{array}$

At $t = \infty$

$i_{1} = 4 A$

at $t = 40 s$

$i_{2} = 4 (1 - e^{- 20000})$

= 4 [1 - \frac{1}{{(e^{2})}^{10000}}]

= 4 [1 - \frac{1}{(7.389)^{10000}}]

\frac{i_{1}}{i_{2}}

is slightly greater than 1 .

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8 months ago

Class 12th NCERT Solutions

Boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of 8% aqueous of non-volatile solute B. The relation between molecular weights of A and B is

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alok kumar singh

Contributor-Level 10

For A : 100 gm solution 2 gm solute A

$∴ M o l a l i t y = \frac{2 / M_{A}}{0.098}$

For B : 100 gm solution 8 gm solute B

$∴ M o l a l i t y = \frac{8 / M_{B}}{0.092}$

$\frac{1}{4.261} = \frac{M_{A}}{M_{B}}$

$∴ M_{B} = 4.261 * M_{A}$

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8 months ago

Class 12th Chemistry

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Activated charcoal adsorbs SO₂ more efficiency than CH₄.

Reason R: Gases with lower critical temperature are readily adsorbed by activated charcoal.

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

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alok kumar singh

Contributor-Level 10

SO₂ is absorbed to a greater extent then CH₄ on activated charcoal under same Gases with higher critical temperature are readily absorbed by activated charcoal.

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Class 12th Chemistry

Given below are two statements.

Statement I: O₂, Cu²⁺, and Fe³⁺ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.

Statement II: NaCl and H₂O are weakly magnetized in opposite direction to magnetic field.

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

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alok kumar singh

Contributor-Level 10

$O_{2}, C u^{2 +} a n d F e^{3 +}$

are paramagnetic

$∴$ Weakly attracted by magnetic field

NaCl and H2O are diamagnetic

$∴$ Weakly repelled by magnetic field.

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8 months ago

Class 12th Maths

The domain of the function f(x) = $s i n^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) i s :$

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alok kumar singh

Contributor-Level 10

$- 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$

$\Rightarrow 0 \leq \frac{2 x^{2} - x + 9}{x^{2} + 2 x + 7} & \frac{- 5 x - 5}{x^{2} + 2 x + 7} \leq 0$

$x \in R & - 1 \leq x < \infty$

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Class 12th Physics

Which of the following physical quantities have the same dimensions?

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Vishal Baghel

Contributor-Level 10

Electric Displacement

$\begin{array}{l} \vec{D} = \in o \vec{E} \\ [D] = [\in_{0} \vec{E}] = [\in_{0} \frac{σ}{\in_{0}}] \\ [D] = [σ] \end{array}$

Electric displacement $(\vec{D})$ and surface change density

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8 months ago

Class 12th Chemistry

250g solution of D-glucose in water contains 10.8% of carbon by weight. The molality of the solution is nearest to

(Given : Atomic weights are, H, 1u ; C, 12u ; O, 16u)

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alok kumar singh

Contributor-Level 10

$C_{6} H_{12} O_{6} \to G l u c o s e$

$\frac{M a s s o f C}{m a s s o f g l u c o s e} = \frac{72}{180}$

% of C = 10.8 = $\frac{M a s s o f C}{m a s s o f s o l u t i o n} * 100$

$\frac{10.8 * 250}{100} m a s s o f C$

$∴ M o l a r i t y = \frac{0.375}{0.1825} = 2.055 = 2.06$

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8 months ago

Class 12th Maths

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} * \vec{b}) . (\vec{b} * \vec{c}) + (\vec{b} * \vec{c}) . (\vec{c} * \vec{a}) + (\vec{c} * \vec{a}) . (\vec{a} * \vec{b})$ = 168, then $| \vec{a} | + | \vec{b} | + | \vec{c} |$ is equal to:

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alok kumar singh

Contributor-Level 10

Given : $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = c o s θ (s a y)$

$| \vec{a} | | \vec{b} | | \vec{c} | = 14$

$(\vec{a} * \vec{b}) \cdot (\vec{b} * \vec{c})$

$= \vec{a} \cdot [(\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}]$

$= (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - {| \vec{b} |}^{2} \vec{a} \cdot \vec{c}$

$= | \vec{a} | {| \vec{b} |}^{2} | \vec{c} | (c o s^{2} θ - c o s θ) = 14 | \vec{b} | (c o s^{2} θ - c o s θ)$

Similarly, $(\vec{b} * \vec{c}) \cdot (\vec{c} * \vec{a})$

= $| \vec{b} | {| \vec{c} |}^{2} | \vec{a} | (c o s^{2} θ - s i n θ) = 14 | \vec{c} | (c o s 2 θ - s i n θ) & (\vec{c} * \vec{a}) \cdot (\vec{a} * \vec{b})$

$= | \vec{c} | {| \vec{a} |}^{2} | \vec{b} | (c o s^{2} θ - s i n θ) = 14 | \vec{a} | (c o s^{2} θ - s i n θ)$

Given : 14 $(c o s^{2} θ - c o s θ) (| \vec{a} | + | \vec{b} | + | \vec{c} |) = 168$

$| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{12}{c o s^{2} θ - c o s θ} = \frac{12}{\frac{1}{4} - (- \frac{1}{2})} = \frac{12}{\frac{3}{4}} = 16$

Given : $\vec{a}, \vec{b}, \vec{c}$ are coplanar & pair wise equal angle.

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