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9 months ago

Class 12th Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-1. The third resonance is observed when the air column has a length of ____________cm.

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Vishal Baghel

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

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9 months ago

Class 12th Chemistry Solutions

A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is:

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alok kumar singh

Contributor-Level 10

$M o l a l i t y = \frac{m o l e s o f s o l u t e * 1000}{w t o f s o l v e n t (g m)}$

$M o l a l i t y = \frac{35 * 1000 * 1.46}{36.5 * 100}$ = 14.0 M

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9 months ago

Class 12th Physics

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω .$ The value of internal resistance of each cell is

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$\frac{2 V}{2 + 2 r} = \frac{V}{2 + \frac{r}{2}}$

$\Rightarrow 2 + 2 r = 4 + r \Rightarrow r = 2 Ω$

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9 months ago

Class 12th Physics

If the electric potential at any point (x, y, z)m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3)m will be:

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Vishal Baghel

Contributor-Level 10

According to relation between field and potential, we can write

^{$\vec{E} = - \frac{d V}{d x} (\hat{i}) = - \frac{d (3 x^{2})}{d x} (\hat{i}) = - 6 x (\hat{i})$}

^{${\vec{E}}_{(1, 0, 3)} = - 6 (\hat{i}) N / C$}

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9 months ago

Class 12th Physics Electric Charge and Field

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

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

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Raj Pandey

Contributor-Level 9

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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9 months ago

Class 12th Physics

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is:

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Raj Pandey

Contributor-Level 9

After switch 'S' is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

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9 months ago

Class 12th Physics Moving Charges and Magnetism

Given below are two statements:

Statement I : The electric force changes the speed of the charged particle and hence charges its kinetic energy; whereas the magnetic force does not change like kinetic energy of the charged particle.

Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerated the moving charged particle along the direction of magnetic field.

In the light of the above statements, chose the most appropriate answer from the option given below:

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Vishal Baghel

Contributor-Level 10

According to Lorentz's Force, we can write

$\vec{F} = {\vec{F}}_{E l e c t r i c} + {\vec{F}}_{M a g n e t i c} = q \vec{E} + q (\vec{v} * \vec{B}), s o$

Statement I is correct but Statement II is incorrect

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9 months ago

Class 12th Vector Algebra

Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If a is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is:

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alok kumar singh

Contributor-Level 10

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For a = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

a = 2

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9 months ago

Class 12th Maths

Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y +2z = 16. Let T be a plane passing through the point Q and contains the line $\vec{r} = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R .$ Then, which of the following points lies on T?

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alok kumar singh

Contributor-Level 10

$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 1}{2} = \frac{- 2 (1 + 4 + 2 - 16)}{1 + 2^{2} + 2^{2}}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$| \begin{matrix} x & y & z + 1 \\ 3 & 6 & 6 \\ 1 & 1 & 2 \end{matrix} | = 0$

->2x – z = 1

option (B) satisfies.

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New answer posted

9 months ago

Class 12th Communication Systems

The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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