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Physics Class 11th

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt[]{2}$ time the speed of sound, then the value of n will be

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Payal Gupta

Contributor-Level 10

As we know that

$v = \sqrt[]{\frac{γ R T}{M}} \Rightarrow$ Speed of sound in a gas, and

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow$ Rms speed of gas molecules, so

$v_{r m s} = \sqrt[]{2} v \Rightarrow \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{2} \sqrt[]{\frac{γ R T}{M}} \Rightarrow γ = \frac{3}{2}$

According formula of specific heat ratio of mixture, we can write

$\Rightarrow$ 5n = 14 n = 2.8

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Physics Physics Oscillations

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $π^{2} m s^{- 2}$ )

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Payal Gupta

Contributor-Level 10

As we know that

$g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}, s o$

$T = 2 π \sqrt[]{\frac{l}{g_{P}}} \Rightarrow l = \frac{g_{p} T^{2}}{4 π^{2}} = \frac{1}{9} m$

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5 months ago

Physics Class 12th

A current of 15 mA flow in the circuit as shown in figure. The value of potential difference between the pointes A and B will be

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Payal Gupta

Contributor-Level 10

$R_{A B} = 5 + (10 / / 5) + 10 = 15 + \frac{10 * 5}{10 + 5} = 15 + \frac{10}{3} = \frac{55}{3} k Ω$

$V_{A B} = R_{A B} l = (\frac{55}{3} * 1 0^{3}) * (15 * 1 0^{- 3}) = 275 V$

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5 months ago

Physics Physics Laws of Motion

A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is:

[Take g = 10m/s^-2]

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alok kumar singh

Contributor-Level 10

$a = μ g = 0.4 * 10 = 4 m / s^{2}$

Slipping stops when V_bag = V_conveyar

0 + at = 2

t = $\frac{2}{a} = \frac{2}{4} = . 5 s e c$ $\frac{}{} \frac{}{}$

So, x = $\frac{1}{2} a t^{2} = \frac{1}{2} * 4 * \frac{1}{4} = 0.5 m$

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5 months ago

Physics Physics Motion in Straight Line

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5ms^-1. The power needed to keep the belt moving with the same velocity will be:

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alok kumar singh

Contributor-Level 10

$P = \vec{F} . \vec{v} = v \frac{d P}{d t} = v^{2} (\frac{d m}{d t}) = 0.5 * 25 = 12.5 W$

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5 months ago

Physics Units and Measurement

A torque meter is calibrated to reference standards of mass, length and time ach with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of:

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alok kumar singh

Contributor-Level 10

$T o r q u e \vec{τ} = \vec{r} * \vec{F}$

$[τ] = [L] [M L T^{- 2}]$

$[τ] = [M L^{2} T^{- 2}]$

as, $τ = m l^{2} t^{- 2}$

$\frac{d τ}{τ} * 100 = | \frac{d m}{m} * 100 | + 2 | \frac{d l}{l} * 100 | + 2 | \frac{d t}{t} * 100 |$

= 5 + 2 * 5 + 2 * 5

= 25 %

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5 months ago

Physics Physics Moving Charges and Magnetism

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 * 10^-4 Wbm^-2. The frequency of revolution of the electron will be

(Take mass of electron = 9.0 * 10^-31kg)

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Payal Gupta

Contributor-Level 10

f = $\frac{q B}{2 π m} = \frac{1.6 * 1 0^{- 19} * 1.0 * 1 0^{- 4}}{2 * 3.14 * 9 * 1 0^{- 31}} = 0.028 * 1 0^{8} H z = 2.8 * 1 0^{6} H z$

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5 months ago

Physics Class 12th

When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on

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Payal Gupta

Contributor-Level 10

This phenomenon works on resonance in AC circuit.

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5 months ago

Physics Physics Electromagnetic Waves

Light wave travelling in air along x-direction is given by Ey = $540 s i n π * 1 0^{4} (x - c t) V m^{- 1} .$ The, the peak value of magnetic field of wave will be (Given c = 3 * 10⁸ms^-1)

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Payal Gupta

Contributor-Level 10

$c = \frac{E_{0}}{B_{0}} \Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{540}{3 * 1 0^{8}} = 18 * 1 0^{- 8} T$

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5 months ago

Physics Class 12th

For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

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Payal Gupta

Contributor-Level 10

When glass slab is not put in between lens and screen

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{12} - \frac{1}{- 240} = \frac{1}{f} . . . . . . (1)$

When glass slab is put in between lens and screen

Shift = $t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} c m$

$\Rightarrow \frac{3}{35} - \frac{1}{- x} = \frac{1}{f} . . . . . . . . . (2)$

With the help of equations (1) and (2), we can write

$\frac{3}{35} - \frac{1}{- x} = \frac{1}{12} - \frac{1}{- 240}$

$\Rightarrow | Δ u | = 560 - 240 = 320 c m = 3.2 m$

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