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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Class 11th Physics Oscillations

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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a year ago

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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Class 11th Physics Laws of Motion

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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a year ago

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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Class 11th Physics Motion in Straight Line

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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a year ago

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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Class 11th Units and Measurement

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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a year ago

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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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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a year ago

For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional forces must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?

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Contributor-Level 10

Let additional force is $\vec{F}$ , so

$\vec{F} + 5 (+ \hat{i}) + 6 (- \hat{i}) + 7 (+ \hat{j}) + 8 (- \hat{j}) = \vec{0}$

$\Rightarrow \vec{F} - (\hat{i} + \hat{j}) = \vec{0} \Rightarrow \vec{F} = \hat{i} + \hat{j} \Rightarrow F = \sqrt[]{2} N, 45 °$

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a year ago

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms^-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?

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Contributor-Level 10

$Δ p$ = Change in momentum of each ball

$Δ p = 2 m v = 2 * 0.05 * 10 = 1 N s$

$\Rightarrow F = \frac{Δ p}{Δ t} = \frac{2 * 0.05 * 10}{0.005} = 200 N$

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a year ago

A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension 7.5 * 10^-4Ncm^-1. The radius of drop in cm will be: (g = 10ms^-2)

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Contributor-Level 10

Since liquid drop is in equilibrium, so

mg = F_B + 2πRT

$\Rightarrow \frac{4}{3} π R^{3} ρ g = \frac{2}{3} π R^{3} σ g + 2 π R T$

$\Rightarrow R = \frac{3 T}{(2 ρ - σ) g} = \frac{15}{\sqrt[]{(2 ρ - σ)}} * 1 0^{- 2} m = \frac{15}{\sqrt[]{(2 ρ - σ)}} c m$

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a year ago

Class 11th Chemistry

200mL of 0.01 M HCl is mixed with 400mL of 0.01 M H₂SO₄. The pH of the mixture is___________.

Given: log2 = 0.30, log 3 = 0.48, log5 = 0.70, log7 = 0.84, log11 = 1.04

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Contributor-Level 10

Total concentration of H⁺ after the mixing of HCl and H₂SO₄

$[H^{+}] = \frac{(200 * 0.01) + (400 * 2 * 0.01)}{600}$

$[H^{+}] = \frac{1}{60}$

$∴ p H = l o g_{10} (\frac{1}{60}) = 1.78$

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a year ago