physics ncert solutions class 11th

The velocity of a small ball of mass 'm' and density d₁, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d₂, then the viscous force acting on the ball, will be:

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Contributor-Level 10

mg = pvg

mg = d₁vg ……. (i)

F_b = weight of fluid displaced = d₂vg …… (ii)

by NLM 1 F_b + F_v = mg

F_V = mg - F_b

$F_{v} = m g (1 - \frac{d_{2}}{d_{1}})$

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Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2}$ . If $\frac{C_{p}}{C_{v}} = γ$ for the gas then a good estimate for $\frac{τ_{2}}{τ_{1}}$ is given by

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Contributor-Level 10

Relaxation time $(τ) \propto \frac{V}{\sqrt[]{T}}$

and $T \propto \frac{1}{V^{γ - 1}}$

$\begin{array}{r} τ \propto V^{1 + \frac{γ - 1}{2}} τ \propto V^{\frac{1 + γ}{2}} \\ \frac{τ_{f}}{τ_{i}} = {(\frac{2 V}{V})}^{\frac{1 + γ}{2}} \frac{τ_{f}}{τ_{i}} = (2)^{\frac{1 + γ}{2}} \end{array}$

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Following statements are given:

(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.

(B) The average kinetic energy of a gas molecule increase with increase in pressure at constant temperature.

(C) The average kinetic energy of a gas molecule decreases with increase in volume.

(D) Pressure of a gas increase with increase in temperature at constant pressure.

(E) The volume of gas decreases with increase in temperature .

Choose the correct answer from the options given below:

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Contributor-Level 10

$K E_{a v g} = \frac{3}{2} K T$

$P = \frac{1}{3} ρ V_{r m s}^{2}$

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It momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:

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Viscosity = Pascal . Second

$P^{X} A^{Y} T^{Z} = [M^{1} L^{- 1} T^{- 1}]$

= ${[M^{1} L^{+ 1} T^{- 1}]}^{x} [L^{2}] y {[T^{1}]}^{z} = M^{- 1} L^{- 1} T^{- 1}$

= x = 1, x + 2y = -1, -x + z = 1

y = -1, Z = 0

Viscosity = [P¹A^-¹T⁰]

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Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is 1 : 4, then

(A) The r.m.s. velocity of gas molecules in two vessels will be the same.

(B) The ratio of pressure in these vessels will be 1 : 4.

(C) The ratio of pressure will be 1 : 1.

(D) The r.m.s. velocity of gas molecules in two vessels will be 1 : 4.

Choose the correct answer form the options given below:

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Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} A s . v_{1} = v_{2}$

$V_{r m s} α \sqrt[]{T}$ $η_{2} = 4 n_{1}$

$T_{1} = T_{2} \Rightarrow V_{r m s - 1} = V_{r m s - 2}$

$P_{1} = \frac{n_{1} R T}{v} P_{z} = \frac{n_{2} R T}{v}$

$\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}} = \frac{1}{4}$

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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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The root mean square speed of smoke particles of mass 5 * 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10^-23 JK^-1]

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m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

&nb

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m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

less

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Two bodies of masses m₁ = 5kg and m₂ = 3kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m₁ will be:

[Take g = 10 ms^-2]

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Contributor-Level 10

m₁ = 5kg

m₂= 3kg

As m₁ is in rest -

T = m_gg = 30

N = m₁ g cosq

T = $m_{1} g s i n θ \Rightarrow 50 s i n θ = 30$

->sinθ $= \frac{3}{5} \Rightarrow θ = 37 °$

$∴ N = 50 c o s θ = 50 c o s 37 ° = 50 * \frac{4}{5} = 40 N$

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Given below are two statements:

Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature.

Statement II : The rms speed of oxygen molecules in a gas is $υ$ . If the temperature is doubles and the oxygen molecules dissociate into oxygen atoms, the rms speeds will become 2 $υ$ . In the light of the above statements, choose the correct answer from the options given below:

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Contributor-Level 10

$V = \sqrt[]{\frac{3 R T}{M}}$ $V' = \sqrt[]{\frac{3 R 2 T}{(\frac{M}{2})}} = 2 \frac{\sqrt[]{3 R T}}{M} = 2 v$

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The force required to stretched a wire of cross-section 1cm² to double its length will be:

(Given Young's modulus of the wire = 2 * 10¹¹ N/m²)

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