physics ncert solutions class 11th

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $?R=?mg=$ 0.1 $*2*9.807=$ 1.96 N

Retardation produced by the frictional force, $a_r$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_n$ = a - $a_r$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_n{t}^2$ = 0 $*10+\left(\frac{1}{2}\right)*2.52*{10}^2$ = 126 m

(a) Work done in 10 s is given by

W = Force $*\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}=7*126=882\mathrm{J}$

(b) Work done by friction in 10 s is given by

W = -f $*s=-1.96*126=$ - 247J

(c) Work done by the net force,

W = (F-f) $*s$ = (7 – 1.96 ) $*126$ = 635 J

(d) Kinetic energy is given by the equation

KE = (1/2)m $v^2$ where v is the final velocity

From the equation v = u + at we get after 10 s, v = 0 + 2.52 $*$ 10 = 25.2 m/s

Final kinetic energy = (1/2) $*2*{25.2}^2$ = 635 J

Initial kinetic energy = (1/2) $*2*0^2$ = 0

Change in kinetic energy = 635 – 0 = 635 J

Hence the work done by the net force = change in kinetic energy

6.1  (a) While the person lifts a bucket out of a well by means of a rope tied to the bucket, the direction of both the force and the displacement are same, hence the work done is positive.

(b) While lifting the bucket, he works against gravity, but the work done by the gravitational force is downward, hence the work done is negative.

(c) The direction of motion of the object is in the opposite direction of the frictional force; hence the work done is negative.

(d) While a body moves on a rough horizontal plane, the frictional forces try to oppose the motion. But since the applied force maintains uniform velocity of the object, the motion of the object and the applied force are in the same direction. Hence the work done is positive.

(e) Since the resistive force of air is trying to bring the vibrating pendulum to rest, the work done is negative.

13.13 According to law of atmospheres, we have

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] ….(i)

where $n_1$ is the number of density at height  $h_1$and $n_2$ is the number of density at height $h_2$

mg is the weight of the particle suspended in the gas column

Density of the medium = $?$

Density of the suspended particle = ${?}^{\text{'}}$

Mass of one suspended particle = m'

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes's principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m'g

= mg- V ${?}^{\text{'}}g$ = mg – ($\frac{m}{?}$ ) ${?}^{\text{'}}g$

= mg(1- $\frac{{?}^{\text{'}}}{?}$ ) ……(ii)

Gas constant R = $k_B$ N

$k_B=\frac{R}{N}$ …(iii)

Substituting from (ii) and (iii) in (i), we get

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

= n1 exp [$\left[-mg\left(-\frac{{?}^{\text{'}}}{?}\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT}\right]$]

= n1 exp[$\left[-mg\left(?-{?}^{\text{'}}\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT?}\right]$]

13.11 Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, $l_o$  = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (76 + 15) = 9 cm

Hence, total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure,  $P_1$ = 76 cm of mercury

Initial volume, $V_1$= 15 ${cm}^3$

Final pressure, $P_2$ = 76 – (76 – h) = h cm of mercury

Final volume, $V_2$ = (24 + h) ${cm}^3$

Since temperature remained constant throughout the process,

 $P_1{V}_1$=$P_2{V}_2$ or 76 $*15$ = h $*(24+h)$$$

Or, $h^2$ + 24h - 76 $*15$ =0

h = $\frac{-24\pm \sqrt{{24}^2+4*76*15}}{2}$=$\frac{-24\pm 71.67}{2}$

So h = 23.84 cm or -47.84 cm

Since height cannot be negative, so h = 23.84 cm

23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 $*1.013*{10}^5$ Pa

Temperature inside the cylinder, T = 17 $?=290K$

Radius of nitrogen molecule, r = 1.0 Å = 1 $*{10}^{-10}$ m

Diameter of nitrogen molecule, d = 2 $*{10}^{-10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28 $*{10}^{-3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence 

The mean free path $\stackrel{}{l}$ is given by

$l ?=\frac{kT}{?2?d^{2}P}$ where k = Boltzmann constant = 1.38$*{10}^{-23}$ kg-$$$m^2{s}^{-2}{K}^{-1}$ 

$\stackrel{?}{l}=\frac{1.38\mathrm{}*{10}^{-23}*290}{?2?{(2\mathrm{}*{10}^{-10})}^2*2\mathrm{}*1.013*{10}^5}$= 1.11$*{10}^{-7}$  m

Collision frequency = $\frac{V_{rms}}{\stackrel{?}{l}}$ = $\frac{508.26}{1.11\mathrm{}*{10}^{-7}}$= 4.57 ${10}^9$ /s

Collision time, T = $\frac{d}{V_{rms}}$  =$\frac{2\mathrm{}*{10}^{-10}\mathrm{}}{508.26}$S= 3.93 $*{10}^{-13}$$\frac{\mathrm{}}{}$ s

Time taken between successive collision T' = $\frac{\stackrel{?}{l}}{V_{rms}}$ = $\frac{1.11\mathrm{}*{10}^{-7}}{508.26}$s = 2.18$*{10}^{-10}$
 s$\frac{\mathrm{T}\mathrm{\text{'}}}{T}$ =$\frac{2.18\mathrm{}*{10}^{-10}}{3.93\mathrm{}*{10}^{-13}}$ = 555.71

Hence, the time taken between successive collisions is 556 times the time taken for a collision.