Physics

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $\beta$ ,a_y= -gcos $\alpha ,t=T$

y=u_yt + $\frac{1}{2}{a}_y{t}^2$

0= V_osin $\beta T+\frac{1}{2}\left(-gcos\alpha \right)T^2$

T[V_osin $\beta -\frac{gcos\alpha }{2}$ T]=0

T = time of flight = $\frac{2V_{o}sin\beta }{gcos\alpha }$

 

Motion along OX

x= L ,u_x= V_ocos $\beta$ , a_x= -gsin $\alpha$

t =T = $\frac{2V_{o}sin\beta }{gcos\alpha }$

x= u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V₀cos $\beta T$ + $\frac{1}{2}(-gsin\alpha )T^2$

L= T[V₀cos $\beta -\frac{1}{2}gsin\alpha T$ ]

L=  $\frac{2V_{o}sin\beta }{gcos\alpha }$ [V_ocos $\beta -\frac{V_{o}sin\alpha sin\beta }{cos\alpha }$ ]

L= $\frac{2v_o^{2}sin\beta }{g{cos}^2\alpha }\mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

Z= sin $\beta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

 = sin $\beta [cos\alpha cos\beta -sin\alpha sin\beta ]$

= $\frac{1}{2}(cos\alpha +sin2\beta -2sin\alpha .{sin}^2\beta )$

= ½ [sin2] $\beta cos\alpha -sin\alpha (1-cos2\beta )$

= $\frac{1}{2}[sin2\beta cos\alpha +cos2\beta .sin\alpha -sin\alpha ]$

= $\frac{1}{2}$  [sin(2 $\beta +\alpha$ )-sin $\alpha$ ]

For z maximum

2 $\beta +\alpha =\frac{\pi }{2}$  , $\beta =\frac{\pi }{4}-\frac{\alpha }{2}$

This is a short answer type question as classified in NCERT Exemplar

Volume occupied by 1 mole = 1mole of the gas at NTP= 22400mL=22400cc

So number of molecules in 1cc of hydrogen= $\frac{6.023*{10}^{23}}{22400}=2.688*{10}^{19}$

H₂ is a diatomic gas, having a total of 5 degrees of freedom

So total degrees possesed by all the molecules

= 5 $*2.688$ $*{10}^{19}=1.344*{10}^{20}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $?x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_o^2}{g}\theta =45$ …………1

Horizontal component of initial velocity = V_ocos $\theta$

Vertical component of initial velocity = -V_osin $\theta$

So h = (-V_osin $\theta$ )t + $\frac{1}{2}gt$ ²………….2

R+ $?x$  = V_ocos $\theta *t$

So t= $\frac{R+?x}{v_{o}cos\theta }$

Substituting value of t in 2 we get

So h = (-V₀sin $\theta$ ) $\left(\mathrm{}\frac{R+?x}{v_{o}cos\theta }\right)+\frac{1}{2}g({\mathrm{}\frac{R+?x}{v_{o}cos\theta })}^2$

H = -(R+ $?x$ )tan $\theta$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^2\theta }\right)$

$\theta =45$ ,  h = -(R+ $?x$ )tan $45$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^{2}45}\right)$

So h = -(R+ $?x$ ) $1$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2\left(\frac{1}{2}\right)}\right)$

So h = -(R+ $?x$ )+ $\frac{{(R+?x)}^2}{R}$

So h = -R- $?x$ +(R+ $\frac{{?x}^2}{R}+2?x$ )

 h= $?x+\frac{{?x}^2}{R}$

This is a short answer type question as classified in NCERT Exemplar

Number of helium = 5

T=7^oC=7+273=280K

(a) hence number of atoms = number of moles $*$ Avogadro's number

= $5*6.023*{10}^{23}=30.015*{10}^{23}=3*{10}^{24}$ atoms

(b) now average kinetic energy per molecule = 3/2 K_BT

= total internal energy

= $\frac{3}{2}{K}_{b}T*$ number of atoms

= $\frac{3}{2}*1.38*{10}^{-23}*280*3*{10}^{24}=1.74*{10}^{4}J$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2gh}$

$=\sqrt[]{2*10*500}=100m/s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x = $\sqrt[]{u^2-u_y^2}=\sqrt[]{{125}^2-{100}^2}=75m/s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*500}{10}}=10s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

This is a short answer type question as classified in NCERT Exemplar

When air is pumped, more molecules are pumped and boyle's law is staed for situation where number of molecules remains constant . in this case as the number of air molecules keep increases, hence mass change. Boyle's law is only applicable in situations, where number of gas molecule of remains fixed.

This is a short answer type question as classified in NCERT Exemplar

Radius = 1A^o

Volume of hydrogen molecules = 4/3 $\pi$ r³

= 4/3 (3.14) (10^-10)³ $\approx 4*{10}^{-30}$ m³

Number of moles of H₂ = mass/molecular mass=0.5/2=0.25

Molecules of H₂ present = number of moles of H₂ present $*6.023*{10}^{23}$

= 0.25 $*6.023*{10}^{23}$

So volume of molecules present = molecule number $*$ volume of each molecules

= 0.25 $*6.0233*{10}^{23}*4*{10}^{23}$

$\approx$ 6 $*{10}^{-7}m$ ³

P_iV_i= P_fV_f

V_{f $\left(\frac{P_i}{p_f}\right)V$} = _i= $\frac{1}{100}(3*{10}^{-2})$ ³

V_f= 2.7 $*{10}^{-7}m$ ³

This is a short answer type question as classified in NCERT Exemplar

(a) The average KE will be the same

M= molar mass of the gas

m=mass of each molecular of the gas

R= gas constant

v_{rms $\propto \frac{1}{\sqrt[]{m}}$}

(b) k = Boltzmann constant

T= absolute temperature

m_A>m_B>m_C

V_rms.Arms.Brms.C

This is a short answer type question as classified in NCERT Exemplar

V₁=2L ,V₂=3L

µ₁ = 4.0and µ₂ =5.2

p₁= 1.00 atm and p₂ = 2.00 atm

p₁V₁= µ₁RT₁

p₂V₂= µ₂RT₂

when the partition is removed the gases get mixed without any loss of energy . the mixture now attains a common equilibrium pressure and total volume of the system is sum of the volume of individual chambers V₁ and V₂

$\mu ={\mu }_1+{\mu }_2$ , V =V₁+V₂

From the kinetic theory of gases pV=2/3 E

For mole 1  ,P₁V₁= 2/3 ${\mu }_1{E}_1$

For mole 2  , P₂V₂= 2/3 ${\mu }_2{E}_2$

Total energy is ( ${\mu }_1{E}_1+{\mu }_2{E}_2$ )= 3/2 ( $P_1{V}_1+P_2{V}_2$ )

PV==2/3E_total = 2/3 $\mu E_{permole}$

P( $V_1+V_2$ )= $\frac{2}{3}*\frac{3}{2}(p_1{V}_1+p_2{V}_2)$

P= $\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}$ = $\frac{1*2+2*3}{2+3}atm$

P=8/5 =1.6atm

This is a short answer type question as classified in NCERT Exemplar

Mean free path l=1/ $\sqrt[]{2}{d}^{2}n$

So n= number of molecules /volume

d = diameter of the molecule

l $\propto \frac{1}{d^2}$

d₁=1A^o, d₂=2A^o

l $\propto \frac{1}{d^2}$

$\frac{l_1}{l_2}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{2}{1}\right)}^2=\frac{4}{1}$

l₁:l₂=4:1