Physics

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

Explanation – r=cos $\theta ?+sin\theta$ ?…….1

$\theta =-sin\theta ?+cos\theta ?$ …….2

Multiplying eq1 by sin $\theta$  and 2 with cos $\theta$ and adding

Rsin $\theta +\theta cos\theta =sin\theta cos\theta ?+{sin}^2\theta j+{cos}^2\theta ?-sin\theta cos\theta ?$

= ?( ${cos}^2\theta +{sin}^2\theta$ )=j

= rsin $\theta +\theta cos\theta =j$

 n(rcos $\theta -\theta sin\theta$ )=i

b)r $\theta =cos\theta ?+sin\theta ?(-sin\theta ?+cos\theta ?)$

 = -cos $\theta sin\theta +sin\theta .cos\theta =0$ $\theta =90$

c)r=cos $\theta ?+sin\theta ?$

dr/dt=d/dt(cos $\theta ?+sin\theta ?$ )=w[-cos $\theta ?+sin\theta ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $\theta r=\theta [\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= dr/dt= $\frac{d\theta }{dt}r+\theta \frac{d}{dt}[\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= $\frac{d\theta }{dr}r+\theta [-\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]\frac{d\theta }{dt}$

= $\frac{d\theta }{dt}r+\theta w\theta =wr+$ w $\theta \theta$

a= $\frac{d}{dt}\left[wr+w\theta \theta \right]=\frac{d}{dt}\left[\frac{d\theta }{dt}r+\frac{d\theta }{dt}\left(\theta \theta \right)\right]$

a= $\frac{d^2\theta }{d{t}^2}r+\frac{d\theta }{dt}$ $\frac{dr}{dt}+\frac{d^2\theta }{d{t}^2}\theta \theta +\frac{d\theta }{dt}\frac{d}{dt}\left(\theta \theta \right)$

= $\frac{d^2\theta }{d{t}^2}r+w^2\theta +\frac{d^2\theta }{d{t}^2}\theta \theta +w^2\theta +w^2\theta (-r)$

This is a multiple choice answer as classified in NCERT Exemplar

(d) In an ideal gas when a molecules collides elastically with a wall, the momentum transferred to each molecule will be twice the magnitude of its normal momentum. For the face EFGH, it transfer only half of that.

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

Explanation- a) for x direction u_x= u+v_ocos $\theta$

u_y=velocity in y direction= v₀sin $\theta$

now tan $\theta =\frac{u_y}{u_x}=\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

$\theta ={tan}^{-1}\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

b) let t be the time flight y =0 u_y=v_osin $\theta ,a_y=-g,t=T$

y= u_yt+1/2 a_yt²

0= v_osin $\theta T$ + $\frac{1}{2}\left(-g{T}^2\right)$

So T = $\frac{2u_{o}sin\theta }{g}$

c) horizontal range R, = (u+v_ocos $\theta$ T= (u+v_ocos $\theta$ ) $\frac{2u_{o}sin\theta }{g}$

d) for range to be maximum dR/d $\theta =0$

$\frac{v_o}{g}\left[2ucos\theta +v_{o}cos2\theta *2\right]=0$

$2ucos\theta +2v_o\left[2{cos}^2\theta -1\right]=0$

4v_ocos^{2 $\theta +2ucos\theta -2v_o=0$}

So cos $\theta$ = $\frac{-u?\sqrt[]{u^2+8v_o^2}}{4v_o}$

$\theta ={cos}^{-1}\left[\frac{-u\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]$

e) cos $\theta$ = $\frac{-v_o?\sqrt[]{u^2+8v_o^2}}{4v_o}=\frac{-1+3}{4}=\frac{1}{2}$

so $\theta =60$

f) if u=0 ${\theta }_{max}={cos}^{-1}\left[\frac{-0\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]={cos}^{-1}\left(\frac{1}{\sqrt[]{2}}\right)=45$ ⁰

This is a multiple choice answer as classified in NCERT Exemplar

(b) As the motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls of the vessel, hence pressure of the gas inside the vessel, as conserved by us, on the ground remains the same.

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

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_r^2+v_s^2}=\sqrt[]{3^2+4^2}=5m/s$

tan so 'N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_s^2-v_r^2}=\sqrt[]{4^2-3^2}=\sqrt[]{7}m/s$

tan $\theta =\frac{v_r}{v}=\frac{3}{\sqrt[]{7}}$

(c) time taken by swimmer t =d/v= d/4s

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1 

 

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

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II 

v_g = 10m/s?

V_rg= V_r - V_g

 = a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =1

So b =-5

Velocity of rain = a? +b?

V_r = 5? -5?

Speed of rain V_r= $\sqrt[]{5^2+{(-5)}^2}=5\sqrt[]{2}m/s$

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

Explanation- y=O, u_y= V_ocos $\theta$

a_y=-gcos $\theta$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2}{a}_y$ t²

0 = V_ocos $\theta T$ +T^{2 $\frac{1}{2}(-gcos\theta )$}

T= $\frac{2V_{o}cos\theta }{gcos\theta }$

T= 2V₀/g

X= L, u_x=V_osin $\theta$  , a_x= gsin $\theta$  , t=T= $\frac{2V_o}{g}$

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

L= V_osin $\theta T+\frac{1}{2}gsin\theta T^2$

L= $\frac{4v_o^2}{g}$ sin $\theta$

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