Physics

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

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.