Physics NCERT Exemplar Solutions Class 11th Chapter Five

              44.4 litres of He $\frac{44.8}{22.4}moles$

            $\eta =2$    moles of He

              Q =   $\eta C_v\Delta T$ (for fixed capacity (v = constant)

              =   $2*\frac{R}{\gamma -1}\Delta T$

              =   $\frac{2*R}{\frac{5}{3}-1}*20$

= $=2*\frac{3R}{2}*20$

 Q = 60R = 60 * 8.3

  Q = 498 J

L₀ = RMV_cm + I_cm   $\omega$

 = RM   $\left(R\omega \right)+\frac{2}{3}M{R}^2\omega$

 L₀ =   $\frac{5}{3}M{R}^2\omega$

    $\frac{5}{3}*1R^2\omega =\frac{5}{3}{R}^2\omega =\frac{a}{3}{R}^2\omega$ =  

   a = 5

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

Explanation- mass of helicopter m =2000kg

Mass of the crew and passengers m= 500kg

Acceleration a =15m/s² and g = 10m/s²

a) force on the floor of the helicopter by the crew and passengers

m (g+a)= 500 (10+15)N= 12500N

b) action of rotor of the helicopter on the surroundings air= (m₁+m₂) (g+a)

= (2000+500) (10+15)= 2500 (25)= 62500N

c) force on the helicopter due to surroundings air

= reaction of force applied by helicopter

= 62500N

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

Explanation -for the box to just starts sliding down mg

sin $\theta =t=\mu N=\mu mgcos\theta$

tan $\theta$ = $\mu$

$\theta ={tan}^{-1}\left(\mu \right)$

b) when angle of inclination is increased to $\alpha >\theta$

F₁= mgsin $\alpha -f=mgsin\alpha -\mu N$

= mg ( $sin\alpha -\mu cos\alpha$ )

c) to keep the box stationary, upward force needed F₂= m $gsin\alpha$ +f= mg ( $sin\alpha +\mu cos\alpha$ )

d) if the box is to move with an upward acceleration a then upward force needed 

F₃=mg ( $sin\alpha +\mu cos\alpha$ )+ma

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

Explanation- on resolving forces into rectangular components in equilibrium forces (F₁+1/ $\sqrt[]{2}$ )N are equal to $\sqrt[]{2}N$  and F₂ is equal to ( $\sqrt[]{2}+1/\sqrt[]{2}$ )N

F₁+1/ $\sqrt[]{2}$ = $\sqrt[]{2}$

F₁= $\sqrt[]{2}-1/\sqrt[]{2}$ = $\frac{2-1}{\sqrt[]{2}}=\frac{1}{\sqrt[]{2}}=0.707N$

F₂= $\sqrt[]{2}+\frac{1}{\sqrt[]{\sqrt[]{2}}}=\frac{2+1}{\sqrt[]{2}}=\frac{3}{\sqrt[]{2}}=2.121N$

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

Explanation – horizontal velocity u_x= v_s

During projectile motion horizontal velocity remains unchanged

V_x=u_x=v_s

In vertical direction v_y²= u_y²+2gH

V_y= $\sqrt[]{2gH}$

Resultant speed of the ball at the bottom

V= $\sqrt[]{v_x^2+v_y^2}$

V= $\sqrt[]{v_s^2+2gH}$

b) when ball is given horizontal velocity and a small downward velocity

in horizontal direction V'_x=u_x=v_s

in vertical direction v_y'²= u_y²+2gH

resultant velocity of the ball at the bottom

v'= $\sqrt[]{v_s^2+u^2+2gH}$

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

Explanation- displacement vector of particle is r (t)=? Acos $wt+$ ? Bsinwt

X=Acoswt

x/A= coswt………1

displacement along y axis is

y=Bsinwt

y/B= sinwt……….2

squaring and then adding eqn1 and 2 we get

x²/A²+y²/B²=cos²wt+sin²wt =1

this is an equation of ellipse. Therefore trajectory of particle is an ellipse.

b)v= dr/dt= id/dt (Acoswt)+jd/dt (Bsinwt)

 ? [A (-sinwt)w]+? [B (coswt).w]

= -? Awsinwt+? Bwcoswt

Acceleration a= dv/dt

So a= -? Awd/dt (sinwt)+? Bw [-sinwt]w

=? Aw²coswt-? Bsinwt

= -w²r

So force acting on the particle f=ma=-mrw²