Physics NCERT Exemplar Solutions Class 11th Chapter Six

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

As the block M is at rest and frictional force =Mgsin $\theta$

The force of friction acting between the blocks and incline opposes the tendency of sliding of the block . since block not in motion therefore no work is done. Hence no dissipation of energy.

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

V= volume of ballon

${\rho }_{air}=$  density of air

${\rho }_{He}=$ density of helium

V( ${\rho }_{air}-{\rho }_{He}$ )g= ma= mdv/dt= upthrust

Integrating with respect to t

V( ${\rho }_{air}-{\rho }_{He}$ )gt=mv

½ mv²= ½ m v²/m² ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

= ½v²/m ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

= if the ballon rises to height h then s= ut +1/2at²

h=1/2at²= ½ $\frac{\mathrm{}\mathrm{V}\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2}{m}$

so from above equation

1/2mv²= [V( ${\rho }_{air}-{\rho }_{He}$ )g][ $\frac{1}{2m}V\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2$ ]

= V( ${\rho }_{air}-{\rho }_{He}$ )gh

So ½ mv²+V ${\rho }_{He}$ gh= ${\rho }_{air}$ hg

KE_ballon+PE_ballon= change in PE of air

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

m =50g = 50 $*{10}^{3}kg$

Side = 1cm = 0.01m

Speed v = 0.1m/s

Young's modulus= 2 $*{10}^{11}N/m$ ²

According to the formula

F/A= Y $\frac{?L}{L}$

And F= K $?L$  where K is the compression in the spring.

K= YA/L = YL

Initial KE= 2 (1/2mv²)= 5 $*{10}^{-4}J$

Final PE= 2 (1/2)K ( $?L$ )²

$?L=\sqrt[]{\frac{KE}{E}}=\sqrt[]{\frac{KE}{YL}}$  = $\sqrt[]{\frac{5*{10}^{-4}}{2*{10}^{11}*0.1}}=1.58*{10}^{-7}m$

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

Let M be the mass of the rocket at any time t and v₁ the velocity of the rocket at the same time t

Let? m = mass of gas ejected in? t time

Relative speed of the gas ejected =u

KE +? t = KE of rocket +KE of gas

= ½ (M-? m) (v+? v)² + ½? m (v-u)²

KE_t= KE of rocket at time t= ½ Mv²

So? K = KE_t+? t -KE_t

= (M? v=? mu)v+1/2? mu²

Since action and reaction forces are equal

M? v/? t=? m/? t|u|

M? v=? m u

So? K= ½? mu²

? K=? W

? W=1/2? mu²

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

(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.

(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

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

(a) work done= increase in PE

= mg (vertical distance travelled)

= mg (s)sin $\theta =1*10*10sin30$ = 50J

(b) work done against friction = fs

= $\mu N\left(s\right)=\mu mgcos\theta$

= 0.1 $*1*10*cos30*10=8.66J$

(c) increase in PE =mgh

 =1 $*10*10*sin30$

$=100\left(\frac{1}{2}\right)=50J$

(d) according to work energy theorem W= change in KE

= -mgh-fs+FS

= -50-8.66+10 (10)

= 41.34J

(e) force f = FS

= 10 (10)= 100J