Physics NCERT Exemplar Solutions Class 11th Chapter Five

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

Explanation – centripetal force, F= mv²/r= f= $\mu N=\mu mg$

V= $\sqrt[]{\mu rg}$

For path ABC path length =3/4 (2 $\pi 2R$ )= $3\pi R=3\pi \left(100\right)=300\pi$

V₁= $\sqrt[]{\mu 2Rg}=\sqrt[]{0.1*100*10}=1.414m/s$

So t= 300 $\frac{\pi }{1.414}=66.6s$

For path DEF  path length = $\frac{1}{4}\left(2\pi R\right)=50\pi$

V₂= $\sqrt[]{\mu Rg}=\sqrt[]{0.1*100*10}=\frac{10m}{s}$

So t₂= 50 $\frac{\pi }{10}$ = 15.7s

For path CD and FA

Path length =R+R= 200m

T= 200/50= 4s

Total time = 66.6+15.7+4= 86.3s

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

Explanation- v_x=2t for 0

= 2 (2-t) for 1

=0 for t>2s

V_y= t for 0

= 1 for t>1s

F_x= ma_x= mdv_x/dt= 1 (2)

F_y = ma_y= mdv_y/dt

= 1 (1) for 0

F= F_x? +F_y?

= 2? +?

=-2?

=0

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

Explanation- as angle is 45

On smooth inclined plane acceleration will be a = gsin $\theta$

So acceleration will become a= g/ $\sqrt[]{2}$

Using equation of motion s =ut +1/2at²

S= $\frac{1}{2}\frac{g}{\sqrt[]{2}}{T}^2$

On rough inclined plane a = g (sin $\theta -\mu cos\theta$ )

= g (sin $45-\mu cos45$ )= $\frac{g(1-\mu )}{\sqrt[]{2}}$

So s=ut +1/2at²

S= 0+ $\frac{1}{2}\frac{g\left(1-\mu \right)}{\sqrt[]{2}}\left(pT\right)$ ²

Comparing two above distance

$\frac{\mathrm{}}{}$$\frac{1}{2}\frac{g\left(1-\mu \right)}{\text{exception 25:}}\left(pT\right)$²= $\frac{1}{2}\frac{g}{\text{exception 25:}}{T}^2$

So after solving we get $$$\mu =(1-1/p^2)$

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

Explanation – as body moving with uniform acceleration a=0

The sum of forces is zero F₁+F₂+F₃=0

a)let F₁, F₂, F₃ be the three forces passing through the point . let F₁and F₂ be in the plane A

so F₃ =- (F₁+F₂) so F₃ is also in plane A.

b)consider the torque of the forces about P . since all the forces pass through P the torque is zero

torque = OP (F₁+F₂+F₃)

since F₁+F₂+F₃=0 so torque =0

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

Explanation -initial speed of coin u= 20m/s

Acceleration of elevator =2m/s² acceleration due to gravity = 10m/s²

Effective acceleration =g+a=12m/s²

We know that v=u +at

0= 20+ (-12)t

So t= 20/12=5/3 s

Time of ascent = time of descent

Total time coin fall back into hand = (5/3+5/3)=10/3s=3.33s

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

Explanation- from the diagram x= t=dx/dt=1m/s

So a_x= 0

From the diagram y =t²

dy/dt=2t  or a_y=d²y/dt²=2m/s²

I_y= ma_y=500 (10^-3) (2) =1N

F_x=ma_x=0

F= $\sqrt[]{F_x^2+F_y^2}$ = 1N

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

Explanation- mass of gun =100kg

Mass of ball =1kg and height of cliff = 500m

So horizontal distance travelled by the ball is x = 400m

So h =1/2gt²

500= $\frac{1}{2}10\left(t\right)$ ² so t= 10s

X=ut then u= 400/10=40m/s

 According to principle of conservation pf momentum

Mv=mu

V= 40/100= 0.4m/s

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

Explanation-given mass of the block =M

Coefficient of friction between block and the wall = $\mu$

In equilibrium  condition vertical and horizontal component balance each other

So f=mg

And F=N but the force of friction f= $\mu N=\mu F$

  $\mu F=Mg$

F= Mg/ μ $\mu$

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

Explanation- in equilibrium, The force mgsin $\theta$ acting on the block A is parallel to the plane should be balanced by the tension in the string

mgsin $\theta =T=F$

and for block B, w=T=F

 from these two above equation 

w=mgsin $\theta$ = 100sin30= 100 (1/2)=50N

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

Explanation- m₁= 5kg, m₂= 3kg

And g= 9.8m/s² and a= 2m/s²

For the upper block T₁-T₂-5g=5a

T₁-T₂= 5 (g+a)

For the lower block T₂-3g =3a

T₂=3 (g+a)=3 (9.8+2)=35.4N

T₁=T₂+5 (g+a)=94.4N