Class 11th

This is a assertion and reason answer type question as classified in NCERT Exemplar

option (i). Both A and R are true and R is the correct explanation of A. 

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

(i)     -    B

(ii)    -     C

(iii)   -     A

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

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 matching answer type question as classified in NCERT Exemplar

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

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

1. At low pressure, the curve of real gas coincides  with that of ideal gas, this shows that the deviation of  behaviour of real gas with respect to ideal gas is small or negligible.

2. At high pressure, the curve of real gas is far apart from  ideal gas, this shows that the deviation of  behaviour of real gas with respect to ideal gas is large.

3. The  pressure p₁ and volume V₁  are  the point where real gas behaves as an ideal gas. 

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

1. According to Boyle's law,  

Pressure of a gas is inversely proportional  to volume of gas at constant temperature. So, the volume decreases with increase in pressure at constant temperature.

2. According to Charles's law,

Volume of a gas is directly proportional to temperature when the pressure is constant. So, the volume of gas increases with increase in temperature.