Class 11th

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

At constant temperature, the pV vs V plot for real gases is not a straight line because there is intermolecular attraction present in real gases which is absent in ideal gases; hence ideal gases form a straight line in the pV vs V plot at constant temperature.

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 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