Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation-A = i+j

 B = i-j

 A.B=|A|B|cos $\theta$

(? +? ). (? -? ) = $\left(\sqrt[]{1+1}\right)\left(\sqrt[]{1+1}\right)$ cos $\theta$

Where $\theta$  is the angle between A and B

Cos $\theta$ = $\frac{1-0+0-1}{\sqrt[]{2\sqrt[]{2}}}$ = $\frac{0}{2}=0$

$\theta$  = 90^o

This is a multiple choice answer as classified in NCERT Exemplar

When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increases the KE of the gas molecule and hence, the collisions between molecules also increases.

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

Explanation- - given |A|=2 and |B|=4

a) |A $*B$ |=AB cos $\theta$ = 0

so 2 $*4$ cos $\theta$ =0

so $\theta =90$  so it matches with ii

b) |A $*$ B|= ABcos $\theta$ =8

2 $*4$ cos $\theta$ =8

So $\theta$ = 0 so it matches with option i

c) |A $*$ B|= ABcos $\theta$ =4

so $\theta$ =60 so it matches with option iv

d) |A $*$ B|= ABcos $\theta$ =-8

so $\theta$ =180 so it matches with option iii

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

Explanation- given |A|=2 and |B|=4

a)|A $*B$ |=AB sin $\theta$ = 0

so 2 $*4$ sin $\theta$ =0

so $\theta =0$ so it matches with iv

B) |A $*$ B|= ABsin $\theta$ =8

2 $*4$ sin $\theta$ =8

So $\theta$ = 90 so it matches with option iii

c) |A $*$ B|= ABsin $\theta$ =4

so $\theta$ =30 so it matches with option i

d) |A $*$ B|= ABsin $\theta$ =4 $\sqrt[]{2}$

so $\theta$ =45 so it matches with option ii

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) pV=nRT

(a) when pressure p =constant

volume is directly proportional to temperature

(b) when T= constant

from PV =constant

so the graph is rectangular hyperbola

(c) when V =constant

from pressure is directly proportional to temperature.

So the graph is straight the passes through the origin.

PV $\propto T$

PV/T=constant

So the graph hence through origin. So d is correct.

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

Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) From fig iii it is clear that c+b=a so a-c=b

(b) From fig I it is clear that b=a+c so b-a =c

(c) From ii it is clear that -c= a+b so a+b+c=0

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) The total energy associated with the molecule is

E= $\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}m{v}_z^2+\frac{1}{2}{I}_x{w}_x^2+\frac{1}{2}{I}_y{w}_y^2$

The number of independent terms in the above expressions is 5. So we can predict velocities of molecule by maxwell's distribution, hence the above expression obeys maxwell 's distribution .

So 2 rotational and 3 translational energies are associated with each molecule

So rotational energy at a given temperature is 2/3 of the translational KE of each molecule.

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

Explanation – a) radius of earth =6400km= 6.4 $*{10}^{6}m$

Time period = 1 day = 24 $*60*60$ = 86400s

Centripetal acceleration a= w²r= R(2 $\pi /T$ )²=4 $\pi$ ²R/T

= $\frac{4*{\left(\frac{22}{7}\right)}^2*6.4*{10}^6}{({24*60*60)}^2}$ = 0.034m/s²

b) time = 1yr=365 $*24*60*60$ days= 365=3.15 $*{10}^{7}s$

centripetal acceleration = Rw²= $\frac{4{\pi }^{2}R}{T^2}$

^{$=\frac{4*{\frac{22}{7}}^2*1.5*{10}^{11}}{({3.15*{10}^7)}^2}=5.97*{10}^{-3}m/s$ 2}

$\frac{a_c}{g}=\frac{5.97*{10}^{-3}}{9.8}=\frac{1}{1642}$

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

Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $*5/18$ = 200m/s

Altitude of the plane P'T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP'=u $*t$ =200t

Vertical distance travelled by the bomb P'T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP'=200 (17.49)m=

tan $\theta$ =P'T/P'P=1500/200 (17.49)=0.49287= tan23⁰12'

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

Explanation – due to air resistance particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper then rise. When we are neglecting air resistance path is parabola when we consider air resistance then path is asymmetric parabola.