Physics

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.

This is a multiple choice answer as classified in NCERT Exemplar

(c) According to kinetic theory, we know walls only exert perpendicular forces on molecule. They do not exert parallel force . so there is only change in translational motion.

So pV=2/3E

Where E is representing only translational part of energy.

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

Explanation – The boy throws the ball at an angle of 60.

Horizontal component of velocity 4cos $?$ = 10cos60

=10 (1/2)

=5m/s.

so horizontal speed of the car is same, hence relative velocity of car and ball in the horizontal direction will be zero.  

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) due to the presence of external positive charge on face ABCD. The usual expression for pressure on the basis of kinetic energy will not be valid as ions would also experience electrostatic force. So presence of positive charge the isotropy is also lost.

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

Explanation – The path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore ball as well car travels equal horizontal distance, due to vertical speed, the ball follows parabolic path.

 

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

Explanation – the direction of (B $*C$ ) will be perpendicular the plane containing B and B by right hand rule. A $*(B*C)$ will lie in the plane of B and C and is perpendicular to vector A.

This is a multiple choice answer as classified in NCERT Exemplar

(d) pV=nRT

So n= PV/RT

As number moles are fixed

$\frac{P_1{V}_1}{{RT}_1}=\frac{P_1{V}_1}{{RT}_1}$

P₂=p₁V₁ (T₂/V₂T₁)

= $\frac{p\mathrm{V\; (}\left(1.1\mathrm{T)}\right)}{1.05\mathrm{V\; (}\left(\mathrm{T)}\right)}$

= P $*\frac{1.1}{1.05}\approx 1.05P$