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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.

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

Explanation- - given |A|=2 and |B|=4

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

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

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

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

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

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

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

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

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

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

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The commission allocated 200 marks for the UPSC IES 2025 personality test or an interview. This is the third stage, or the final stage, of the UPSC IES exam. The first stage is the preliminary examination, and the second stage is the Mains examination.

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

Explanation- given |A|=2 and |B|=4

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

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

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

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

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

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

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

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

d) |A $\times$ 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

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

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

Time period = 1 day = 24 $\times 60\times 60$ = 86400s

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

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

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

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

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

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

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

Let $\theta$

Speed of the plane =720km/h = 720 $\times 5/18$ = 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 $\times t$ =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’

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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.

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BBA applicants must bring key documents such as their Class 12 marksheet, transfer certificate, valid ID proof, recent passport-size photographs, and other supporting academic documents. Since document requirements may occasionally vary, students should refer to the latest list provided on the official NCAS website before attending the verification session.

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.