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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.  

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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.

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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.

After completing the BBA registration process, NCAS reviews all received applications and prepares a merit list based on academic performance in Class 12. Only students who meet the eligibility criteria and feature on this list are shortlisted for admission. Being listed is essential to move forward with further admission formalities.

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 $\times \frac{1.1}{1.05}\approx 1.05P$

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

Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

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

Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

This is a multiple choice answer as classified in NCERT Exemplar

(b) For a function f (v), the number of molecules n=f (v) which are having speeds between v and v+dv

So f₁ (v) and f₂ (v) will obey’s maxwell distribution law

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

Explanation- vsin $\theta$ = vertical component

V_x= horizontal component of velocity =vcos $\theta$ = constant

V_y = vertical component of velocity =vsin $\theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

This is a multiple choice answer as classified in NCERT Exemplar

(d) When number of molecules breaks then number of moles would become double.

P $\propto nT$ where n is the number of moles and T is temperature of ideal gas

As gases breaks number of moles becomes twcice of initial so n₂ =2n₁

$\frac{P_2}{P_1}=\frac{n_2{T}_2}{n_1{T}_1}=\frac{2n_1\left(3000\right)}{n_1\left(300\right)}$ = 20

p₂=20p₁

so 20 times

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

Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

This is a multiple choice answer as classified in NCERT Exemplar

(a) Pressure is inversely proportional to slope

So $\frac{p_1}{}$$\frac{p_{}}{}$$\frac{2_{}}{}=\frac{m_2}{m_1}\mathrm{<1}$

So p₁ > p₂