Class 11th

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- a

Explanation- x (t)= pt+qt²+rt³

V= dx/dt=p+2qt+3rt²

So a= o+2q+6rt

At t=2s a= 2q+6 $*$ 2 (r)

= 2q+12r

= 2 (4)+12 (5)=68m/s

So F =ma= 2 (68)= 136N

This is a multiple choice answer as classified in NCERT Exemplar

option (ii). High pressure and

(iii). Low temperature

The real gases obey the ideal gas equation, PV= nRT at low pressure and high temperature. So, a gas does not obey the ideal gas equation or behave ideally at high pressure and low temperature.

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- c

Explanation –consider the adjacent diagram

 OA= p₁= initial momentum of player northward

AB= p₂= final momentum of player towards west

OB= OA+AB

Change in momentum = P₂-P₁= AB-OA so we can say that AR will be along south west.

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- d

Explanation-we know that for a system F_ext=dp/dt

If F_ext=0, dp =0,

P = constant

Hence, momentum of a system will remain conserve if external force on the system is zero.

In case of collision between particles, equal and opposite forces will act on individual particles by newton's third law.

according to newton's 2^nd law F=dp/dt

If F =0 then p = constant so momentum remains constant if external force is zero.

This is a multiple choice answer as classified in NCERT Exemplar

Option (i).  16 grams of gas and

(iv). 11.2 litres of gas 

1 mole of dioxygen at STP  occupies   =  22.4 L    

molar mass of dioxygen  =   32 g 

contain  6.023*10²³ molecules of  oxygen gas 

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- c

Explanation- change in momentum $?p$ =- [0.9? +1.2? ]

Magnitude = $\sqrt[]{{0.9}^2+{1.2}^2}$

$=\sqrt[]{0.81+1.44}=$ 1.5 kg m s^-1       

This is a Multiple Choice  type Questions as classified in NCERT Exemplar

Answer- c

Explanation- given u= (3? +4? )m/s

And v= - (3? +4? )m/s, and mass of ball is 150g=0.15kg

Change in momentum = final momentum – initial momentum

 = mv-mu

= m (v-u) = 0.15 [- (3? +4? )- (3? +4? )]

= - [0.15 $*6i+0.15*8j$ ]

= - [0.9? +1.2? ]

$?$ P =- [0.9? +1.2? ]

This is a multiple choice answer as classified in NCERT Exemplar

Option (ii). Complete disorder of molecules and

(iii). Random motion of molecules 

In a gaseous state of matter, molecules are arranged in complete disorder and  they move in random motion in all directions.

This is a Multiple Choice type Questions as classified in NCERT Exemplar

Answer-b

Explanation- to solve this question we have to apply newton's law of motion, in terms of force and change in momentum.

as we know F= dp/dt

As body moving uniform velocity so dp= 0

So F=0

As all part of scale is moving with uniform velocity and total force is zero . hence torque will also be zero.

This is a multiple choice answer as classified in NCERT Exemplar

Option (ii).  Decreases

When the temperature increases, kinetic energy of the molecule increases which leads to decrease in the intermolecular forces and hence, decreases the surface tension.