Class 11th

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- v_x=2t for 0

= 2 (2-t) for 1

=0 for t>2s

V_y= t for 0

= 1 for t>1s

F_x= ma_x= mdv_x/dt= 1 (2)

F_y = ma_y= mdv_y/dt

= 1 (1) for 0

F= F_x? +F_y?

= 2? +?

=-2?

=0

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- as angle is 45

On smooth inclined plane acceleration will be a = gsin $\theta$

So acceleration will become a= g/ $\sqrt[]{2}$

Using equation of motion s =ut +1/2at²

S= $\frac{1}{2}\frac{g}{\sqrt[]{2}}{T}^2$

On rough inclined plane a = g (sin $\theta -\mu cos\theta$ )

= g (sin $45-\mu cos45$ )= $\frac{g(1-\mu )}{\sqrt[]{2}}$

So s=ut +1/2at²

S= 0+ $\frac{1}{2}\frac{g\left(1-\mu \right)}{\sqrt[]{2}}\left(pT\right)$ ²

Comparing two above distance

$\frac{\mathrm{}}{}$$\frac{1}{2}\frac{g\left(1-\mu \right)}{\text{exception 25:}}\left(pT\right)$²= $\frac{1}{2}\frac{g}{\text{exception 25:}}{T}^2$

So after solving we get $$$\mu =(1-1/p^2)$

This is a matching answer type question as classified in NCERT Exemplar

This is a matching answer type question as classified in NCERT Exemplar

This is a short answer type question as classified in NCERT Exemplar

1. At low pressure, the curve of real gas coincides  with that of ideal gas, this shows that the deviation of  behaviour of real gas with respect to ideal gas is small or negligible.

2. At high pressure, the curve of real gas is far apart from  ideal gas, this shows that the deviation of  behaviour of real gas with respect to ideal gas is large.

3. The  pressure p₁ and volume V₁  are  the point where real gas behaves as an ideal gas. 

This is a short answer type question as classified in NCERT Exemplar

1. According to Boyle's law,  

Pressure of a gas is inversely proportional  to volume of gas at constant temperature. So, the volume decreases with increase in pressure at constant temperature.

2. According to Charles's law,

Volume of a gas is directly proportional to temperature when the pressure is constant. So, the volume of gas increases with increase in temperature.

This is a short answer type question as classified in NCERT Exemplar

The increase in temperature increases the kinetic energy of the molecules which decreases the intermolecular forces operating between its particles and hence, the viscosity of a liquid decreases. So, the viscosity of a liquid  decreases if its temperature is increased.

This is a short answer type question as classified in NCERT Exemplar

Hexane is a  nonpolar molecule which has london force between the molecules, which is a weak force. 

Water and glycerine have  O atoms which is an electronegative atom that forms H bonding between the molecules along with dipole-dipole interaction. 

Glycerine has three  O atoms, so it forms more H bonding and hence, has stronger  intermolecular forces.

So the increasing order of intermolecular forces is Hexane   <   Water  <  Glycerin. Stronger the  intermolecular forces, the greater is the viscosity, so the increasing order of their viscosities is: 

Hexane   <   Water  <  Glycerin

This is a short answer type question as classified in NCERT Exemplar

The two phenomena that can be explained on the basis of surface tension are:

1. Spherical shape of rain droplets.

2. Capillary action due to which the liquid in capillary rises and falls.

This is a short answer type question as classified in NCERT Exemplar

Unit of 'P'  =  N m^-2

Unit of 'a'   =  N m^-2  X   (m³)²  / (mol)²

= N-m⁴ mol^-2

Unit of 'a' when pressure is in atm, and volume in dm³   

Unit  of 'P' =  atm

Unit of 'a'   =  atm  X  (dm³)²  / (mol)²

= atm-dm⁶ mol^-2