Class 11th

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. 

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

Answer- c

Explanation- in a uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant . in this situation body A will be in translatory motion.

This is a multiple choice answer as classified in NCERT Exemplar

Option (iii) Decrease 

The increase in temperature increases the kinetic energy which can overcome intermolecular forces of attraction due to which the viscosity decreases and the liquid starts flowing. 

This is a multiple choice answer as classified in NCERT Exemplar

Option (i). B only 

For all ideal gas PV = constant. Only B has no change with the change in PV, so only line B represents the curve of ideal gas.

This is a multiple choice answer as classified in NCERT Exemplar

option (i). Shimla 

The temperature at which the liquid boils when the vapour pressure becomes equal to the atmospheric pressure  is known as the boiling point of the liquid. When the  atmospheric pressure is low, the boiling point of the liquid will be low. As Shimla has the lowest atmospheric pressure, it will boil first.

This is a multiple choice answer as classified in NCERT Exemplar

option (ii) Nsm^-2 

viscosity coefficient (η) = $\frac{Force}{Area*VelocityGradient}$

SI unit of (η)= Nsm^-2

This is a multiple choice answer as classified in NCERT Exemplar

Option (iv) O₂, N₂, H₂, He

Higher the critical temperature of gases more easily, the gas will be liquefied. So, the order of liquefying of gases will be  O₂ > N₂ > H₂ >He. 

This is a multiple choice answer as classified in NCERT Exemplar

Option (i) increases

Gay Lussac's law shows the direct relationship between the pressure and the temperature of a fixed amount of gas at constant volume.

Gay Lussac's law states that at constant volume, the pressure of a fixed amount of a gas is directly proportional to the temperature.

Therefore,  

P∝T   ……….at constant volume

Thus, pressure increases if the temperature is increased at a constant volume.

To elaborate more on this topic, we will see a real-life example.

This is a multiple choice answer as classified in NCERT Exemplar

option (iii) 8 * 10⁴  Nm^-2 

Partial pressure of oxygen, Po₂= Xo₂   $*$ P_total

Mole fraction of O₂= $\frac{Moleofoxygen}{Molesofoxygen+Molesofhydrogen}$

= $\frac{4}{4+1}$ = $\frac{4}{5}$

Po₂= $\frac{4}{5}$ $*$ 1 = $\frac{4}{5}$  (?P_total = 1 atm)

1 atm =  1.1034 * 10⁵ Nm^-2 or Pa

Partial pressure of dioxygen = 0.8 *10⁵ = 8*10⁴  Nm^-2