Class 11th

1. Since the ordered pairs are equal, the corresponding elements are equal.

$\therefore \frac{x}{3}+1=\frac{5}{3}.$ and $y-\frac{2}{3}=\frac{1}{3}$

$\frac{x}{3}=\frac{5}{3}-1$ $y=\frac{1}{3}+\frac{2}{3}$

$\Rightarrow \frac{x}{3}=\frac{5-3}{3}$ $\Rightarrow y=\frac{1+2}{3}$

$\Rightarrow \frac{x}{3}=\frac{2}{3}$ $\Rightarrow y=\frac{3}{3}$

$\Rightarrow x=2$ y = 1.

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

Explanation- mass of helicopter m =2000kg

Mass of the crew and passengers m= 500kg

Acceleration a =15m/s² and g = 10m/s²

a) force on the floor of the helicopter by the crew and passengers

m (g+a)= 500 (10+15)N= 12500N

b) action of rotor of the helicopter on the surroundings air= (m₁+m₂) (g+a)

= (2000+500) (10+15)= 2500 (25)= 62500N

c) force on the helicopter due to surroundings air

= reaction of force applied by helicopter

= 62500N

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

i. The  boiling points of liquid A  is approximately 315 K and of liquid B is approximately 345 K (shown below graphically).

ii. Liquid C in a closed vessel will not boil as the pressure keeps on increasing.

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

i. CO₂ will exist in a gaseous state between the points a and b at temperature T₁. 

ii. At point b, CO₂ will start liquefying when temperature is T₁

iii. At point g,   CO₂ will be completely liquefied when temperature is  T₂

iv. No, the condensation will not take place  when the temperature is T₃ because T₃ > T_c.

v. Between  b and c of the isotherm at T₁ , represents the liquid and  gaseous CO₂ at equilibrium.

This is a assertion and reason answer type question as classified in NCERT Exemplar

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

Explanation -for the box to just starts sliding down mg

sin $\theta =t=\mu N=\mu mgcos\theta$

tan $\theta$ = $\mu$

$\theta ={tan}^{-1}\left(\mu \right)$

b) when angle of inclination is increased to $\alpha >\theta$

F₁= mgsin $\alpha -f=mgsin\alpha -\mu N$

= mg ( $sin\alpha -\mu cos\alpha$ )

c) to keep the box stationary, upward force needed F₂= m $gsin\alpha$ +f= mg ( $sin\alpha +\mu cos\alpha$ )

d) if the box is to move with an upward acceleration a then upward force needed 

F₃=mg ( $sin\alpha +\mu cos\alpha$ )+ma

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i)  Both A and R are true and R is the correct explanation of A.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Both A and R are true and R is the correct explanation of A.

This is a assertion and reason answer type question as classified in NCERT Exemplar

option (iii). A is true but R is false. 

At high altitude, the atmospheric pressure is low. 

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

Explanation- on resolving forces into rectangular components in equilibrium forces (F₁+1/ $\sqrt[]{2}$ )N are equal to $\sqrt[]{2}N$  and F₂ is equal to ( $\sqrt[]{2}+1/\sqrt[]{2}$ )N

F₁+1/ $\sqrt[]{2}$ = $\sqrt[]{2}$

F₁= $\sqrt[]{2}-1/\sqrt[]{2}$ = $\frac{2-1}{\sqrt[]{2}}=\frac{1}{\sqrt[]{2}}=0.707N$

F₂= $\sqrt[]{2}+\frac{1}{\sqrt[]{\sqrt[]{2}}}=\frac{2+1}{\sqrt[]{2}}=\frac{3}{\sqrt[]{2}}=2.121N$