Class 11th

7. i. Let the points be P (0, 7, –10), Q (1, 6, –6) and R (4, 9, –6)

So,

6. Here, a_n=n $\frac{x^2+5}{4}$

Putting n=1,2,3,4,5 we get,

$a_1=1*\frac{\left(1^2+5\right)}{4}=1*\frac{6}{4}=\frac{3}{2}$

$a_2=2*\left(\frac{2^2+5}{4}\right)=2*\frac{(4+5)}{4}=\frac{9}{2}$

$a_3=3*\frac{\left(3^2+5\right)}{4}=3*\frac{(9+5)}{4}=3*\frac{14}{4}=\frac{21}{2}$

$a_4=4*\frac{\left(4^2+5\right)}{4}=4*\frac{(16+5)}{4}=4*\frac{21}{4}=21$

$a_5=5*\frac{\left(5^2+5\right)}{4}=5*\frac{(25+5)}{4}=5*\frac{30}{4}=\frac{75}{2}$

Hence, the first five terms are $\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$ .

3. Here a_n=2ⁿ

Substituting n=1,2,3,4,5 we get,

a₁=2¹=2

a₂=2²=4

a₃=2³=8

a₄=a⁴=16

a₅=2⁵=32.

Hence the first five terns are 2,4,16,32 and 64.

2. Here, a₁= $\frac{n}{n+1}$

Substituting n=1,2,3,4,5 we get,

$a_1=\frac{1}{1+1}=\frac{1}{2}$

$a_2=\frac{2}{2+1}=\frac{2}{3}$

$a_3=\frac{3}{3+1}=\frac{3}{4}$

$a_4=\frac{4}{4+1}=\frac{4}{5}$

$a_5=\frac{5}{5+1}=\frac{5}{6}$ .

Hence the first five terns are $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}$ and $\frac{5}{6}$ .

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