Physics NCERT Exemplar Solutions Class 11th Chapter Eight

We know

Energy on any orbit is given by,

$E_n=\frac{-13.6}{n^2}{z}^2$

${}_{}\frac{}{}$  $\Rightarrow E_1=\frac{-13.6}{12}*9$ [n1 = 1] (1)

For 3rd orbit

$$ $E_3=\frac{-13.6}{9}*$   [n2 = 3]

E3 = 13.6ev

$\Delta E=E_3-E_1$

= 13.6 – (13.6 * 9)

$\Delta E=8*13.6ev=photonenergy$

$8*13.6ev=\frac{hc}{\lambda }$

$\Rightarrow 8*13.6ev=\frac{1242ev}{\lambda }nm$

$\lambda =11.415*10^9$

= 114.15 * 1010m

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\frac{}{}$ $\Rightarrow \frac{dy}{D}=\left(2n+1\right)\frac{\lambda }{2}$   [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{dy}{D}=\frac{\lambda }{2}$

$\lambda =\frac{2d}{D}y$

$$ $=\frac{2d}{D}*\frac{d}{2}[y=\frac{d}{2},Given$  first dark fringe is observed on the screen directly opposite to one of the slits]

$\lambda =\frac{2*0.6*0.6}{800*2}$

$\lambda =450mm$

for lamp,   $P=\frac{v^2}{R}$

$\Rightarrow R=\frac{25*25}{5}=125\Omega$              

$l_{max}=\frac{v}{R}=\frac{25}{125}=\frac{1}{5}A$    

$\Rightarrow I_{max}=\frac{220}{125+R}=\frac{1}{5}$

$\Rightarrow 1100=125+R$        

R = 975  $\Omega$

T₂ = 200 k

$T_1=527°C+273=800K$      

w = 12000 kJ = 12 * 10⁶ J

Q₁ =?

  $x=1-\frac{T_2}{T_1}=\frac{w}{Q_1}=1-\frac{200}{800}=\frac{12*10^6}{Q_1}$              

$\Rightarrow \frac{3}{4}=\frac{12*10^6}{Q_1}$      

$\Rightarrow Q_1=16*10^{6}J$        

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_y=20sin\alpha ,V_y=vsin\beta$

$V_y=V_y+a_{y}t$

$$ $Vsin\beta =20sin\alpha -10*10$  ……. (ii)

$\Rightarrow (ii)÷(i)$

$\Rightarrow tan\beta =tan\alpha -5sec\alpha$

1°  ® 60'

              Þ 60' ® 10°

              L = 1.5 * 10¹¹ m

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- for small objects  sizes less than 100 m centre of mass very close with the centre of gravity of the body. But when the size of object increases it weight changes and its CM and CG become far from each other.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c

Explanation- a geostationary satellite is having same sense of rotation as that of earth i.e west east direction.

A polar satellite goes around earth's pole in north south direction.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a,c,d

Explanation- F₁=-F₂=- $\frac{r_{12}}{r_{12}^3}G{M}_o^2\left(\frac{m_1{m}_2}{M_o^2}\right)$ ⁿ

r₁₂=r₁-r₂

acceleration due to gravity ,g = F/mass= $\frac{G{M}_o^2({m_1{m}_2)}^n}{r_{12}^2{\left(M_o\right)}^{2n}}*\frac{1}{mass}$

as g is constant hence constant of proportionality will not be constant in kepler's third law.

Hence kepler's law will not be valid.

For negative n, g= $\frac{G{M}_o^2{\left(m_1{m}_2\right)}^{-n}}{r_{12}^2{\left(M_o\right)}^{-2n}}*\frac{1}{mass}$

= $\frac{G{M}_o^{2(1+n)}}{r_{12}^2}\frac{{{(m}_1{m}_2)}^{-n}}{mass}$

M_o>m₁ or m₂ so objects lighter than water will sink

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c, d

Explanation- Gravitation force F= $\frac{GMm}{r^2}$   where M is mass of sun and m is mass of earth.

When G decrease force will also decrease so according to this it will not in the fixed circular orbits around the sun. as when radius increases the force will decreases and soon earth leave the solar system.