Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation – because alpha and beta particle have some charge and mass so their energy level will change but in case of gamma particle it does not contain any charge.

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

Explanation- as the refractive index of the class $\mu$ , the path difference will be calculated as $?x$ =2dsin $\theta$ +( $\mu -1$ )L

For principal maxima ,(path difference is zero)

2dsin $\theta$ ₀+( $\mu -1$ )L=0

Sin $\theta$ ₀= - $\frac{L(\mu -1)}{2d}$ = $\frac{-L\left(0.5\right)}{2d}$

Sin $\theta$ ₀=-1/16

OP=Dtan $\theta$ ₀= Dsin $\theta$ ₀=-D/16

For pat $?$ h difference $?\frac{\lambda }{2}$

2dsin $\theta$ ₁+0.5L= $?\frac{\lambda }{2}$

Sin $\theta$ ₁= $\frac{?\frac{\lambda }{2}-0.5L}{2d}$ = $\frac{?\frac{\lambda }{2}-\frac{d}{8}}{2d}$

= $\frac{\frac{\lambda }{2}-\frac{\lambda }{8}}{2\lambda }$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$  and- $\frac{1}{4}-\frac{1}{16}$ = $-\frac{5}{16}$

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

Total current i=i1+i2

Vmsinwt=Ri1

I1= $\frac{V_{m}sinwt}{R}$

If q2 is charge on the capacitor at any time t then for series combination of C and L

Applying KVL in below circuit

So q2=qmsin(wt $\frac{q_2}{C}+\frac{Ld{i}_2}{dt}-V_{m}sinwt=0$ )……….1

qm[ $+\varnothing$ sin(wt+ $\frac{q_2}{C}+\frac{L{d}_2{q}_2}{d{t}^2}=V_{m}sinwt$ )]= Vmsinwt

if $\frac{d{q}_2}{dt}=-q_m{w}^2\mathrm{s}\mathrm{i}\mathrm{n}?(wt+\varnothing )$

qm= $usethisvalueineqn1$

from above equation i2= $\frac{1}{C}+L(-w^2)$

i2= $\varnothing$ when $\varnothing =0$

so total current I = i1+i2

I= $\frac{V_m}{(\frac{1}{c}-L{w}^2)}$

I= i1+i2= Ccos $\frac{d{q}_2}{dt}=w{q}_m\cos ?\left(wt+\varnothing \right)$ +Csin $\frac{w{V}_m\mathrm{c}\mathrm{o}\mathrm{s}?(wt+\varnothing )}{\frac{1}{c}-L{w}^2}$

I= Csin(wt+ $\varnothing =0,i_2=\frac{V_{m}coswt}{\frac{1}{wC}-Lw}$ )

C= $\frac{V_{m}sinwt}{R}+\frac{V_{m}coswt}{\frac{1}{wC}-Lw}$

$\varnothing sinwt$ 1/2

$\varnothing coswt$ = tan-1 $\varnothing$

$\sqrt[]{A^2+B^2}$ 1/2

This is the expression for impedance.2

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

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

Power drawn, p= 2KW= 2000W

tan $\varnothing$ =-3/4

P=V2/Z

Z=V2/P= $\frac{223*223}{2*{10}^3}$ =25

Z = $\sqrt[]{R^2+({X_L-X_C)}^2}$

25= $\sqrt[]{R^2+({X_L-X_C)}^2}$

625 = $R^2+({X_L-X_C)}^2$ ………… (1)

tan $\varnothing$ = $\frac{X_L-X_C}{R}$ = ¾

XL-XC= 3R/4

Use this in eqn 1

625= R2+ (3R/4)2 = R2+9R2/16

625 = 25R2/16, R= 20ohm

XL-XC=15ohm

Im= $\sqrt[]{2}$ I= $\sqrt[]{2}V/Z$ = $\frac{223*\sqrt[]{2}}{25}$ = 12.6A

If R, XL and Xc are all doubled, tan $\varnothing$ does not change . Z is doubled, current is halved. So power is also halved.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- F=GMm/r²

M=effective mass of hydrogen atom=mass of electron +mass of proton -B²/c

B.E of hydrogen atom = 13.6eV

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer (c)

Explanation- it is a process in which metal decays spontaneously . so during a span of 1 year it will decay almost half of its original value so it will come close to 5000.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- a nuclide 1 is said to be mirror isobar of nuclide 2, if Z₁=N₂ and Z₂= N₁

Now ₁₁Na²³, ₁₂Mg²³ for which Z₂=12=N₁ and N₂ =23-12=11=Z₁

(b) As ₁₂Mg²³ contain even number of protons against ₁₁Na²³ which has odd number of protons (11), therefore Mg has greater binding energy than sodium.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Each particle (neutron and proton) present inside the nucleus is called nucleon .

Let $\lambda ={10}^{-15}m$

$\lambda =\frac{h}{p}$

and KE= PE

$E=hc/\lambda$ = $\frac{6.6*{10}^{-34}*3*{10}^8}{{10}^{-15}*1.6*{10}^{-19}}$  eV

K.E= 10⁹eV