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NEET 2022 Physics Questions with Answers Key Solutions
Solution:
XCM=20×1020+10=203m
Solution:
(a) Radio wave (ii) ≈102m (ii)
(b) Microwave ≈ (iii) 10−2m (iii)
(c) Infrared radiations ≈(iv)10−4m (iv)
(d) X-ray (i) ≈A˚ (i)
(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Solution:
E=P \times t=100 \times 10^{3} \times 3600
=36 \times 10^{7} J
Solution:
1 : Isochoric
2 : Adiabatic
3 : Isothermal
4 : Isobaric
Solution:
Plane angle and solid angle are dimensionless but have units.
Q6. In half wave rectification, if the input frequency is 60 \,Hz, then the output frequency would be:
Solution:
In half wave rectification
f _{ in}= f _{ out }
\Rightarrow f _{\text {out }}=60 \,Hz
Solution:
hv = W +\frac{ v _{0}}{2} e
\frac{ h v}{2}= W + v _{0} e
on solving we get, W =3 / 2 h v
h _{ o }=3 / 2 hv
v _{0}=3 / 2 v
Solution:
Velocity is slope of x-t graph
V =\frac{ dx }{ dt }=\tan \theta
\frac{ V _{1}}{ V _{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}
=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}=\frac{1}{\sqrt{3}}
Solution:
\left[ MLT ^{-2} A ^{-2}\right]= Magnetic permeability
Solution:
Peak voltage is \sqrt{2} times rms voltages in ac.
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Solution:
S _{ nth }= u +\frac{ a }{2}(2 n -1)
=0+\frac{ a }{2}(2 n -1)
S _{ nth } \propto(2 n -1)
\Rightarrow S _{1 st }, S _{2 nd }, S _{3 rd }, S _{4 th }
=[2(1)-1]:[2(2)-1]:[2(3)-1]:[2(4)-1]
=1: 3: 5: 7
Solution:
I_{g}=\frac{F}{m}
=\frac{3}{60 \times 10^{-3}}=50 \,N / kg
Solution:
{ }_{11}^{22} Na \longrightarrow X +e^{+}+v
This is \beta^{+}-decay
{ }_{11}^{22} Na \longrightarrow{ }_{10}^{22} Ne + e ^{+}+ v
Solution:
P = P _{0}+\frac{4 T }{ R }
\Rightarrow R increases and P decreases
Solution:
As both resistors are in parallel combination so potential drop ( V ) across both are same.
P =\frac{ V ^{2}}{ R } \Rightarrow P \propto \frac{1}{ R }
\frac{ P _{1}}{ P _{2}}=\frac{ R _{2}}{ R _{1}}=\frac{200}{100}=\frac{2}{1}
=2: 1
Solution:
V =\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{ Q }{ R }
\frac{1}{4 \pi \epsilon_{0}}= constant
Q = same (Given)
\therefore V \propto \frac{1}{ R }
\therefore Potential is more on smaller sphere.
Solution:
\omega =\omega_{0}+\alpha t
\alpha =\frac{\omega-\omega_{0}}{ t }
=\frac{(3120-1200)}{16 s } rpm
=\frac{1920}{16} \times \frac{2 \pi}{60} rad / s ^{2}
=4 \pi rad / s ^{2}
Solution:
n=\sqrt{\epsilon_{ r } u_{r}}
n=\frac{c}{v} \Rightarrow v=\frac{c}{n}
v=\left(\frac{c}{\sqrt{\epsilon_{r} \mu_{r}}}\right)
Solution:
B =\mu_{0} ni =\mu_{0} \frac{ N }{\ell} i
\therefore B =4 \pi \times 10^{-7} \times \frac{100}{10^{-3}} \times 1
=12.56 \times 10^{-2} T
Solution:
By conservation of momentum :
m(0)=\frac{2 m}{5}(-v \hat{i})+\frac{2 m}{5}(-v \hat{j})+\frac{m}{5} \vec{v}^{\prime}
\Rightarrow \vec{v}^{\prime}=2 v \hat{i}+2 v \hat{j}
\Rightarrow v^{\prime}=\sqrt{(2 v)^{2}+(2 v)^{2}}
=2 \sqrt{2} v
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Solution:
For conductors \alpha is (+)ve
For semiconductors & Insulators \alpha is (-)ve
Solution:
R _{1}= R _{2}=20\, cm =0.2\, m
\mu=\frac{3}{2}
P =\frac{1}{ f }=(\mu-1)\left(\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right)
P =\left(\frac{3}{2}-1\right)\left(\frac{1}{0.2}+\frac{1}{0.2}\right)
P =\frac{1}{2}\left(\frac{2}{0.2}\right)=\frac{10}{2}=+5 D
Solution:
dB =\frac{\mu_{0}( Id \vec{\ell} \times \vec{r})}{4 \pi r ^{3}}
As per Biot Savart law, the expression for magnetic field depends on current carrying element Id \vec{\ell}, which is a vector quantity, therefore, statement-I is correct and statement-II is wrong.
Solution:
B=0.5 T
Angle between \vec{B}\, \& \,\vec{A} is zero
\phi = B.A. \cos 0
=0.5 \times(1) \times 1
=0.5 \,Wb
Solution:
Constant velocity \Rightarrow a =0
\Rightarrow T = W + f
=20000+3000
=23000 \,N
\Rightarrow Power = Tv
=23000 \times 1.5
=34500 watts
Solution:
Method (i)
By Snell's law
1 \sin 60^{\circ}=\sqrt{3} \sin r
\frac{\sqrt{3}}{2}=\sqrt{3} \sin r
\sin r=\frac{1}{2}
r=30^{\circ}
Angle between refracted and reflected ray is 90^{\circ}
Method (ii)
Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is 90^{\circ}
\tan i_{p}=\mu=\sqrt{3}
i_{p}=60^{\circ}= i
Solution:
Radius of wire =\frac{10^{-2}}{\sqrt{\pi}}
Cross sectional area A=\pi r ^{2}=10^{-4} m ^{2}
j =\frac{ i }{ A }=\left(\frac{ V }{ R }\right) \cdot \frac{1}{ A }=\frac{ E \ell}{ RA } R =\frac{\rho \ell}{ A }
j =\frac{10 \times 10}{10 \times 10^{-4}}=10^{5} A / m ^{2}
or
J =\sigma E \Rightarrow \frac{E}{\rho}=\frac{ E \ell}{ RA }=\frac{10 \times 10 \times \pi}{10 \times 10^{-4} \times \pi}
\Rightarrow 10^{5} A / m ^{2}
Solution:
In (a) & (c) circuits, both the junctions are in same biasing conditions so offers equal resistances.
Since both are in series, therefore equal potential will drop across the junction.
Solution:
v \propto \sqrt{\text { Tension }}
\frac{ v _{ i }}{ v _{ f }}=\sqrt{\frac{ T _{ i }}{ T _{ f }}}
\frac{ v _{ i }}{ v _{ f }}=\sqrt{\frac{ T }{2 T }}
\frac{ v _{ i }}{ v _{ f }}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}
Solution:
k =\sqrt{\frac{ I }{ m }}
\Rightarrow \frac{ k _{1}}{ k _{2}}=\sqrt{\frac{ I _{1}}{ I _{2}}}
=\sqrt{\frac{ mR ^{2} / 2}{ mR ^{2} / 4}}=\sqrt{2}: 1
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Solution:
y=(n \lambda)\left(\frac{D}{d}\right)
n _{1} \lambda_{1}= n _{2} \lambda_{2}
(8) (600 \,nm )= n _{2}(400)
n _{2}=12
Solution:
Electric field is always perpendicular to EPS.
Solution:
Initially speed is zero, then increases & after some time it becomes constant.
Acceleration (slope of v/t curve) of ball first decreases and after some time it becomes zero.
Solution:
First excited state \Rightarrow n =2
T _{1}=-13.6 \frac{ z ^{2}}{ n ^{2}}=-\frac{13.6}{4} eV
Second excited state \Rightarrow n =3
T _{2}=-13.6 \frac{ z ^{2}}{ n ^{2}}=-\frac{13.6}{9} eV
T _{1}: T _{2}=\frac{1}{4}: \frac{1}{9}=9: 4
Solution:
\lambda=\frac{ h }{ p }
Graph will be hyperbolic
Solution:
( n ) T _{\ell}=( n +1) T _{ s }
( n ) 2 \pi \sqrt{\frac{1.21}{ g }}=( n +1) 2 \pi \sqrt{\frac{1}{ g }}
( n )(1.1)=( n +1)
0.1( n )=1
n =10
No. of oscillation of smaller one
= n +1
=10+1
=11
Solution:
V = (no. of moles) (22.4 litre)
=\frac{\text { mass }}{\text { molar mass }}\left(22.4 \times 10^{-3} m ^{3}\right. )
=\frac{4.5 \times 10^{3}}{18} \times 22.4 \times 10^{-3} m ^{3}
=5.6 \,m ^{3}
Solution:
Nuclear Radius :
R = R _{0}( A )^{1 / 3}
\frac{ R (125)}{ R (64)}=\frac{ R _{0}(125)^{1 / 3}}{ R _{0}(64)^{1 / 3}}=\frac{5}{4}
Solution:
In stretching of a spring shape charges therefore shear modulus is used.
Y _{\text {copper }}< Y _{\text {steel }}
Solution:
We have the initial energy as \frac{1}{2} CV ^2=\frac{1}{2} \times 9 \times 10^{-10} \times 100^2=4.5 \times 10^{-6} J
Now the capacitor is disconnected and is connected to another 900 pF capacitor.
In the steady situation, the two capacitors have their positive plates at the same potential.
Let the common potential difference be V '. The charge on each capacitor is Q ^{\prime}= CV '.
By charge conservation, Q^{\prime}=Q / 2. This implies V^{\prime}=V / 2.
The total energy of the system is =2 \times(1 / 2) Q ^{\prime} V '=(1 / 4) QV =2.25 \times 10^{-6} J
Thus we get the electrostatic energy stored by the system as \frac{1}{2}\left(C_1+C_2\right) V^2= \frac{1}{2}\left(2 \times 9 \times 10^{-10}\right) \times 50^2=2.25 \times 10^{-6}
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Solution:
Area = Length \times Breadth
=55.3 \times 25
=1382.5
=14 \times 10^{2}
Resultant should have 2 significant figures.
Solution:
C=\overline{A \cdot B} \cdot \overline{\bar{A}} \cdot B
using De-Morgan Theorem
C=\overline{ A \cdot B +\overline{ A } \cdot B }
C =\overline{ B ( A +\overline{ A })}=\overline{ B }
Therefore
Solution:
\mu=\frac{ C }{ u } \Rightarrow u \propto \frac{1}{\mu}
Critical angle
\operatorname{Sini}_{ c }=\frac{\mu_{ R }}{\mu_{ D }}=\frac{ u _{ D }}{ u _{ R }}=\frac{1.5}{2}=\frac{3}{4}
i _{ c }=\sin ^{-1}\left(\frac{3}{4}\right)
\sin \dot{i}_{ c }=\frac{\mu_{ R }}{\mu_{ D }}=\frac{ u _{ D }}{ u _{ R }}
i _{ c }=\sin ^{-1}\left(\frac{3}{4}\right)
Solution:
i _{\max }=\frac{ E _{\max }}{ R }=\frac{ NBA \omega }{ R }
i _{\max }=\frac{1000 \times 2 \times 10^{-5} \times \pi\left(10^{2}\right) \times 2}{12.56}
i _{\max }=1 A
Solution:
Resistance of P \& Q should be approx. equal as it decreases error in experiment.
Solution:
It is electric dipole at large distance electric field intensity
E=\frac{K P}{R^{3}} \sqrt{1+3 \cos ^{2} \theta}
\therefore E \propto \frac{1}{R^{3}}
Solution:
Gravitational constant =\left[ M ^{-1} L ^{3} T ^{-2}\right]
Gravitational potential energy =\left[ ML ^{2} T ^{-2}\right]
Gravitational potential =\left[ L ^{2} T ^{-2}\right]
Gravitational intensity =\left[ LT ^{-2}\right]
Solution:
\omega=100
v=\frac{\omega}{2 \pi}=\frac{100}{2 \pi}=\frac{50}{\pi} Hz
Resonance frequency
v_{0}=\frac{1}{2 \pi \sqrt{ LC }}=\frac{1}{2 \pi} \sqrt{\frac{1}{10 \times 10 \times 10^{-6}}}
=\frac{50}{\pi} Hz
Solution:
At highest point only horizontal component of velocity remains
\Rightarrow u_{x}=u \cos \theta
u _{ x }= u \cos \theta =10 \cos 30^{\circ}
=5 \sqrt{3} \,ms ^{-1}
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