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NEET 2021 Physics Questions with Answers Key Solutions
Solution:
$B=\frac{\mu_{0} i^{i}}{2 \pi r}$
$F=e v B=\frac{e v \times \mu_{0} i}{2 \pi r}$
$=\frac{\left(1.6 \times 10^{-19}\right) \times 10^{5} \times 4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.2} \,N$
$=\frac{1.6 \times 2 \times 5}{2} \times 10^{-20} \,N$
$\Rightarrow F=8 \times 10^{-20} \,N$
Solution:
Potential energy of a particle executing SHM is given by
$U =\frac{1}{2} m \omega^{2} x ^{2}$
So, if the frequency of oscilation is $n$,
then frequency of its potential energy will be $2 n$.
Solution:
${ }_{z} X ^{A} \rightarrow{ }_{z-1} B \left(\beta^{+}\right.$decay $)$
${ }_{z-1} B \rightarrow_{z-3} C (\alpha-$ decay $)$
${ }_{z-3} C \rightarrow{ }_{z-2} D \left(\beta^{-}\right.$decay $)$
Solution:
$V_{e}=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G \frac{4}{3} \pi R^{3} \rho}{R}}$
$ \Rightarrow V_{e} \propto R$
$\therefore \frac{V'}{V}=\frac{4 R}{R}$
$ \Rightarrow V^{'}=4 V$
Solution:
$A=\frac{A_{0}}{2^{t / T_{1 / 2}}} \Rightarrow \frac{A}{A_{0}}$
$=2^{-t / T_{1 / 2}}=2^{-\frac{150}{100}}$
$=2^{-3 / 2}=\frac{1}{2 \sqrt{2}}$
Solution:
Point $F$ is the second focal length of $A$ and first focal length of $B$.
$\therefore d=(20-5) \,cm =15\, cm$
Solution:
$I_d=V_o \omega C \cos \omega t$
Given $V=V_0 \sin \omega t$
Now, displacement current $i_d$ is given by,
$
I_d=C \frac{d V}{d t}=C \frac{d}{d t}\left(V_0 \sin \omega t\right)
$
$= C \left( V _0 \omega\right) \cos \omega t$
$=V_0 \omega C \cos \omega t$
Solution:
$S _{ n }=0+\frac{1}{2} a (2 n -1)$
$S _{ n +1}=0+\frac{1}{2} a (2( n +1)-1)$
$=\frac{1}{2} a (2 n +1)$
$\frac{ S _{ n }}{ S _{ n +1}}=\frac{(2 n -1)}{(2 n +1)}$
Solution:
Conserving energy,
$K + U = mgs$
$ \Rightarrow 3 U + U = mgs $
$\Rightarrow 4 mgh = mgs $
$\Rightarrow h = S / 4$
$\therefore K=3 m g \frac{S}{4}$
$\Rightarrow \frac{1}{2} m v^{2}=\frac{3}{4} m g s$
$ \Rightarrow v=\sqrt{\frac{3 g s}{2}}$
Solution:
Potential gradient $=\frac{1.5}{36}$ [Initially]
Finally $\frac{2.5}{I_{2}}=\frac{1.5}{36}$
$\Rightarrow I _{2}=\frac{36 \times 2.5}{1.5}=60\, cm$
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Solution:
$(\vec{ E } \times \vec{ B })$ gives the direction of propagation
$\vec{ E }=-\hat{ j }+\hat{ k }, \vec{ B }=-\hat{ j }-\hat{ k }$
$\vec{E} \times \vec{B}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & 1 \\ 0 & -1 & -1\end{vmatrix}=2 \hat{i}$
Also $\vec{E} \cdot \vec{B}=(-\hat{j}+\hat{k})(-\hat{j}-\hat{k})=0$
Solution:
$F_{D}+B=M g$
$\Rightarrow F_{D}+\left(\frac{d}{2}\right)\left(\frac{M}{d}\right) g=M g $
$\Rightarrow F_{D}=\frac{M g}{2}$
Solution:
$A \rightarrow Q$
$B \rightarrow P$
$C \rightarrow S$
$D \rightarrow R$
Solution:
$
h=60 m
$
$
\begin{array}{l}
\frac{ m }{ t }=15 kg / s \\
g =10 m / s ^{2}
\end{array}
$
Loss due to friction force is $10 \%$, only $90 \%$ input is used to generate power,
$
P = gh \frac{ m }{ t } \times \frac{90}{100}
$
$
P=10 \times 60 \times 15 \times 0.9
$
$
P=8.1 kW
$
Solution:
All of the above
Solution:
$I_{N}=$ Current in N-type semiconductor
$=\left(\mu_{ e } N _{ e }+\mu_{ n } n _{ n }\right) eAE$
$I_{P}=$ Current in P-type semiconductor
$=\left(\mu_{n} N_{n}+\mu_{e} n_{e}\right) e A E$
Now $N_{e}=N_{h}$ and $n_{e}=n_{h}$
Also $\mu_{e}>\mu_{n}$ and $N_{e}>>n_{e}, N_{n}>>n_{e}$
Where $N_{e}=$ electron concentration in $N$-type
$N_{h}=$ Hole concentration in P-type and $N _{ e }= N _{ h }$
$n _{ h }=$ hole concentration in N-type
$n _{ e }=$ Electron concentration in P-type
$\therefore I _{ N }- l _{ p }=\left[\left(\mu_{ e }-\mu_{ h }\right) N _{ e }+\left(\mu_{ h }-\mu_{ e }\right) n _{ e }\right] eAE$
$=\left(\mu_{ e }-\mu_{ h }\right)\left( N _{ e }- n _{ e }\right) eAE$
Since $\mu_{ e }>\mu_{ h }$ and $N _{ e }> n _{ e }$
We conclude $I_{N}>I_{P}$
Solution:
Final total binding energy
$=2 \times 8.5 \times 120 \,MeV =2040\, MeV$
Initial total binding energy
$=240 \times 7.6\, MeV =1824\, MeV$
Gain $=(2040-1824) \,MeV =216\, MeV$
Solution:
$B \propto r(0 \leq r \leq R)$
$B \propto \frac{1}{r}(r \geq R)$
Solution:
$\frac{ Q _{1}}{4 \pi \varepsilon_{0} R _{1}}=\frac{ Q _{2}}{4 \pi \varepsilon_{0} R _{2}}= V$
$\therefore \frac{ Q _{1}}{ Q _{2}}=\frac{ R _{1}}{ R _{2}}$
$\therefore \frac{\sigma_{1}}{\sigma_{2}}=\frac{ Q _{1} / 4 \pi R _{1}^{2}}{ Q _{2} / 4 \pi R _{2}^{2}}$
$=\frac{ Q _{1}}{ Q _{2}}\left(\frac{ R _{2}}{ R _{1}}\right)^{2}=\left(\frac{ R _{1}}{ R _{2}}\right)\left(\frac{ R _{2}}{ R _{1}}\right)^{2}$
$\Rightarrow \frac{\sigma_{1}}{\sigma_{2}}=\frac{ R _{2}}{ R _{1}}$
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Solution:
Dimensional formula of energy
$
[E]=\left[M^{1} L^{2} T^{-2}\right] \ldots(I)
$
Dimensional formula of gravitational constant
$
[G]=\left[M^{-1} L^{3} T^{-2}\right] \ldots \text { (II) }
$
Dividing (I) by (II), we get:
Hence, $E / G==\left[M^{2} L^{-1} T^{0}\right]$
Solution:
$K =\frac{10}{5 \times 10^{-2}} N / m =200 N / m$
$\therefore T =2 \pi \sqrt{\frac{ m }{ K }}=2 \pi \sqrt{\frac{2}{200}} s =\frac{\pi}{5} s$
$T =0.628 \,s$
Solution:
$j = neV _{ d }=\sigma E$
$\sigma= j / E $
$\Rightarrow \rho= E / j$
$( A ) \rightarrow( R )$
$( B ) \rightarrow( S )$
$( C ) \rightarrow( P )$
$( D ) \rightarrow( Q )$
Solution:
As the electric field will loose its strength in the rightward direction. So, the force on $(+q)$ will be more then the force on $(-q)$
$\left|\vec{F}_{1}\right|>\left|\vec{F}_{2}\right|$
Hence the dipole will move towards right as in the direction of electric field potential reduces
Solution:
When Zener diode is forward biased, it behaves like ordinary diode and when reverse biased, it will be a voltage stabilizer. Also, at room temperature, the voltage across the depletion layer for silicon is about 0.6−0.7V and for germanium is about 0.3−0.35V Hence, statement A is correct but statement B is wrong.
Solution:
Here, pitch of the screw gauge, $p =1 mm$
Number of circular divisions, $n =100$
Thus, least count $LC = p / n =1 / 100=0.01 mm =0.001 cm$
Diameter of the wire $= MSR +( CSR \times LC )=0+(52 \times 0.001 cm )=0.052 cm$
Solution:
Given:
$V_{L}=40$ volt; $V_{R}=40$ volt; $V_{C}=10$ volt
Now,
$V _{ RMS }=\sqrt{ V _{ R }^{2}+\left( V _{ L }- V _{ C }\right)^{2}}$
$=\sqrt{(40)^{2}+(40-10)^{2}}=50 V$
$I _{ RMS }=\frac{ I _{ O }}{\sqrt{2}}=\frac{10 \sqrt{2}}{\sqrt{2}}=10 A$
$ \because V _{ RMS }= I _{ RMS } \times Z$
$\therefore Z=\frac{V_{ RMS }}{ l _{ RMS }}=\frac{50}{10}=5 \Omega$
Solution:
Energy density $u=\frac{1}{2} \varepsilon_{0} E^{2}$
Total energy stored, $U=u \times(A d)$
$=\frac{1}{2} \varepsilon_{0} E^{2} A d$
Solution:
$K _{\text {max }}=\frac{ hc }{\lambda}$
$\Rightarrow \frac{ p ^{2}}{2 m }=\frac{ hc }{\lambda}$
$ \Rightarrow p =\sqrt{\frac{2 mhc }{\lambda}}$
$\Rightarrow \lambda_{ d }=\frac{ h }{ p }= h \sqrt{\frac{\lambda}{2 mhc }}=\sqrt{\frac{\lambda h }{2 mc }} $
$\Rightarrow \lambda=\left(\frac{2 mc }{ h }\right) \lambda_{ d }^{2}$
Solution:
$
\begin{array}{l}
r_{1}+r_{2}=A=30^{\circ} \\
r_{2}=30^{\circ}\left(r_{1}=0^{\circ}\right)
\end{array}
$
From Snell's law,
$\sqrt{3} \sin r_{2}=1 x$ sine
$
\begin{array}{l}
\sqrt{3} \sin 30^{\circ}=\text { sine } \\
e=60^{\circ}
\end{array}
$
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Solution:
Two capacitors are in parallel combination.
Hence equivalent capacitance $C_{\text {eq }}=C_{1}+C_{2}=C+C=2 C$
Solution:
Energy $= F . S = F \frac{ S }{ T ^{2}} T ^{2}=\left[ FAT ^{2}\right]$
Solution:
$\frac{\left(\frac{80-60}{t'}\right)}{\left(\frac{90-80}{t}\right)}=\frac{\left(\frac{80+60}{2}-20\right)}{\left(\frac{90+80}{2}-20\right)}$
$\Rightarrow \frac{20}{t'} \times \frac{t}{10}=\frac{50}{65} $
$\Rightarrow t=2 t \times \frac{65}{50}=\frac{13}{5} t$
Solution:
If the wires are connected in parallel then
$R _{ eq }= R / 4=0.25$
$\Rightarrow R =1 \Omega$
So, after they are connected in series
$R _{ eq }=4 R =4 \Omega$
Solution:
The number of protons delivered per second is given by
$\frac{ n }{ t }=\frac{ IA \lambda}{ hc }=\frac{ P \lambda}{ hc } $
$\Rightarrow \frac{ n }{ t }=\frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{3 \times 10^{8} \times 6.6 \times 10^{-34}}$
$\frac{ n }{ t }=10^{16}$
Solution:
$\frac{ i _{2}}{ i _{3}}=\frac{ r _{3}}{ r _{2}} $
and $ i _{2}+ i _{3}= i _{1}$
$\frac{ i _{2}}{ i _{3}}+1=\frac{ r _{3}+ r _{2}}{ r _{2}}$
$\frac{ i _{2}+ i _{3}}{ i _{3}}=\frac{ r _{2}+ r _{3}}{ r _{2}}$
$\frac{ i _{1}}{ i _{3}}=\frac{ r _{2}+ r _{3}}{ r _{2}}$
$\frac{ i _{3}}{ i _{1}}=\frac{ r _{2}}{ r _{2}+ r _{3}}$
Solution:
Position of the image formed by lens.
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{60}=\frac{1}{30}$
$\frac{1}{v}=\frac{1}{30}-\frac{1}{60}=\frac{1}{60}$
$v=60 \,cm$
So, position of the image is $(60-40)=20\, cm$ behind the plane mirror, it acts as a virtual object. So final image is a real image $20\, cm$ from the plane mirror.
This real image will act as an object for the lens and final image is
$\frac{1}{v}+\frac{1}{20}=\frac{1}{30}$
$\frac{1}{v}=\frac{-1}{60}$
$v=-60\, cm$ from lens
i.e., $20\, cm$ behind the plane mirror.
Solution:
The logic operation
$Y=A B+\overline{B C} $
$=A B+\bar{B}+\bar{C}$
Now input
A
B
C
Y
0
0
1
1
1
0
1
1
0
1
0
1
1
1
0
1
| A | B | C | Y |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 |
Solution:
$V_{p}=220\, V$
$ V_{s}=11 \,V $
$P=44 \,W$
As it is an ideal transformer
$V_{P} i_{P}=V_{S} i_{S}=44 W$
$220 \times i_{P}=44$
$i_{P}=\frac{44}{220}=\frac{1}{5}=0.2 A$
Solution:
(i) If it is an equilateral triangle of side '$a$' no. of triangular loops formed $\frac{12 a}{3 a}=4$
and magnetic moment $4 \cdot \frac{\sqrt{3} a^{2}}{4} l=\sqrt{3} l a^{2}$
(ii) Side of the square if it is a.
No. of square formed $\frac{12 a}{4 a}=3$
Magnetic moment of the square loop is $3 a^{2} I$.
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Solution:
$\vec{ F }= q (\vec{ V } \times \vec{ B })$
$\vec{ F } \cdot \vec{ B }=0$
$\vec{ V } \times \vec{ B }=\hat{ i }\left[4 B _{0}-6 B \right]+\hat{ j }\left[6 B -2 B _{0}\right]+\hat{ k }[2 B -4 B ]$
$=\hat{ i }\left[4 B _{0}-6 B \right]+\hat{ j }\left(6 B -2 B _{0}\right)-\hat{ k }(2 B )$
$4 B _{0}-6 B =4\ldots(1)$
$6 B -2 B _{0}=-20 \ldots(2)$
$-2 B =12\ldots(3)$
Solving (1), (2) and (3) we get
$B =-6$ and $B _{0}=-8$
Solution:
Velocity of the particle $v=\frac{2 \pi R}{T}$
When projected at angle $\theta$
$ H=$ maximum height $=\frac{v^{2} \sin ^{2} \theta}{2 g}=4 R$
$=\frac{4 \pi^{2} R^{2}}{T^{2}} \frac{\sin ^{2} \theta}{2 g}=4 R$
$\sin ^{2} \theta=\frac{2 g R T^{2}}{\pi^{2} R^{2}} $
$\sin ^{2} \theta=\frac{2 g T^{2}}{\pi^{2} R} $
$\sin \theta=\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2} $
$\theta=\sin ^{-1}\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2}$
Solution:
$\begin{array}{l} Q =\frac{\omega}{\Delta \omega}=\frac{\omega L }{ R } \\ \Rightarrow \Delta \omega= R / L =\frac{50}{4}=8 rad / sec \\ \omega_0=\frac{1}{\sqrt{ LC }}=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=50 rad / sec \\ \omega_{\min }=\omega_0-\frac{\Delta \omega}{2}=46 rad / sec \\ \omega_{\max }=\omega_0-\frac{\Delta \omega}{2}=54 rad / sec \end{array}$
Solution:
If $90^{\circ}$ arc is removed, remaining part is $270^{\circ}$ and mass of the remaining part is $3 / 4 M$ and moment of inertia is
$\frac{3}{4} M R^{2}=K M R^{2}$
$K=\frac{3}{4}$
Solution:
From principle of moments
$2 \times 20=0.5 \times 60+ m \times 120$
$2=1.5+6 \,m $
$0.5=6 \,m$
$m =\frac{1}{12} \,kg$
Solution:
If each drop has a charge ' $q$ ' and radius ' $r$ '.
Then from conservation of charge, charge on the big drop is $n q=27 q(n=27)$
from conservation of volume $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$
$R=n^{1 / 3} r$
Now potential of the small drop $V=\frac{q}{4 \pi \epsilon_{0} r}=220 V$
Potential of the big drop,
$V=\frac{n q}{4 \pi \in_{0} R}=\frac{n q}{4 \pi \in_{0} n^{1 / 3} r}=n^{2 / 3} \frac{q}{4 \pi \in_{0} r} $
$V=(27)^{2 / 3} \times 220 V$
$=9 \times 220=1980 V$
Solution:
Velocity of the car at $t =4 \sec$ is
$V _{ x }= at =4 \times 5=20 \,m / s$
So horizontal velocity $=20 \,m / s$ (remain constant)
Vertical velocity at $t =6 \sec$
i.e. after $2 sec$ of free fall
$V_{y}=g t=20\, m / s$
So net velocity $=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \,m / s$
and once it starts falling acceleration is only ' $g$ '
i.e., $10 \,m / s ^{2}$
Solution:
As the particle is projected with a velocity less than escape velocity, it will go to a maximum height and come back.
From conservation of energy
$\frac{m g R h}{R+h}=\frac{1}{2} m\left(k V_{e}\right)^{2}$
$\frac{2 g R h}{R+h}=k^{2}(2 g R)$
$\frac{h}{R+h}=k^{2}$
$\frac{R+h}{h}=\frac{1}{k^{2}}$
$\frac{R}{h}+1=\frac{1}{k^{2}}$
$\frac{R}{h}=\frac{1}{k^{2}}-1$
$h=\frac{R k^{2}}{1-k^{2}}$
Solution:
Velocity while hitting the ground
$v =\sqrt{2 \times 10 \times 10}=10 \sqrt{2}$
As it goes to the same height it will return with same speed.
So change in velocity $v-(-v)=2 v$
Change in momentum or impulse
$=2 \,mv $
$=2 \times 0.15 \times 10 \sqrt{2}$
$=3 \sqrt{2}=4.2\, kg\,m / s$
Solution:
Magnetic field due to $R_{1}$ at the center
$B=\frac{\mu_{0} i}{2 R_{1}}$
Flux linked with $R_{2}$
$\phi= BA _{2}=\frac{\mu_{0} i }{2 R _{1}} \times \pi R _{2}^{2}= Mi$
$M=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}} $
$M \propto \frac{R_{2}^{2}}{R_{1}}$
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