NEET 2020 Physics Questions with Answers Key Solutions

Q1. For which one of the following, Bohr model is not valid?

A

Hydrogen atom

B

Singly ionised helium atom $(He^+)$

C

Deuteron atom

D

Singly ionised neon atom $(Ne^+)$

Solution:

Bohr model is applicable for only single electron species. Singly ionised neon atom has more than one electron in orbit. Hence, Bohr model is not valid

Q2. The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)

A

c : 1

B

1 : 1

C

1 : c

D

$1 : c^2$

Solution:

In an electromagnetic wave, half of the intensity is provided by the electric field and half by the magnetic field. Hence, required ratio should be $1 : 1$

Q3. The Brewsters angle $i_b$ for an interface should be:

A

$0°< i_b < 30°$

B

$30°< i_b < 45°$

C

$45°< i_b < 90°$

D

$i_b = 90°$

Solution:

The brewster angle for any interface should lie between:
$ 45< i _{ b }<90 . $

Q4. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature $27^\circ\,C$. Its density is :$(R = 8.3\,J\,mol^{-1}K^{-1}$)

A

$0.5 \,kg/m^3$

B

$0.2 \,kg/m^3$

C

$0.1 \,kg/m^3$

D

$0.02 \,kg/m^3$

Solution:

We can express, an ideal gas equation as
$\frac{P}{\rho}=\frac{R.T}{Mw}$
$\Rightarrow \rho=\frac{pMw}{RT}$
$=\frac{249 \times 10^{3}\times 2 \times 10^{-3}}{8.314 \times 300}$
$=0.199$
$\Rightarrow \rho=0.2 kg/m^{3}$

Q5. A ray is incident at an angle of incidence $i$ on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$. then the angle of incidence is nearly equal to:

A

$\frac {A}{2\mu}$

B

$\frac {2A}{\mu}$

C

$\mu A$

D

$\frac {\mu \,A}{2}$

Solution:

Light ray emerges from another surface normally, hence e(angle of emergence) = 0

$\Rightarrow r_{2}=0$
$r_{1}+r_{2}=A$
$\Rightarrow r_{1}=A$
Applying Snell’s law on first surface,
$1. sin\,i=\mu\,sin\,r_{1}$
For small angle , $sin\,\theta\, \approx\,\theta$ Hence,
$i=\mu\,A$

Q6. Two cylinders $A$ and $B$ of equal capacity are connected to each other via a stop clock. A contains an ideal gas at standard temperature and pressure. $B$ is completely evacuated. The entire system is thermally insulated. The stop clock is suddenly opened. The process is:

A

isothermal

B

adiabatic

C

isochoric

D

isobaric

Solution:

The entire system is completely insulated and the free expansion of the gas will take place, the temperature of the gas will remain constant. So the process will be adiabatic.

Q7. The energy equivalent of 0.5 g of a substance is:

A

$4.5 \times 10^{16} J$

B

$4.5 \times 10^{13} J$

C

$1.5 \times 10^{13} J$

D

$0.5 \times 10^{13} J$

Solution:

From mass-energy equivalence,
$E=mc^{2}$
$=0.5\times 10^{-3}\times 9\times 10^{16}$
$=4.5 \times 10^{13}\,J$

Q8. A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

A

48 N

B

32 N

C

30 N

D

24 N

Solution:

$w_{s}=mg_{s}=72\,N$
$w_{h}=mg_{h}=\frac{mg_{s}}{\left(1+\frac{h}{R}\right)^{2}}$
$=\frac{72\,N}{\left(1+\frac{R /2}{2}\right)}$
$=\frac{72}{9 /4}$
$w_{h}=32\,N$

Q9. The solids which have negative temperature coefficient of resistance are :

A

metals

B

insulators only

C

semiconductors only

D

insulators and semiconductors

Solution:

For metals temperature coefficient of resistance is positive while for insulators and semiconductors, temperature coefficient of resistance is negative

Q10. The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

A

$\pi$ rad

B

$\frac {3\pi}{2}$ rad

C

$\frac {\pi}{2}$ rad

D

$zero$

Solution:

Displacement of the particle executing SHM $x = A$ sin wt Acceleration of the particle $a =\frac{ d ^2 x }{ dt ^2}$
$ \Longrightarrow a =-A w ^2 \sin w t =A w ^2 \sin (\pi+ wt ) $
Thus phase difference between displacement and acceleration of the particle is $\pi$ radian.

Q11. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is:

A

0.01 mm

B

0.25 mm

C

0.5 mm

D

1.0 mm

Solution:

L.C. $=\frac{\text{Pitch}}{\text{Number of division on circular scale}}$
$\Rightarrow 0.01\,mm=\frac{\text{pitch}}{50}$
$\Rightarrow $ pitch $=0.5\,mm$

Q12. In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6\,Hz$. When tension in B is slightly decreased, the be at frequency increases to $7\,Hz$. If the frequency of A is $530\,Hz$, the original frequency of B will be :

A

523 Hz

B

524 Hz

C

536 H z

D

537 H z

Solution:

It is given, the difference of $f_{A}$ and $f_{B}$ is $6\,HZ$
Guitar string i.e. string is fixed from both ends

Frequency $\propto \sqrt{\text{Tension}}$
If tension in B slightly decrease then frequency of B decreases
If B is 536 Hz, as the frequency decreases, beats with A also decreases
If B is 524 Hz, as the frequency decreases, beats with A increases
If tension decreases, $f_{B}$ decreases and becomes $f'_{B}$
Now, difference of $f_{A}$ and $f'_{B}=7\,Hz$ (increases)
So, $f_{A}=f_{B}$
$f_{A}-f_{B}=6\,Hz$
$f_{A}=530\,Hz$
$f_{B}=524\,Hz$ (original)

Q13. Two particles of mass $5\,kg$ and $10\,kg$ respectively are attached to the two ends of a rigid rod of length 1m with negligible mass.
The centre of mass of the system from the $5\,kg$ particle is nearly at a distance of :

A

33 cm

B

50 cm

C

67 cm

D

80 cm

Solution:

Let position of centre of mass be ${(x_{c.m}, 0)}$

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$
$=\frac{5 \times 0+100 \times 10}{5+10}$
$=\frac{200}{2}$
$=66.66\,cm$
$x_{cm}=67\,cm$

Q14. Find the torque about the origin when a force of $3 \,\hat{j}\, N$ acts on a particle whose position vector is $2\, \hat {k}\, m$ .

A

$6 \hat{i}\, Nm $

B

$6 \hat{j} \,Nm $

C

$-6 \hat{i} \,Nm $

D

$6 \hat{k} Nm $

Solution:

Given: $\vec{F}=3 \hat{j}\,N$,
$\vec{r}=2\hat{k}$
We know, $\vec{\tau}=\vec{r}\times\vec{F}=2\hat{k}\times3\hat{j}$
$=6\left(\hat{k} \times\hat{j}\right)=6\left(-\hat{i}\right)$
$\vec{\tau}=-6 \hat{i}\,Nm$

Q15. Light with an average flux of $20\,\Omega/cm^2$ falls on a non-reflecting surface at normal incidence having surface area $20\,cm^2$. The energy received by the surface during time span of $1\,minute$ is :

A

$10 \times 10^3J$

B

$12 \times 10^3J$

C

$24 \times 10^3J$

D

$48 \times 10^3J$

Solution:

Energy received = Intensity $\times$ Area $\times$ Time
$I=\frac{E}{A}$
$E=IAt$
$=\frac{20}{10^{-4}}\times 20\times 10^{-4}\times 60$
$=24 \times 10^{3}\,J$

Q16. A spherical conductor of radius 10cm has a charge of $3.2 \times 10^{-7} C$ distributed uniformly. What is the magnitude of electric field at a point $15\,cm$ from the centre of the sphere ?
$(\frac {1}{4\pi \epsilon_o} = 9 \times 10^9 Nm^2/C^2)$

A

$1.28 \times 10^4\,N/C$

B

$1.28 \times 10^5\,N/C$

C

$1.28 \times 10^6\,N/C$

D

$1.28 \times 10^7\,N/C$

Solution:

Electric field outside a conducting sphere
$E=\frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r^{2}}$
$=\frac{9\times10^{9}\times3.2\times10^{-7}}{225\times10^{-4}}$
$=0.128\times10^{6}$
$=1.28\times10^{5} N /C$

Q17. In a certain region of space with volume $0.2 \,m^3$, the electric potential is found to be $5\,V$ throughout. The magnitude of electric field in this region is:

A

zero

B

0.5 N/C

C

1 N/C

D

5 N/C

Solution:

Since electric potential is found throughout constant, hence electric field,
$E=-\frac{dV}{dt}=0$

Q18. The increase in the width of the depletion region in a p-n junction diode is due to :

A

forward bias only

B

reverse bias only

C

both forward bias and reverse bias

D

increase in forward current

Solution:

In reverse bias, external battery attracts majority charge carriers
So, width of the depletion region increases

Q19. A $40\, \mu\,F$ capacitor is connected to a $200\,V$,$50\,Hz$ ac supply. The rms value of the current in the circuit is, nearly :

A

1. 7 A

B

2. 05 A

C

2. 5 A

D

25. 1 A

Solution:

The rms value of the current
$i_{\text{rms}}=C\,\omega\,\varepsilon_{\text{rms}}$
$C=40\times10^{-6}F $
$\omega=2\pi\,f =100\,\pi$
$\varepsilon_{\text{rms}}=200\,V $
$\therefore i_{\text{rms}}=200\times40\times10^{-6}\times2\pi\times50$
$=2.5\,A$

Q20. The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as :

A

$\frac {1}{\sqrt{2} n \pi d}$

B

$\frac {1}{\sqrt{2} n \pi d^2}$

C

$\frac {1}{\sqrt{2} n^2 \pi d^2}$

D

$\frac {1}{\sqrt{2} n^2 \pi^2 d^2}$

Solution:

Mean free path for a gas sample $\lambda_{m}=\frac{1}{\sqrt{2}\pi d^{2} n}$
where d is diameter of a gas molecule and n is molecular density

Q21. For transistor action, which of the following statements is correct ?

A

Base, emitter and collector regions should have same doping concentrations.

B

Base, emitter and collector regions should have same size.

C

Both emitter junction as well as the collector junction are forward biased.

D

The base region must be very thin and lightly doped.

Solution:

For Bi-polar junction transistor,
Length profile is $L_{C}>\,L_{E}>\,L_{B}$
and doping profile is $E>\,C>\,B$

Q22. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

A

doubled

B

four times

C

one-fourth

D

zero

Solution:

$K_{1}=1.5 \,hv_{0}-\phi_{0}=0.5\,hv_{0}$
$K_{2}=\frac{1.5}{2}hv_{0}-hv_{0}=-0.25\,hv_{0}$
$\because$ Kinetic energy can never be negative, so no emission and $i=0$

Q23. When a uranium isotope $^{235}_92 U$ is bombarded with a neutron, it generates $^{89}_{36} Kr$, three neutrons and :

A

$^{144}_{56} Ba$

B

$^{91}_{40} Zr$

C

$^{101}_{36} Kr$

D

$^{103}_{36} Kr$

Solution:

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr+3 ^{1}_{0}n+X_{X}^{A}$
$92+0=36+Z$
$\Rightarrow Z=56$
$235+1=89+3+A$
$\Rightarrow A=144$
So, $^{144}_{56}Ba$ is generated

Q24. The energy required to break one bond in DNA is $10^{-20}J $. This value in eV is nearly

A

6

B

0.6

C

0.06

D

0.006

Solution:

$1\,eV=1.6\times 10^{-19}\\,J$
$1\,J=\frac{1}{1.6\times 10^{-19}}eV$
$10^{-20}\,J=\frac{10^{-20}}{1.6\times 10^{-19}}eV$
$=0.06\,eV$

Q25. Two bodies of mass $4\,kg$ and $6\,kg$ are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

A

g

B

g/2

C

g/5

D

g/10

Solution:

The acceleration of the system $a=\frac{\left(m_{2}-m_{1}\right)g}{m_{1}+m_{2}}$ where $m_{2}>\,m_{1}$
$a=\frac{(6-4)g}{6+4}$
$=\frac{2g}{10}$
$a=\frac{g}{5}$

Q26. A wire of length $L$, area of cross section $A$ is hanging from a fixed support. The length of the wire changes to $L_1$ when mass $M$ is suspended from its free end. The expression for Young’s modulus is:

A

$\frac {MgL_1}{AL}$

B

$\frac {Mg(L_1-L)}{AL}$

C

$\frac {MgL}{AL_1}$

D

$\frac {MgL}{A(L_1-L)}$

Solution:

When the new mas is hanged to the wire, the force exerted on the wire is:
$F = mg$
The initial length of the wire is $L$ and the new length is $L_{1}$.
We know that the Young's modulus is given as:
$Y =\frac{ FL _{ o }}{ A \Delta l }$
Here, $L_{o}$ is the initiallength of wire and $\Delta l$ is the chnage in length of wire.
Substitute the values:
$Y =\frac{ mg \times L }{ A \times\left( L _{1}- L \right)}$
So, $Y=\frac{m g L}{A\left(L_{1}-L\right)}$

Q27. The average thermal energy for a mono-atomic gas is : ($k_B$ is Boltzmann constant and T, absolute temperature)

A

$\frac {1}{2} k_BT$

B

$\frac {3}{2} k_BT$

C

$\frac {5}{2} k_BT$

D

$\frac {7}{2} k_BT$

Solution:

Average thermal energy $=\frac{3}{2}K_{B}T$
(where 3 is translational degree of freedom)
For monoatomic gas total degree of freedom
$f= 3$ (translational degree of freedom)

Q28. Which of the following graph represents the variation of resistivity $(\rho)$ with temperature (T) for copper ?

A

B

C

D

Solution:

For some metals like copper, resistivity is nearly proportional to temperature although a non linear region always exists at very low temperature

Q29. The color code of a resistance is given below :
The values of resistance and tolerance, respectively, are :

A

470 $k\Omega,5\%$

B

47 $k\Omega,10\%$

C

4.7 $k\Omega,5\%$

D

470 $\Omega,5\%$

Solution:

From colour coding information for electric resistances, we have
$\begin{matrix}{\text{Yellow}}&{\text{Violet}}&{\text{Brown}}&{\text{Gold}}\\ 4&7&1&5\%\end{matrix}$
So, $R=47 \times 10^{1} \pm\,5\%$
$R=470\,\Omega\,\pm\,5\%$

Q30. In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

A

double

B

half

C

four times

D

one-fourth

Solution:

Fringe width, $\beta=\frac{\lambda\,D}{d}$
$\beta' = \frac{\lambda\,D'}{d'}$
Now, $d'=\frac{d}{2}$ and
$D'=2\,D$
So, $\beta'=\frac{\lambda \times 2 D}{d / 2}$
$=\frac{4 \,\lambda\,D}{d}$
$\beta'=4\,\beta$
Fringe width becomes 4 times

Q31. The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes 30 $\mu F.$ The permittivity of the medium is :
$(\epsilon_0 = 8.85 \times 10^{-12} C^2N^{-1}m^{-2}$)

A

$0.44 \times 10^{-13} C^2 N^{-1}m^{-2}$

B

$1.77 \times 10^{-12} C^2 N^{-1}m^{-2}$

C

$0.44 \times 10^{-10} C^2 N^{-1}m^{-2}$

D

$5.00 \,C^2 N^{-1}m^{-2}$

Solution:

$C=KC_{0}$
$K=\frac{C}{C_{0}}$
$=\frac{30}{6}=5$
$K=\frac{\varepsilon}{\varepsilon_{0}}$
$\varepsilon=K\,\varepsilon_{0}$
$=5 \times 8.85\times 10^{-12}$
$=0.44 \times 10^{-10}\,C^{2}\,N^{-1}\,m^{-2}$

Q32. Dimensions of stress are:

A

$ [MLT^{-2}]$

B

$ [ML^2T^{-2}]$

C

$=\left[ ML ^{-1} T ^{-2}\right]$

D

$ [ML^{-2} T^{-2}]$

Solution:

We know that the formula for the stress is:
$ \text { stress }=\frac{\text { F orce }}{\text { Area }} $
$ =\frac{\left[ MLT ^{-2}\right]}{\left[ L ^{2}\right]} $
$ =\left[ ML ^{-1} T ^{-2}\right] $

Q33. Assume that light of wavelength $600\, nm$ is coming from a star. The limit of resolution of telescope whose objective has a diameter of $2\,m$ is :

A

$3.66 \times 10^{-7} \,rad$

B

$1.83 \times 10^{-7} \,rad$

C

$7.32 \times 10^{-7} \,rad$

D

$6.00 \times 10^{-7} \,rad$

Solution:

Limit of resolution,
$\theta_{R}=1.22 \frac{\lambda}{d}$;
$\lambda=600 \times 10^{-9}\,m$
$d=2\,m$
$\theta_{R}=\frac{1.22 \times 600 \times 10^{-9}}{2}$
$\theta_{R}=3.66 \times 10^{-7}$ rad

Q34. A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\frac {\pi}{3}$ . If instead C is removed from the circuit, the phase difference is again $\frac {\pi}{3}$ between current and voltage. The power factor of the circuit is :

A

zero

B

0.5

C

1.0

D

-1.0

Solution:

When L is removed,
$tan\,\phi=\frac{\left|X_{C}\right|}{R}$
$\Rightarrow tan \frac{\pi}{3}=\frac{X_{c}}{R} \ldots\left(i\right)$
When C is removed,
$tan \,\phi=\frac{\left|X_{L}\right|}{R}$
$\Rightarrow tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
Since $X_{L}=X_{C}$, the circuit is in resonance
$Z=R$
Power factor $=cos\,\phi=\frac{R}{Z}=1$

Q35. A short electric dipole has a dipole moment of $16 \times 10^{-9} C\, m$. The electric potential due to the dipole at a point at a distance of $0.6 \,m$ from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is :
$(\frac {1}{4\pi \epsilon_o}= 9 \times 10^9 N\,m^2/C^2)$

A

$50\,V$

B

$200\,V$

C

$400\,V$

D

zero

Solution:

The electric potential due to the dipole,
$V=\frac{kp\,cos\,theta}{r^{2}}$
$=\frac{9\times 10^{9}\times 16\times 10^{-9}}{(0.6)^{2}}\times \frac{1}{2}$
$V=200\,V$

Q36. An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A $m^{-1}$. The permeability of the material of the rod is :
$(\mu_o = 4\pi \times 10^{-7} T\,m\,A^{-1})$

A

$2.4 \pi \times 10^{-4} T\,m\,A^{-1}$

B

$8.0 \pi \times 10^{-5} T\,m\,A^{-1}$

C

$2.4 \pi \times 10^{-5} T\,m\,A^{-1}$

D

$2.4 \pi \times 10^{-7} T\,m\,A^{-1}$

Solution:

Given, $\chi_{m}=599$
Also, $\mu_{r}=1+\chi_{m}=600$
We know, $\mu=\mu_{r}\mu_{o}$
$\mu=600\times4\pi\times10^{-7}$
$\mu=2400\pi\times10^{-7}$
$\mu=2.4\pi\times10^{-4}\,T\,m\,A^{-1}$

Q37. A long solenoid of $50\,cm$ length having $100\,turns$ carries a current of $2.5\,A$. The magnetic field at the centre of the solenoid is :
$(\mu_o = 4\pi \times 10^{-7} T\,m\,A^{-1})$

A

$6.28 \times 10^{-4}T$

B

$3.14 \times 10^{-4}T$

C

$6.28 \times 10^{-5}T$

D

$3.14 \times 10^{-5}T$

Solution:

Magnetic field at centre of solenoid $=\mu_{0}nI$
$n=\frac{N}{L}$
$=\frac{100}{50 \times 10^{-2}}$
$=200$ tums/m
$I = 2.5 \,A$
On putting the values,
$B=4\pi\times 10^{-7}\times 200 \times 2.5$
$=6.28 \times 10^{-4}\,T$

Q38. A charged particle having drift velocity of $7.5 \times 10^{-4} \,m\,s^{-1}$ in an electric field of $3 \times 10^{-10}Vm^{-1}$, has a mobility in $m^2V^{-1}s^{-1}$ of:

A

$2.25 \times 10^{15}$

B

$2.5 \times 10^{6}$

C

$2.5 \times 10^{-6}$

D

$2.25 \times 10^{-15}$

Solution:

Mobility, $\mu =\frac{v_{d}}{E}$
$=\frac{7.5 \times 10^{-4}}{3 \times 10^{-10}}$
$=2.5 \times 10^{6}$

Q39. The quantities of heat required to raise the temperature of two solid copper spheres of radii $r_1$ and $r_2$ $(r_1 = 1.5 \,r_2$) through 1 K are in the ratio :

A

$\frac {27}{8}$

B

$\frac {9}{4}$

C

$\frac {3}{2}$

D

$\frac {5}{3}$

Solution:

Heat supplied $\Delta\,Q =MS\,\Delta\,T$
For same material ‘s’ remains the same
$\Delta\,Q \propto\,M$ and $M=\frac{4}{3}\pi\,r^{3}\,\rho$
$\Rightarrow \frac{\Delta\,Q_{1}}{\Delta\,Q_{2}}$
$=\left(\frac{r_{1}}{r_{2}}\right)^{3}$
$=\left(\frac{1.5}{1}\right)^{3}$
$=\frac{27}{8}$

Q40. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}nm$, the potential difference is :

A

$10$ V

B

$10^2$ V

C

$10^3$ V

D

$10^4$ V

Solution:

$\lambda = 1.227 \times 10^{-2}\,m$
$=0.1227 \, \mathring{A}$
$\lambda=\frac{12.27}{\sqrt{V}} \mathring{A}$
$0.1227 = \frac{12.27}{\sqrt{V}} \mathring{A}$
$\sqrt{V}\times10^{2} $
$\Rightarrow V=10^{4}$

Q41. Taking into account of the significant figures, what is the value of $9.99 \,m - 0.0099 \,m$ ?

A

9.9801 m

B

9.98 m

C

9.980 m

D

9.9 m

Solution:

$9.99 m -0.0099 m =9.9801 m$
But our answer, Should have same no. of significant figures, as the least accurate number.
$\therefore $ Ans $=9.98 m$.

Q42. A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower. It hits the ground after some time with a velocity of 80m/s. The height of the tower is : (g = $10m/s^2$)

A

360 m

B

340 m

C

320 m

D

300 m

Solution:

Using, $v^{2}=u^{2}+2gh$
Given, $V=80m/s$ and $u=20 m/s$
$h=\frac{v^{2}-u^{2}}{2g}$
$=\frac{6400-400}{20}$
$=300\,m$

Q43. A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5\,g$. Another capillary tube of radius $2\,r$ is immersed in water. Themass of water that will rise in this tube is :

A

2.5 g

B

5.0 g

C

10.0 g

D

20.0 g

Q44. A resistance wire connected in the left gap of a metre bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5\,m$, then the length of 1 $\Omega$ of the resistance wire is:

A

$1.0 \times 10^{-2} m$

B

$1.0 \times 10^{-1} m$

C

$1.5 \times 10^{-1} m$

D

$1.5 \times 10^{-2} m$

Solution:

Initially, $\frac{P}{10}=\frac{l_{1}}{l_{2}}=\frac{3}{2}$
$\Rightarrow P=\frac{30}{2}=15\Omega$
Now resistance, $R=\frac{\rho\,l}{A}$
$\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}$: Length of $15\,\Omega$ resistance wire is 1.5 m
$\Rightarrow \frac{15}{1}=\frac{1.5}{l_{2}}$
$\Rightarrow l_{2}=0.1\,m$
$=1.0 \times 10^{-1}\,m$
$\therefore $ Length of $1\,\Omega$ resistance wire is $1.0\times 10^{-1}\,m$

Q45. For the logic circuit shown, the truth table is :

A

B

C

D

Solution:

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NEET 2020 Physics Questions with Answers Key Solutions

Q1. For which one of the following, Bohr model is not valid?

Hydrogen atom

Singly ionised helium atom $(He^+)$

Deuteron atom

Singly ionised neon atom $(Ne^+)$

Bohr model is applicable for only single electron species. Singly ionised neon atom has more than one electron in orbit. Hence, Bohr model is not valid

Q2. The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)

c : 1

1 : 1

1 : c

$1 : c^2$

In an electromagnetic wave, half of the intensity is provided by the electric field and half by the magnetic field. Hence, required ratio should be $1 : 1$

Q3. The Brewsters angle $i_b$ for an interface should be:

$0°< i_b < 30°$

$30°< i_b < 45°$

$45°< i_b < 90°$

$i_b = 90°$

The brewster angle for any interface should lie between: $ 45< i _{ b }<90 . $

Q4. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature $27^\circ\,C$. Its density is :$(R = 8.3\,J\,mol^{-1}K^{-1}$)

$0.5 \,kg/m^3$

$0.2 \,kg/m^3$

$0.1 \,kg/m^3$

$0.02 \,kg/m^3$

We can express, an ideal gas equation as $\frac{P}{\rho}=\frac{R.T}{Mw}$ $\Rightarrow \rho=\frac{pMw}{RT}$ $=\frac{249 \times 10^{3}\times 2 \times 10^{-3}}{8.314 \times 300}$ $=0.199$ $\Rightarrow \rho=0.2 kg/m^{3}$

Q5. A ray is incident at an angle of incidence $i$ on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$. then the angle of incidence is nearly equal to:

$\frac {A}{2\mu}$

$\frac {2A}{\mu}$

$\mu A$

$\frac {\mu \,A}{2}$

Light ray emerges from another surface normally, hence e(angle of emergence) = 0 $\Rightarrow r_{2}=0$ $r_{1}+r_{2}=A$ $\Rightarrow r_{1}=A$ Applying Snell’s law on first surface, $1. sin\,i=\mu\,sin\,r_{1}$ For small angle , $sin\,\theta\, \approx\,\theta$ Hence, $i=\mu\,A$

Q6. Two cylinders $A$ and $B$ of equal capacity are connected to each other via a stop clock. A contains an ideal gas at standard temperature and pressure. $B$ is completely evacuated. The entire system is thermally insulated. The stop clock is suddenly opened. The process is:

isothermal

adiabatic

isochoric

isobaric

The entire system is completely insulated and the free expansion of the gas will take place, the temperature of the gas will remain constant. So the process will be adiabatic.

Q7. The energy equivalent of 0.5 g of a substance is:

$4.5 \times 10^{16} J$

$4.5 \times 10^{13} J$

$1.5 \times 10^{13} J$

$0.5 \times 10^{13} J$

From mass-energy equivalence, $E=mc^{2}$ $=0.5\times 10^{-3}\times 9\times 10^{16}$ $=4.5 \times 10^{13}\,J$

Q8. A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

48 N

32 N

30 N

24 N

$w_{s}=mg_{s}=72\,N$ $w_{h}=mg_{h}=\frac{mg_{s}}{\left(1+\frac{h}{R}\right)^{2}}$ $=\frac{72\,N}{\left(1+\frac{R /2}{2}\right)}$ $=\frac{72}{9 /4}$ $w_{h}=32\,N$

Q9. The solids which have negative temperature coefficient of resistance are :

metals

insulators only

semiconductors only

insulators and semiconductors

For metals temperature coefficient of resistance is positive while for insulators and semiconductors, temperature coefficient of resistance is negative

Q10. The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

$\pi$ rad

$\frac {3\pi}{2}$ rad

$\frac {\pi}{2}$ rad

$zero$

Displacement of the particle executing SHM $x = A$ sin wt Acceleration of the particle $a =\frac{ d ^2 x }{ dt ^2}$ $ \Longrightarrow a =-A w ^2 \sin w t =A w ^2 \sin (\pi+ wt ) $ Thus phase difference between displacement and acceleration of the particle is $\pi$ radian.

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Q11. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

0.01 mm

0.25 mm

0.5 mm

1.0 mm

L.C. $=\frac{\text{Pitch}}{\text{Number of division on circular scale}}$ $\Rightarrow 0.01\,mm=\frac{\text{pitch}}{50}$ $\Rightarrow $ pitch $=0.5\,mm$

Q12. In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6\,Hz$. When tension in B is slightly decreased, the be at frequency increases to $7\,Hz$. If the frequency of A is $530\,Hz$, the original frequency of B will be :

523 Hz

524 Hz

536 H z

537 H z

Q13. Two particles of mass $5\,kg$ and $10\,kg$ respectively are attached to the two ends of a rigid rod of length 1m with negligible mass. The centre of mass of the system from the $5\,kg$ particle is nearly at a distance of :

33 cm

50 cm

67 cm

80 cm

Let position of centre of mass be ${(x_{c.m}, 0)}$ $x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$ $=\frac{5 \times 0+100 \times 10}{5+10}$ $=\frac{200}{2}$ $=66.66\,cm$ $x_{cm}=67\,cm$

The brewster angle for any interface should lie between:
$ 45< i _{ b }<90 . $

We can express, an ideal gas equation as
$\frac{P}{\rho}=\frac{R.T}{Mw}$
$\Rightarrow \rho=\frac{pMw}{RT}$
$=\frac{249 \times 10^{3}\times 2 \times 10^{-3}}{8.314 \times 300}$
$=0.199$
$\Rightarrow \rho=0.2 kg/m^{3}$

Light ray emerges from another surface normally, hence e(angle of emergence) = 0

$\Rightarrow r_{2}=0$
$r_{1}+r_{2}=A$
$\Rightarrow r_{1}=A$
Applying Snell’s law on first surface,
$1. sin\,i=\mu\,sin\,r_{1}$
For small angle , $sin\,\theta\, \approx\,\theta$ Hence,
$i=\mu\,A$

From mass-energy equivalence,
$E=mc^{2}$
$=0.5\times 10^{-3}\times 9\times 10^{16}$
$=4.5 \times 10^{13}\,J$

$w_{s}=mg_{s}=72\,N$
$w_{h}=mg_{h}=\frac{mg_{s}}{\left(1+\frac{h}{R}\right)^{2}}$
$=\frac{72\,N}{\left(1+\frac{R /2}{2}\right)}$
$=\frac{72}{9 /4}$
$w_{h}=32\,N$

Displacement of the particle executing SHM $x = A$ sin wt Acceleration of the particle $a =\frac{ d ^2 x }{ dt ^2}$
$ \Longrightarrow a =-A w ^2 \sin w t =A w ^2 \sin (\pi+ wt ) $
Thus phase difference between displacement and acceleration of the particle is $\pi$ radian.

Q11. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is:

L.C. $=\frac{\text{Pitch}}{\text{Number of division on circular scale}}$
$\Rightarrow 0.01\,mm=\frac{\text{pitch}}{50}$
$\Rightarrow $ pitch $=0.5\,mm$

Q13. Two particles of mass $5\,kg$ and $10\,kg$ respectively are attached to the two ends of a rigid rod of length 1m with negligible mass.
The centre of mass of the system from the $5\,kg$ particle is nearly at a distance of :

Let position of centre of mass be ${(x_{c.m}, 0)}$

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$
$=\frac{5 \times 0+100 \times 10}{5+10}$
$=\frac{200}{2}$
$=66.66\,cm$
$x_{cm}=67\,cm$

Given: $\vec{F}=3 \hat{j}\,N$,
$\vec{r}=2\hat{k}$
We know, $\vec{\tau}=\vec{r}\times\vec{F}=2\hat{k}\times3\hat{j}$
$=6\left(\hat{k} \times\hat{j}\right)=6\left(-\hat{i}\right)$
$\vec{\tau}=-6 \hat{i}\,Nm$

Energy received = Intensity $\times$ Area $\times$ Time
$I=\frac{E}{A}$
$E=IAt$
$=\frac{20}{10^{-4}}\times 20\times 10^{-4}\times 60$
$=24 \times 10^{3}\,J$

Q16. A spherical conductor of radius 10cm has a charge of $3.2 \times 10^{-7} C$ distributed uniformly. What is the magnitude of electric field at a point $15\,cm$ from the centre of the sphere ?
$(\frac {1}{4\pi \epsilon_o} = 9 \times 10^9 Nm^2/C^2)$

Electric field outside a conducting sphere
$E=\frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r^{2}}$
$=\frac{9\times10^{9}\times3.2\times10^{-7}}{225\times10^{-4}}$
$=0.128\times10^{6}$
$=1.28\times10^{5} N /C$

Since electric potential is found throughout constant, hence electric field,
$E=-\frac{dV}{dt}=0$

In reverse bias, external battery attracts majority charge carriers
So, width of the depletion region increases

The rms value of the current
$i_{\text{rms}}=C\,\omega\,\varepsilon_{\text{rms}}$
$C=40\times10^{-6}F $
$\omega=2\pi\,f =100\,\pi$
$\varepsilon_{\text{rms}}=200\,V $
$\therefore i_{\text{rms}}=200\times40\times10^{-6}\times2\pi\times50$
$=2.5\,A$

Mean free path for a gas sample $\lambda_{m}=\frac{1}{\sqrt{2}\pi d^{2} n}$
where d is diameter of a gas molecule and n is molecular density

For Bi-polar junction transistor,
Length profile is $L_{C}>\,L_{E}>\,L_{B}$
and doping profile is $E>\,C>\,B$

$K_{1}=1.5 \,hv_{0}-\phi_{0}=0.5\,hv_{0}$
$K_{2}=\frac{1.5}{2}hv_{0}-hv_{0}=-0.25\,hv_{0}$
$\because$ Kinetic energy can never be negative, so no emission and $i=0$

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr+3 ^{1}_{0}n+X_{X}^{A}$
$92+0=36+Z$
$\Rightarrow Z=56$
$235+1=89+3+A$
$\Rightarrow A=144$
So, $^{144}_{56}Ba$ is generated

$1\,eV=1.6\times 10^{-19}\\,J$
$1\,J=\frac{1}{1.6\times 10^{-19}}eV$
$10^{-20}\,J=\frac{10^{-20}}{1.6\times 10^{-19}}eV$
$=0.06\,eV$