NEET PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Thermodynamics

A spherical body of emissivity e = 0.6, placed inside a perfectly black body is maintained at

temperature T then the energy radiated per second will be

(1)E = 0.4 eAT⁴ (2) E = 0.8 eAT⁴

(3) E = 0.6 eAT⁴ (4) E = 1.0eAT⁴

Option 1

Option 2

Option 3 Solution:

Option 4

2 Thermodynamics

4.0 g of a gas occupier 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 J/K mol. If the speed of sound in this gas at NTP is 952 1/ ms, then the heat capacity at constant pressure is (Take gas constant R = 8.3 J/K mol)

1) 8.5 J/K mol
2) 8.0 J/K mol
3) 7.5 J/K mol
4) 7 J/K mol

Option 1

Option 2 Solution:

Solution

given, mass(m)= 4 g

Volume(V) = 22.4 liters

Specific heat capacity of gas at constant volume ($${ C }_{ v }$$) = 5.0 J/K mol

Velocity of sound ($$v_{ sound }$$)=952 m/s

As we know that,

Velocity of sound, $$v_{ sound }=\sqrt { \dfrac { \gamma PV }{ m } } $$

$$\Rightarrow \gamma =\dfrac { m{ { v }_{ sound } }^{ 2 } }{ PV } $$ .......................equation(1)

Also,

$$\gamma =\dfrac { { C }_{ p } }{ C_{ v } } $$

So, equation (1) becomes,

$$\dfrac { { C }_{ p } }{ { C }_{ v } } =\dfrac { m{ { v }_{ sound } }^{ 2 } }{ PV } $$

$$\Rightarrow { C }_{ p }={ C }_{ v }\dfrac { m{ { v }_{ sound } }^{ 2 } }{ PV } $$

by substituting the values we get,

$$\Rightarrow { C }_{ p }=\dfrac { 5\times 4\times { 10 }^{ -3 }\times { 952 }^{ 2 } }{ { 10 }^{ 5 }\times 22.4\times { 10 }^{ -3 } } $$

$$\Rightarrow {C}_{p}$$=8.09 J/K mol

$$\Rightarrow {C}_{p}$$ $$\approx$$8.00 J/K mol

Option 3

Option 4

3 Ray Optics

A beam of monochromatic ligh first travels through glass of R.I. 1.5 and then through water (R.I. =4/3). If the difference in their wavelength in the two media is 40nm, then the wavelength of the monochromatic light in vacuum will be

1) 4000 Å
2) 4400 Å
3) 4800 Å
4) 5200 Å

Option 1

Option 2

Option 3 Solution:

Solution

$$\displaystyle{5200}Å$$
$$\displaystyle\therefore\lambda_{{w}}-\lambda_{{g}}=\lambda_{{a}}{\left[\frac{{{3}}}{{{4}}}-\frac{{{2}}}{{{3}}}\right]}=\frac{{\lambda_{{a}}}}{{{12}}}={40}\times{10}^{{-{9}}}$$

∴λa=12×40×10⁻⁹= 4800Å

Option 4

4 Ray Optics

A commonly used lens coating material is MgF₂ with n=1.38.. Find the thickness of non-reflective coating, one shall have for 550nm light if it is applied to glass of n=1.52.

1) 400nm
2) 200nm
3) 300nm
4) 100nm

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

$$550nm$$ wavelength in $$MgF_2$$ will be $$\lambda=\displaystyle \frac{\displaystyle \lambda_{air}}{n}=\displaystyle \frac{550}{1.38}=400nm$$
$$l=\displaystyle \frac{\lambda}{4}=100nm$$

5 Dual Nature of Matter and Radiation

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

1) 4.81eV
2) 3.74eV
3) 2.65eV
4) 1.07eV

Option 1

Option 2

Option 3 Solution:

Solution

Energy of incident light

$$E\left( eV \right) =\dfrac { 12375 }{ 3320 } =3.72eV$$

We know that, $$E={ W }_{ 0 }+{ eV }_{ 0 }$$

$${ V }_{ 0 }=\dfrac { \left( E-{ W }_{ 0 } \right) }{ e } =\dfrac { 3.72eV-1.07eV }{ e } $$

$${ V }_{ 0 }=2.65$$ Volt

Option 4

6 Dual Nature of Matter and Radiation

When a certain metallic surface is illuminated with monochromatic light of wavelength $$\lambda$$, the stopping potential for photoelectric current is $$3V_0$$ and when the same surface is illuminated with light of wavelength $$2\lambda$$, the stopping potential is $$V_0$$. The threshold wavelength of this surface for photoelectric effect is

1) $$6\lambda$$
2) $$4\lambda /3$$
3) $$4\lambda$$
4) $$8\lambda$$

Option 1

Option 2

Option 3 Solution:

Solution

Let threshold wavelength be $$ \lambda_0 $$

Therefore work function for the surface will be $$ W= \dfrac{hc}{\lambda_0} $$

Stopping potential $$ V=\dfrac{E-W}{e} $$ where $$ E $$ is the energy of the incident beam ($$ \dfrac{hc}{\lambda} $$) and $$ e $$ is the charge on an electron.
Given

$$ 3eV_0=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0} $$

$$ eV_0=\dfrac{hc}{2\lambda}-\dfrac{hc}{\lambda_0} $$

Multiplying the second eqaution by 3 and subtracting it from the first we get

$$ \dfrac{2hc}{\lambda_0}-\dfrac{hc}{2\lambda}=0 $$

Solving we get $$ \lambda_0=4\lambda $$

Option 4

7 Properties of Bulk Matter

A Cylindrical tank is filled with water to a level of 3m. A hole is opened at a height of 52.5 cm from bottom. The ratio of the area of hole to that at cross sectional area at cylinder is 0.1. Find the square of velocity with which water is coming out (g = 10m/sec²)

(1) 50 (m/sec)² (2) 40 (m/sec)²

(3) 51.5 (m/sec)²(4) 50.5 (m/sec)²

Option 1 Solution:

Solution

Suppose I be the area of cross section of tank, a be the area of hole, ve be the velocity of efflux, h be the height of liquid above the hole, Let v be the speed with which the level decreases in the container. Using equation of continuity, we get av_e = Av or v = av_e/A Using Bernoulli's theorem, we have

Option 2

Option 3

Option 4

8 Atoms and Nuclei

An element with atomic no. z = 11 emits Ka - X ray of wavelength l, then the atomic no. of the element which emits µ - K X ray of wavelength 4l is

(1)11
2) 44
(3) 6
(4) 5

Option 1

Option 2

Option 3 Solution:

Option 4

9 Dual Nature of Matter and Radiation

Electromagnetic radiations described by the equation: $$E = 100 \left[ \sin \left( 2 \times 10 ^ { 15 } \right) t + \sin \left( 6.28 \times 10 ^ { 15 } \right) t \right] V / m$$ are incident on a photosensitive surface having work function $$1.5\ eV$$. The maximum kinetic energy of photoelectrons will be

1) 0.125 eV
2) 2.625 eV
3) 4.125 eV
4) 6.125 eV

Option 1

Option 2 Solution:

Solution

Standard form will be like $$E=E_0[sin(\omega _1t) +sin (\omega _2 t)]$$

so we have $$\omega _1 =2\times 10^{15}rad/s$$ and $$\omega _2 =6.28\times 10^{15}rad/s$$

corresponding frequency will be $$\nu _1=\omega _1/2\pi=\dfrac{2\times 10^{15}}{6.28}$$

and $$\nu _2=\omega _2/2\pi=\dfrac{6.28\times 10^{15}}{6.28}=10^{15}Hz$$

We can see that $$\nu _2 $$ is $$higher $$ so it will provide $$maximum $$ kinetic energy.

Energy of the $$\nu_2 $$ photon will be $$E=h\nu _2=6.6\times 10^{-34}Js\times 10^{15} Hz=6.6\times 10^{-19} Joule=\dfrac{6.6\times 10^{-19}Joule}{1.6\times 10^{-19}J/eV}$$

or $$E=4.125eV$$

Consumed energy in work function will be $$W=1.5eV$$

So remaining energy that is maximum kinetic energy will be $$KE=E-W=(4.125-1.5)eV=2.625eV$$

Option 3

Option 4

10 Electromagnetic Waves

The electric field of an electromagnetic wave travelling through vacuum is given by the equation E = E−0 sin (Kx - ωt). the quantity that is independent of wavelength is

1) kω
2) $$\dfrac{K}{\omega}$$
3) $$K^2\omega$$
4) $$\omega$$

Option 1

Option 2 Solution:

Solution

Here $$k = \dfrac{2\pi}{\lambda} , \, \omega \, = \, 2\pi \upsilon$$
$$\therefore \, \dfrac {k}{\omega} \, = \, \dfrac{2\pi / \lambda}{2\pi \upsilon} \, = \, \dfrac{1}{\upsilon \lambda} \, = \, \dfrac{1}{c} \,\, (\because \, c \, = \, u \lambda)$$
where c is the speed of electromagnetic wave in vacuum. It is a constant whose value is $$3 \, \times \, 10^8 \, ms^{-1}$$

Option 3

Option 4

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