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Question 1
1 / -0

The ratio of wavelengths of proton and deuteron accelerated through the same potential difference will be

A
1 : 2

B
2 : 1

C

$\sqrt{2} \cdot 1$

D

$1 \sqrt{2}$

Solution

$\lambda=\frac{h}{\sqrt{2 n V_{ m }}}$
Since both proton and deuteron have same charge and accelerated through the same potential difference.
$\therefore \frac{\lambda_{p}}{\lambda_{4}}=\sqrt{\frac{m_{a}}{m_{p}}}=\sqrt{\frac{2 m_{p}}{m p}}=\sqrt{2}$
Concepts:
Main Concept:
De Broglie Wavelength of A Particle 1. De Broglie first used Einstein's famous equation relating matter and energy:
$E =\operatorname{me}^{2}$
with
$E =$ energy
$m =$ mass
$c =$ speed of light
2. Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation:
$E = hv$
with
$E =$ energy
$h =$ Plank's constant $\left(6.62607 \times 10^{-34} Js \right)$
$v =$ frequency
3. since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal.
$mc ^{2}=h v$

4. Because real particles do not travel at the speed of light, De Broglie submitted velocity ( $v$ ) for the speed of light (c).
$m v^{2}=h v$
5. Through the equation $\lambda$, de Broglie substituted $v / \lambda$ for $v$ and arfived at the final expression that relates wavelength and particle with speed.
$m v^{2}=\frac{h v}{\lambda}$
Hence
$\lambda=\frac{h v}{m v^{2}}=\frac{h}{m v}$
A majority of Wave particle Duality problems are simple plug and chug via equation 5 with some variation of canceling out units.
De Broglie' Wavelength of An Accelerated Charge
de-Broglie wavelength : According to de-Broglie theory, the wavelength of de-Broglie wave is given by
Where $h=$ Plank's constant, $m=$ Mass of the particle, $v=$ Speed of the particle, $E=$ Energy of the particle.
The smallest wavelength whose measurement is possible is that of $\gamma$ -rays.

The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron $\alpha$ particle etc. is of the order of $10^{-10} m$. Examples on De Broglie Wave Equation
What is the wavelength of De Broglie wave of electron with kinetic energy E?
Step 1: Manipulate the $K.E =\frac{1}{2} mv ^{2}$ to get the velocity of the electron. $v =\sqrt{\frac{2 E }{ m }}$
Step 2 : Find the momentum of electron using the velocity $p=m v$
Step 3 : Fill in the De Broglie Wave Equation the momentum and other constants. $\lambda=\frac{b}{p}$
Step 4 : Calculate the wavelength.
The question may contain a different source to extrapolate the velocity rather than kinetic energy, the rest of steps remain the same.

The deBroglie Equation: Example Problems
Example 1: What is the wavelength of an electron (Mass $=9.11 \times 10^{-32} kg$ ) traveling
1) The first step in the solution is to calculate the kinetic energy of the electron:
$KE =\left(\frac{1}{2}\right) mv ^{2}$
$x=\left(\left(\frac{1 / 1}{2)}\right)(9.11 \times 10)^{-31} kg 5.31 \times 10^{6} m / s ^{2}\right.$
$x=1.28433 \times 10^{-17} kg m ^{2} s ^{-2}$ (। kept some guard digits)
When I use this value just below, I will use J (for Joules).
2) Next, we will use the de Broglie equation to calculate the wavelength:
$\lambda=\frac{ h }{ p }$
$\lambda=\frac{ h }{\sqrt{(2 Em )}}$
Just to be sure about two things:
(1) the unit on Planck's Constant is Joule-seconds, both are in the numerator and
(2) there are three values following the radical in the denominator. All three of them are under the radical sign.
The answer:
$x=1.37 \times 10^{-10} m$
Td like to compare this wavelength to ultraviolet light, if I may. Let's use $A=4000 \times 10^{-4} cm =4 \times 10^{-7} m$
Our electron's wavelength is almost 3000 times shorter than our ultraviolet example and its wavelength puts it in the X-ray region of the electromagnetic spectrum.

This turned out to be very important because one could then take a beam of electrons and perform experiments with detectable results. You canit do that with the short wavelengths of heavier particles (see examples below).
In 1926 , de Broglie predicted that matter had wave like properties. In 1927 , experiments were done that showed electrons behaved as a wave (by showing the property of diffraction and interference patterns). In 1937 , the Nobel Prize in Physics was awarded to Clint Davisson and George Thomson (son of J.J. Thomson) for this work.

Example 2 : What is the wavelength in meters of a proton traveling at $255,000,000 m / s$ (which is $85 \%$ of the speed of light)? (Assume the mass of the proton to be $1.673 \times 10^{-27} kg$ )
1) Calculate the kinetic energy of the proton:
$KE =\left(\frac{1}{2}\right) mv ^{2}$
$x =\left(\frac{1 /}{2}\right)(1.673 \times 10)^{-27} kg 2.55 \times 10^{9} m / s ^{2}$
$x=5.43934 \times 10^{-11} J$
2) Use the de Broglie equation:
$\lambda=\frac{h}{p}$
$\lambda=\frac{ h }{\sqrt{(2 Em )}}$
$x=1.55 \times 10^{-15} m$
This wavelength is comparable to the radius of the nuclei of atoms, which range from $1 \times 10^{-15} m$ (or 1 to 10 fm).
Example 3 : Calculate the wavelength (in nanometers) of a H atom
(Mass $=1.674 \times 10^{-27} kg$ ) moving at $698 cm / s$
1) Convert $cm / s$ to $m / s$
$698 cm / s =6.98 m / s$

2) Calculate the kinetic energy of the proton:
$KE =\left(\frac{1}{2}\right) mv ^{2}$
$x =\left(\frac{1 / 1}{2}\right)(1.674 \times 10)^{-27} kg 6.98 m / s ^{2}$
$x=5.84226 \times 10^{-27} J$
Hence, the correct option is (C)
Question 2
1 / -0

Two radio antennas radiating waves in phase are located at point A and B, 200 m part (Figure). The radio waves have a frequency of 6 MHz. A radio receiver is moved out from point B along a line perpendicular to the line connecting A and B ( line BC shown in figure). At what distances from B will there be destructive interference ?

A
787.5, 229m, 97.5m, 26.7m

B
789.5, 230m, 98.5m, 27.7m

C
788.5, 228m, 98.5m, 28.7m

D
777.5, 239m, 87.5m, 36.7m

Solution

$\mathrm{C}=\mathrm{n} \lambda \quad \Rightarrow \lambda=\frac{\mathrm{C}}{\mathrm{n}}$
$\lambda=\frac{3 \times 10^{8}}{6 \times 10^{6}} \quad \Rightarrow \lambda=50 \mathrm{m}$
path difference at $P=A P-B P$
path difference at $P=\sqrt{(4 \lambda)^{2}+x^{2}}-x \ldots \ldots \ldots \ldots . .(1)$
For destructive interference at $P$
path difference $=(2 n-1) \frac{\lambda}{2} \ldots \ldots$
From (1) and (2)
$\sqrt{(4 \lambda)^{2}+\mathrm{x}^{2}}-\mathrm{x}=(2 \mathrm{n}-1) \frac{\lambda}{2}$
$\sqrt{(4 \lambda)^{2}+\mathbf{x}^{2}}=(2 \mathbf{n}-1) \frac{\lambda}{2}+\mathbf{x}$
For first minima $\mathrm{n}=1$
$\mathrm{x}_{1}=\frac{63 \lambda}{4} \quad \Rightarrow \mathrm{x}_{1}=787.5 \mathrm{m}$
For second minima $n=2$,
$\mathrm{x}_{2}=\frac{55 \lambda}{12} \Rightarrow \mathrm{x}_{2}=229 \mathrm{m}$
For third minima $n=3$,
$\mathrm{x}_{3}=\frac{39 \lambda}{20} \Rightarrow \mathrm{x}_{3}=97.5 \mathrm{m}$
Hence, the correct option is (A)
Question 3
1 / -0

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically.The tension in the string is equal to :

A
$\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}$

B
m√g²+a²-ma

C
m(g+a)

D
m√g²+a²+ma

Solution

(Force diagram in the frame of the car)

Applying Newton's law perpendicular to string
$mg sin θ = ma cos θ$

$tan θ = \frac{a/}{g}$

Applying Newton's law along string

$\mathrm{T}-\mathrm{m} \sqrt{g^{2}+a^{2}}=m a$

$\mathrm{T}=\mathrm{m} \sqrt{\mathrm{g}^{2} \mathrm{a}^{2}} \mathrm{ma}$
Hence, the correct option is (D)
Question 4
1 / -0

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency . If L is doubled and C is changed to 4C, the frequency will be

A
$f / 2$

B
$\frac{f}{4}$

C
8f

D
$\frac{f}{2 \sqrt{2}}$

Solution

Frequency of LC oscillation =12πLC√

or, f1f2=(L2C2L1C1)1/2=(2L×4CL×C)1/2=(8)1/2

$\therefore \quad \frac{f_{1}}{f_{2}}=2 \sqrt{2} \Rightarrow f_{2}=\frac{f_{1}}{2 \sqrt{2}}$ or $f_{2}=\frac{f}{2 \sqrt{2}}\left(\because f_{1}=f\right)$
Hence, the correct option is (D)

Question 5

1 / -0

The combination of 'NAND' gates shown here (in figure), are equivalent to

An OR gate and an AND gate respectively

An AND gate and a NOT gate respectively

An AND gate and an OR gate respectively

An OR gate and a NOT gate respectively

Solution

$\mathrm{C}=\overline{\mathrm{A} \bar{B}} \quad$ (if a NAND is given only 1 input it s behaves as NOT gate, as it is $\overline{\mathrm{AA}}=\overline{\mathrm{A}}$ (one input is divided in two and then passed through NAND)

A	B	$\underset{A}{\to}$	$\underset{B}{\to}$	$\underset{A B}{\to}$	$\overline{\bar{A} \bar{B}}$
1	0	0	0	0	1
0	1	1	0	0	1
1	0	0	1	0	1
0	0	1	1	1	0

i.e. OR gate.

For second circuit C = Not of (A NAND B)

i.e. AND only
Hence, the correct option is (A)

Question 6
1 / -0

Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity $σ$ and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is :

A

$\frac{4 \pi \sigma}{\ell n(b / a) V}$

B

$\frac{4 \pi \sigma}{(b+a) V}$

C

$\frac{2 \pi \sigma}{(b+a) V}$

D

$\frac{2 \pi \sigma}{\ell n(b / a) V}$

Solution

$I=I / A=\sigma E$
$E=\frac{I}{2 \pi \sigma L} \frac{1}{s} \hat{r}$
$V=\int_{a}^{b} E \cdot d s=\int_{a}^{b}\left(\frac{I}{2 \pi \sigma L}\right) \frac{1}{s} d s$
$=\frac{I}{2 \pi \sigma L} \ln (b / a)$
$I=\frac{(2 \pi \sigma L) V}{\ln (b / a)}$
Hence, the correct option is (D)
Question 7
1 / -0

A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. ( $γ$ is universal gravitational constant )

A
$\frac{3}{8}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}$

B
$\frac{2}{8}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}$

C
$\frac{8}{3}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}$

D
None of these

Solution

We consider a spherical concentric shell of radius x and thickness dx.

The mass of considered element is dm = (4π x²dx)ρ.

Gravitational field at a point in the shell is

$\begin{aligned} \overrightarrow{\mathrm{g}}=-\frac{4 \pi}{3} \gamma \rho \vec{x} & \\ \therefore \quad d F=\overrightarrow{\mathrm{g}} \quad d m=\frac{4 \pi}{3} \gamma \rho x d m=\frac{4 \pi}{3} \gamma \rho x\left(4 \pi x^{2} d x\right) \rho \end{aligned}$

The pressure in the element is
\mathrm{dP}=\frac{\mathrm{dF}}{A}=\frac{\left(\frac{4 \pi}{3} \gamma \rho x\right)\left(4 \pi x^{2} \mathrm{dx}\right) \rho}{4 \pi x^{2}}
$\therefore$
Pressure $P=\frac{4 \pi \rho^{2} \gamma}{3} \int_{r}^{R} \quad x \quad d x=\frac{4 \pi \rho^{2} \gamma}{3}\left[\frac{x^{2}}{2}\right]_{r}^{R}=\frac{4 \pi \rho^{2} \gamma}{3}\left[\frac{R^{2}}{2}-\frac{r^{2}}{2}\right]$
But $\quad \rho=\frac{M}{\frac{4}{3} \pi R^{3}}$
By putting the value of $\rho,$ we get,
\mathbf{P}=\frac{3}{8}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}
Hence, the correct option is (A)
Question 8
1 / -0

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is

A
V

B
2V

C
4V

D
– 2V

Solution

The electric field in between the shell and sphere is $E \cdot \pi x^{2}=\frac{Q_{e n}}{\epsilon_{0}}=\frac{Q}{\epsilon_{0}} \quad($ using Gauss's law $)$
$E=\frac{Q}{4 \pi \epsilon_{0} x^{2}}$
The potential difference between the shells is $d V=V_{r}-V_{R}=\int_{r}^{R} E d x=\int_{r}^{R} \frac{Q}{4 \pi \epsilon_{0} x^{2}} d x=$
$\frac{Q}{4 \pi \epsilon_{0}}(1 / r-1 / R)$
Thus, $V=\frac{Q}{4 \pi \epsilon_{0}}(1 / r-1 / R)$
As the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere, since the two values have not changed, potential difference does not change. Hence the potential difference remains $V$.
Hence, the correct option is (A)
Question 9
1 / -0

A wooden cylinder of mass 20 g and area of cross-section 1 cm², having a piece of lead of mass 60 g attached to its bottom floats in water. The cylinder is depressed and then released.

A
0.557 s^-1

B
0.556 s^-1

C
0.545 s^-1

D
0.513 s^-1

Solution

Suppose that the loaded wooden block sinks upto a height h. Then,
weight of water displaced by the block
$=$ weight of the block with lead
If a is area of cross-section of the block and $\sigma$, the density of water, then
$a h \sigma \mathrm{g}=(\mathrm{M}+\mathrm{m}) \mathrm{g}$
or $h=\frac{M+m}{a \sigma}$
$=\frac{20+60}{1 \times 1}=80 \mathrm{cm} \quad\left(\because\right.$ density of water $\left.=1 \mathrm{gcm}^{-3}\right)$
Now for small displacement ' $x$ ' from equilibrium
$\sigma g a(x+h)-(M+m) g=-(M+m) A$
$\sigma g a x=-(M+m) A$
$A=-\frac{\sigma g a}{M+m} x$
$f=\frac{1}{2 \pi} \sqrt{\frac{g}{h}}$
$=\frac{1}{2 \pi} \sqrt{\frac{980}{80}}$
$=0.557 s^{-1}$
Hence, the correct option is (A)
Question 10
1 / -0

Estimate the maximum magnitudes of the electric and magnetic fields of the light that is incident on this page because of the visible light coming from your desk lamp. Treat the bulb as a point source of electromagnetic radiation, i.e., about 5% efficient in converting electrical energy to visible light.
(HINT: assume 60W bulb and a distance of 0.3 m)

A
45 V/m, 1.5 × 10^-7T

B
4.6 V/m, 1.5 × 10^-7T

C
0.35 V/m, 1.5 × 10^-7T

D
0.045 V/m, 1.5 × 10^-2T

Solution

$I=\frac{P}{4 \pi r^{2}},$ where $P=5 \%$ of $60 W$ $\begin{array}{l}\text { \& } \\ I\end{array}=\frac{1}{2} \varepsilon_{0} E_{0}^{2} \times c$
$=\frac{1}{2 \mu_{0}} B_{0}^{2} \times c$
$\Rightarrow E_{0}=\sqrt{\frac{2 I}{\varepsilon_{0} c}}$
$\& B_{0}=\sqrt{\frac{2 \mu_{0} I}{c}}$
Hence, the correct option is (A)

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Physics Test - 29

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Physics Test - 29

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Question 1

1 : 2

2 : 1

\(\sqrt{2} \cdot 1\)

\(1 \sqrt{2}\)

Solution

Question 2

787.5, 229m, 97.5m, 26.7m

​789.5, 230m, 98.5m, 27.7m

​788.5, 228m, 98.5m, 28.7m

​777.5, 239m, 87.5m, 36.7m

Solution

Question 3

\(\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}\)

m√g2+a2-ma

m(g+a)

m√g2+a2+ma

Solution

Question 4

f/2

f4

8f

f22

Solution

Question 5

An OR gate and an AND gate respectively

An AND gate and a NOT gate respectively

An AND gate and an OR gate respectively

An OR gate and a NOT gate respectively

Solution

Question 6

\(\frac{4 \pi \sigma}{\ell n(b / a) V}\)

\(\frac{4 \pi \sigma}{(b+a) V}\)

\(\frac{2 \pi \sigma}{(b+a) V}\)

\(\frac{2 \pi \sigma}{\ell n(b / a) V}\)

Solution

Question 7

\(\frac{3}{8}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}\)

\(\frac{2}{8}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}\)

\(\frac{8}{3}\left(1-\frac{r^{2}}{R^{2}}\right) \frac{\gamma M^{2}}{\pi R^{4}}\)

None of these

Solution

Question 8

V

2V

4V

– 2V

Solution

Question 9

0.557 s-1

0.556 s-1

0.545 s-1

0.513 s-1

Solution

Question 10

45 V/m, 1.5 × 10-7T

4.6 V/m, 1.5 × 10-7T

0.35 V/m, 1.5 × 10-7T

0.045 V/m, 1.5 × 10-2T

Solution

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789.5, 230m, 98.5m, 27.7m

788.5, 228m, 98.5m, 28.7m

777.5, 239m, 87.5m, 36.7m

m√g²+a²-ma

m√g²+a²+ma

$f / 2$

$\frac{f}{4}$

$\frac{f}{2 \sqrt{2}}$

0.557 s^-1

0.556 s^-1

0.545 s^-1

0.513 s^-1

45 V/m, 1.5 × 10^-7T

4.6 V/m, 1.5 × 10^-7T

0.35 V/m, 1.5 × 10^-7T

0.045 V/m, 1.5 × 10^-2T