Physics Test

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Physics Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Radius of gyration is denoted by ————

A

R

B

G

C

K

D

I

Solution

The radius of gyration is denoted by the alphabet ‘K’.

A radius of gyration in general is the distance from the center of mass of a body at which the whole mass could be concentrated without changing its moment of rotational inertia about an axis through the center of mass.
Question 2
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Consider an excited hydrogen atom in state n moving with a velocity v (v<< c). It emits a photon in the direction of its motion and changes its state to a lower state m. Apply momentum and energy conservation principle to calculate the frequency v of the emitted radiation. Compare this with the frequency v₀ emitted if the atom were at rest.

A
\(\nu_{0}=\nu\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

B
\(\nu_{0}=\nu\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)\)

C
\(\nu=\nu_{0}\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

D
\(\nu=\nu_{0}\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)\)

Solution

Let \(\mathrm{E_{n}}\) and \(\mathrm{E_{m}}\) be the energies of electron in \(n^{\text {th }}\) and \(m^{\text {th }}\) states.

Then, \(\mathrm{E_{n}-E_{m}=h v_{0}} \ldots(1)\)

In the second case when the atom is moving with a velocity \(\mathrm{v}\). Let \(\mathrm{v^{\prime}}\) be the velocity of atom after emitting the photon. Applying conservation of linear momentum,

\(\mathrm{mv}=\mathrm{mv}^{\prime}+\frac{\mathrm{h\nu}}{\mathrm{c}}\) ( \(\mathrm{m}\) = mass of hydrogen atom)

\(\Rightarrow \mathrm{v}^{\prime}=\left(\mathrm{v}-\frac{\mathrm{h\nu}}{\mathrm{mc}}\right) \ldots(2)\)

Applying conservation of energy

\(\mathrm{E}_{\mathrm{n}}+\frac{1}{2} \mathrm{mv}^{2}=\mathrm{E}_{\mathrm{m}}+\frac{1}{2} \mathrm{mv'}^{2}+\mathrm{h}\nu\)

\(\Rightarrow \mathrm{h} \nu=\left(\mathrm{E}_{\mathrm{n}}-\mathrm{E}_{\mathrm{m}}\right)+\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{v'}^{2}\right)\)

From equation (1) and (2)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\left(\mathrm{v}-\frac{\mathrm{h} \nu}{\mathrm{mc}}\right)^{2}\right]\)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\mathrm{v}^{2}-\frac{\mathrm{h}^{2} \nu ^{2}}{\mathrm{~m}^{2} \mathrm{c}^{2}}+\frac{2 \mathrm{h\nu v}}{\mathrm{mc}}\right]\)

\(=\mathrm{h} \nu_{0}+\frac{\mathrm{h} \nu \mathrm{v}}{\mathrm{c}}-\frac{\mathrm{h}^{2} \nu^{2}}{2 \mathrm{mc}^{2}}\)

Here the term is \(\frac{\mathrm{h}^{2}\nu^{2}}{2 \mathrm{mc}^{2}}\) is very small. So, can be neglected.

\(\therefore \mathrm{h \nu=h \nu_{0}+\frac{h \nu v}{c}}\)

\(\Rightarrow \nu=\nu_0 +\frac{\nu\mathrm{v}}{c}\)

\(\Rightarrow \nu_{0}=\nu\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)
Question 3
1 / -0

The depletion layer in a \(p-n\) junction diode is \(10^{-6} ~m\) wide and its knee potential is \(0.5 ~V\). What is the inner electric field in the depletion region?

A

\(5 \times(10)^{6}\) \(V/m\)

B

\(5 \times(10)^{-7}\) \(V/m\)

C

\(5 \times(10)^{5}\) \(V/m\)

D
None of these

Solution

In both forward biasing and reverse biasing, applied potential establishes an internal electric field which acts against or towards the potential barrier. This internal electric field is weakened or stronger at the junction. In forward biasing knee voltage is the forwards voltage at which the current through the junction starts to increase rapidly. Once the applied forward voltage exceeds the knee voltage, the current starts increasing rapidly.
In forward biasing condition, the inner electric field is given by \(E=-\frac{\Delta V}{\Delta r}\)
or
\(|E|=\frac{\Delta V}{\Delta r}=\frac{5 ×10^{-1}}{10^{-6}}\)
\(=5 × 10^{5} ~V / m\)
Question 4
1 / -0

Two thin equiconvex lenses, each of focal length \(0.2\) m, are placed coaxially with their optic centres \(0.5\) m apart. What is the focal length of the combination?

A
\(-0.4\) m

B
\(0.4\) m

C
\(-0.1\) m

D
\(0.1\) m

Solution

Equivalent focal length \((F)\) of two lens separated by distance \(d\) is given by

\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}\)

\(=\frac{1}{0.2}+\frac{1}{0.2}-\frac{0.5}{(0.2)(0.2)}\)

\(=5+5-0.5 \times 5 \times 5\)

\(=10-12.5\)

\(=-2.5\)

\(\therefore F=-\frac{1}{2.5}=-0.4\) m
Question 5
1 / -0

A body of \(5\) kg is moving with a velocity of \(20 \) m/s. If a force of \(100\) N is applied on it for \(10\) s in the same direction as its velocity, what will now be the velocity of the body?

A
\(200\) m/s

B
\(220\) m/s

C
\(240\) m/s

D
\(260\) m/s

Solution

Given,

\(u = 20\) m/s

\(t = 10\) s

\(F = 100\) N

\(m = 5\) kg

By the first law of motion,

\(v=u+a t\)

\(\Rightarrow v=u+\left(\frac{F}{m}\right) t\)

\(\Rightarrow v=20+\left(\frac{100}{5}\right) \times 10\)

\(\Rightarrow v=220 \) m/s
Question 6
1 / -0

As per given figure \(A, B\) and \(C\) are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths (. i.e. \(\frac{\lambda_1}{\lambda_2}\) ) is \(\frac{7}{4 \mathrm{n}}\), then the value of \(\mathrm{n}\) will be_________.

A
5

B
8

C
2

D
6

Solution

For \(A, n=2\)

\(\mathrm{B}, \mathrm{n}=3 \)

\(\mathrm{C}, \mathrm{n}=4 \)

\( \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \)

\(\frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \)

\(\frac{1}{\lambda_2}=\frac{7 \mathrm{R}}{144} \ldots(1) \)

\(\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \)

\( \frac{1}{\lambda_1}=\frac{5 \mathrm{R}}{36} \ldots(2)\)

(1) and (2)

\(\frac{\lambda_1}{\lambda_2}=\frac{7}{20}=\frac{7}{4 \times 5}\)

\( n=5\)
Question 7
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If \(v=\frac{A}{t}+B t^{2}+C t^{3}\) where \(v\) is velocity, \(t\) is time and \(A, B\) and \(C\) are constants, then the dimensional formula of \(B\) is:

A
\(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\)

B
\(\left[\mathrm{ML}^{0} \mathrm{~T}^{0}\right]\)

C
\(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]\)

D
\(\left[\mathrm{M}^{0} \mathrm{LT}^{-3}\right]\)

Solution

Given,

\(v=\frac{A}{t}+B t^{2}+C t^{3}\)

Where,

v = Velocity

t = Time

A, B and C = Constants

\(\mathrm{As}, {v}=\frac{\mathrm{A}}{\mathrm{t}}+\mathrm{B} \mathrm{t}^{2}+\mathrm{C} \mathrm{t}^{3}\)

So, we can write the dimensional equation as:

\(\operatorname{dim}(\mathrm{v})=\operatorname{dim}\left(\mathrm{B} \mathrm{t}^{2}\right)\)

\(\therefore \operatorname{dim}(\mathrm{B})=\frac{\operatorname{dim}(\mathrm{v})}{\operatorname{dim}\left(\mathrm{t}^{2}\right)}\)

\(=\frac{\left[\mathrm{LT}^{-1}\right]}{\left[\mathrm{T}^{2}\right]}\)

\(=\left[\mathrm{LT}^{-3}\right]\)

\(\Rightarrow \operatorname{dim}(\mathrm{B})=\left[\mathrm{M}^{0} \mathrm{L} \mathrm{T}^{-3}\right]\)
Question 8
1 / -0

A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly

A
12.6 × 10^–6 m³/s

B
8.9 × 10^–6 m³/s

C
2.23 × 10^–6 m³/s

D
6.4 × 10^–6 m³/s

Solution

Rate of flow liquid

\(Q=a u=a \sqrt{2 g h} \)

\(=2 \times 10^{-6} m ^{2} \times \sqrt{2 \times 10 \times 2} ~m / s\)

\(=2 \times 2 \times 3.16 \times 10^{-6} m ^{3} / s\)

\(=12.64 \times 10^{-6} m ^{3} / s\)

\(=12.6 \times 10^{-6} m ^{3} / s\)
Question 9
1 / -0

The magnitude of the magnetic field at the center of an equilateral triangular loop of side 1m which is carrying a current of 10 A is :

\(\left[\right.\) Take \(\mu_0=4 \pi \times 10^{-7} \mathrm{NA}^{-2}\) ]

A
\(18 \mu \mathrm{T}\)

B
\(9 \mu \mathrm{T}\)

C
\(3 \mu \mathrm{T}\)

D
\(1 \mu \mathrm{T}\)

Solution

\(\begin{aligned} r & =\left(\frac{1}{3}\right)(\mathrm{a} \sin 60) \\ \mathrm{r} & =\frac{\mathrm{a}}{3} \times \frac{\sqrt{3}}{2}=\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right) \\ \mathrm{B}_0 & =3\left[\frac{\mu_0 1}{4 \pi \mathrm{r}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)\right] \\ & =\frac{3 \mu_0 1}{4 \pi\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)} \times(2)\left(\frac{\sqrt{3}}{2}\right)=\frac{9}{2}\left(\frac{\mu_0 1}{\pi \mathrm{a}}\right) \\ & =\frac{9 \times 2 \times 10^{-7} \times 10}{1}=18 \mu \mathrm{T}\end{aligned}\)
Question 10
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A body cools from \(80^{\circ} C\) to \(60^{\circ}\) in 5 minutes. The temperature for the surrounding is \(20^{\circ} C\). The time it takes to cool from \(60^{\circ} C\) to \(40^{\circ} C\) is

A
\(450 s\)

B
\(420 s\)

C
\(500 s\)

D
\(\frac{25}{3} s\)

Solution

The formula to calculate the rate of cooling of the object is given by

\(\frac{d T}{d t}=K\left[\frac{T_f+T_i}{2}-T_0\right]\)

For the first case, it can be written, using equation (1) that

\(\frac{80-60}{5} =K\left[\frac{80+60}{2}-20\right] \)

\(4 =50 K \ldots(2)\)

If \(t\) is the required time for the second case, from equation (1), it can be written that

\(\frac{60-40}{t} =K\left[\frac{60+40}{2}-20\right] \)

\(\frac{20}{t} =30 K \ldots(3)\)

Divide equation (2) by equation (3) and solve to calculate the required time.

\(\frac{4}{\frac{20}{t}} =\frac{50 K}{30 K} \)

\( \Rightarrow \frac{t}{5}=\frac{5}{3} \)

\( \Rightarrow t=\frac{25}{3} \min =500 s\)

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Re-Attempt Weekly Quiz Competition

Physics Test - 3

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Physics Test - 3

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SHARING IS CARING

Question 1

R

G

K

I

Solution

Question 2

\(\nu_{0}=\nu\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

\(\nu_{0}=\nu\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)\)

\(\nu=\nu_{0}\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

\(\nu=\nu_{0}\left(1+\frac{\mathrm{v}}{\mathrm{c}}\right)\)

Solution

Question 3

\(5 \times(10)^{6}\) \(V/m\)

\(5 \times(10)^{-7}\) \(V/m\)

\(5 \times(10)^{5}\) \(V/m\)

None of these

Solution

Question 4

\(-0.4\) m

\(0.4\) m

\(-0.1\) m

\(0.1\) m

Solution

Question 5

\(200\) m/s

\(220\) m/s

\(240\) m/s

\(260\) m/s

Solution

Question 6

5

8

2

6

Solution

Question 7

\(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\)

\(\left[\mathrm{ML}^{0} \mathrm{~T}^{0}\right]\)

\(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right]\)

\(\left[\mathrm{M}^{0} \mathrm{LT}^{-3}\right]\)

Solution

Question 8

12.6 × 10–6 m3/s

8.9 × 10–6 m3/s

2.23 × 10–6 m3/s

6.4 × 10–6 m3/s

Solution

Question 9

\(18 \mu \mathrm{T}\)

\(9 \mu \mathrm{T}\)

\(3 \mu \mathrm{T}\)

\(1 \mu \mathrm{T}\)

Solution

Question 10

\(450 s\)

\(420 s\)

\(500 s\)

\(\frac{25}{3} s\)

Solution

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12.6 × 10^–6 m³/s

8.9 × 10^–6 m³/s

2.23 × 10^–6 m³/s

6.4 × 10^–6 m³/s