Physics Test

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Physics Test - 39

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Question 1
1 / -0

A rod AD, consisting of three segments AB, BC and CD joined together, is hanging vertically from a fixed support at A. The lengths of the segments are respectively 0.1 m, 0.2 m and 0.15 m. The cross - section of the rod is uniform 10^-4m². A weight of 10 kg is hung from D. Calculate the displacements of point D if Y_AB = 2.5 x 10¹⁰ N/m², Y_BC = 4 x 10¹⁰N/m² and Y_CD = 1 x 10¹⁰ N/m². (Neglect the weight of the rod.)

A
23.5 x 10^-6 m

B
4.90 x 10^-6 m

C
14.7 x 10 ^-6 m

D
8.82 x 10^-6 m

Solution

By definition of Young's modulus,
$$
\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}}=\frac{\mathrm{MgL}}{\mathrm{YA}} \quad[\mathrm{as} \mathrm{F}=\mathrm{Mg}]
$$
So for rod $\mathrm{AB}, \Delta \mathrm{L}_{1}=\frac{10 \times 9.8 \times 0.1}{2.5 \times 10^{10} \times 10^{-4}}$
$$
=3.92 \times 10^{-6} \mathrm{m}
$$
and for rod $\mathrm{BC}, \Delta \mathrm{L}_{2}=\frac{10 \times 9.8 \times 0.2}{4 \times 10^{10} \times 10^{-4}}$
$$
=4.90 \times 10^{-6} \mathrm{m}
$$
and for rod $\mathrm{CD}, \Delta \mathrm{L}_{3}=\frac{10 \times 9.8 \times 0.15}{1 \times 10^{10} \times 10^{-4}}$
$$
=14.7 \times 10^{-6} \mathrm{m}
$$
So displacement of $\mathrm{B}=\Delta \mathrm{L}_{1}$
$$
\begin{aligned}
&=3.92 \times 10^{-6} \mathrm{m} \\
\text { displacement of } \mathrm{C} &=\Delta \mathrm{L}_{1}+\Delta \mathrm{L}_{2} \\
&=8.82 \times 10^{-6} \mathrm{m}
\end{aligned}
$$
and displacement of $\mathrm{D}=\Delta \mathrm{L}_{1}+\Delta \mathrm{L}_{2}+\Delta \mathrm{L}_{3}$
$$
=23.5 \times 10^{-6} \mathrm{m}
$$
Hence, the correct option is (A)
Question 2
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Energy from the sun falls on the earth at a rate of 1353 W/m², which is known as solar constant, i.e., the power incident per unit area per second at the top of atmosphere. Find the r.m.s values of the electric and magnetic fields in the sunlight reaching the top of the atmosphere.

A
714.0 V/m, 2.38 × 10^–6 T

B
7.14 V/m, 2.38 × 10^-8 T

C
71.4 V/m, 2.38 × 10^–7 T

D
None of these

Solution

For an electromagnetic wave of sinusoidal form $\vec{E}=E_{0} \cos (k x-\omega t) \hat{j} \quad$ and $\quad \vec{B}=B_{0} \cos (k x-\omega t) \hat{\mathbf{k}}$

Energy flux is given by $\widehat{\mathrm{S}}=\frac{E_{0}^{2}}{\mu_{0} c} \cos ^{2}(k x-\omega t) . \quad\left\{B_{0}=\frac{E_{0}}{c}\right\}$

The mean value of energy flux is intensity

$I=(S)=\frac{E_{0}^{2}}{2 \mu_{0} c}$

[The mean value of $\cos ^{2} \theta$ over on cycle is $\left.1 / 2\right]$ Hence

$E_{\mathrm{r} \ldots \mathrm{m}}=\sqrt{\mu_{0} \mathrm{CI}}$

$=\sqrt{4 \pi \times 10^{-7} \times 3 \times 10^{8} \times 1353}$

$=714.0 \mathrm{V} / \mathrm{m}$

$B_{\mathrm{r} . \mathrm{m.s}}=\frac{E_{\mathrm{rm}}}{c}=\frac{714.0}{2.998 \times 10^{5}}$

$=2.381 \times 10^{-6} \mathrm{T}$

Hence, the correct option is (A)
Question 3
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50 V battery is supplying a steady current of 10 amp when connected to an external resistor. If the efficiency of the battery at this current is 25%, then internal resistance of battery is:

A
2.5 Ω

B
1.25 Ω

C
5 Ω

D
3.75 $Ω$

Solution

$50=10[R+r]$
$\left\{\begin{array}{l}\mathrm{R}=\text { external resistance } \\ \mathrm{r}=\text { internal resistance }\end{array}\right\}$
$\mathrm{R}+\mathrm{r}=5 \Omega$
$\eta=\frac{\text { Pout }}{\text { Pouer of cell }}$
$\eta=\frac{1}{4}=\frac{i^{2} R}{\epsilon i}=\frac{10 R}{50}$
$\therefore \mathrm{R}=\frac{5}{4}=1.25 \Omega$
Form $(i)$
$\therefore \quad 5-R=3.75 \Omega$
Hence, the correct option is (D)
Question 4
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A lens (μ = 1.5) is coated with a thin film of refractive index 1.2 in order to reduce the reflection from its surface at $λ = 4800 Å$ . Find the minimum thickness of the film which will minimise the intensity of the reflected light. (Assume near normal incidence)

A
10^-7 m

B
0.5x10^-7 m

C
2x10^-7 m

D
1.5x10^-7 m

Solution

Path difference $=\frac{\lambda}{2}$

$2 \mu_{2} t=\frac{\lambda}{2}$
$t=\frac{\lambda}{4 \mu_{2}}$
$\Rightarrow t=\frac{4800 }{4 \times 1.2}$
$t=1000 \mathrm{A}$
$\Rightarrow t=10^{-7} \mathrm{m}$
Hence, the correct option is (A)
Question 5
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A block of mass m & charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform, horizontal and out of the plane of paper as shown. Find the time from start when block loses contact with the surface.

A

B

C

D

Solution

Block will loose contact with surface when force due to magnetic field will become equal to $m g \cos \theta$
$q v B=m g \cos \theta$
$v=\frac{m g \cos \theta}{q B}=u+a t$ (along the inclined plane)
$v=\frac{m g \cos \theta}{q B}=(g \sin \theta) t$
$t=\frac{m \cot \theta}{q B}$
Hence, the correct option is (C)
Question 6
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In the figure shown the acceleration of A is, $\overrightarrow{\mathrm{a}_{\mathrm{A}}}=1515$,then the acceleration of B is : (A remains in contact with B)

A

B

C

D

Solution

$\left(\mathbf{a}_{\mathbf{A}}\right)_{\perp}=\left(\mathbf{a}_{\mathbf{B}}\right)_{\perp}$

$\mathbf{a}_{\mathrm{A}} \mathrm{x} \cos 53^{\circ}-\mathrm{a}_{\mathrm{AY}} \cos 37^{\circ}=\mathrm{a}_{\mathrm{B}} \cos 53^{\circ}$
$\mathrm{a}_{\mathrm{B}}=-5 \mathrm{m} / \mathrm{s}$
$\overrightarrow{\mathrm{a}_{\mathrm{B}}}=-5 \hat{\mathrm{i}}$
Hence, the correct option is (D)
Question 7
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Binding energy per nucleon versus mass number curve for nuclei is shown in figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is:

A
$Y \rightarrow 2 Z$

B
$\mathrm{W} \rightarrow \mathrm{XZ}$

C
$\mathrm{W} \rightarrow 2 \mathrm{Y}$

D
$x \rightarrow Y Z$

Solution

Energy is released in a process when total binding energy of the nucleus (= binding energy per nucleon x number of nucleons) is increased or we can say, when total binding energy of products is more than the reactants.

Binding energy of reactants = 120 x 7. 5 = 900 MeV

and binding energy of products = 2 (60 x 8. 5)

= 1020 MeV > 900 MeV.

Hence, the correct option is (C)
Question 8
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Two bodies of same mass tied with an inelastic string of length l lie together on a horizontal surface. One a horizontal surface of them is projected vertically upwards with velocity $\sqrt{6 g l} .$ Find the maximum height up to which the centre of mass of system of the two masses rises.

A
31/4

B
1/2

C
31/2

D
1

Solution

Using $v^{2}=u^{2}+2 a s$
$\Rightarrow v^{2}=(6 g l)+2(-g)(l)$
$\Rightarrow v^{2}=4 g l$
$\Rightarrow v=\sqrt{4 g l}=2 \sqrt{g l}$
Using conservation of momentum $m v^{\prime}+m v^{\prime}=m v+0$
$\Rightarrow v^{\prime}=\frac{v}{2}=\sqrt{g l}$
As velocity of both the blocks is $v^{\prime}$ therfore velocity of $\mathrm{CM}$ will also be $v^{\prime}$ $\therefore h_{\max }=($ Initial height of $\mathrm{CM})+\frac{v_{C M}^{2}}{2 g}\left(a_{C M}=-g\right)$
$=\frac{l}{2}+\frac{g l}{2 g}=l$

Hence option D is the correct answer.
Question 9
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A battery of internal resistance 4 $Ω$ is connected to the network of resistances as shown in figure. In order that the maximum power can be delivered to the network, the value of R should be :

A
$\frac{4}{9} \Omega$

B
$\frac{8}{3} \Omega$

C
18 $Ω$

D
$2 \Omega$

Solution

The given circuit is a balanced Wheatstone's bridge.

Thus, no current will flow across 6R of the side CD. The given circuit will now be equivalent to

For maximum power, net external resistance

= Total internal resistance

or 2R = 4

or R = 2Ω.
Hence, the correct option is (D)
Question 10
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The speed of sound in hydrogen gas at N.T.P. is 1,328 m s^-1. What will be its value in air at N.T.P., if density of hydrogen is 1/16th that of air?

A
340 m s^-1

B
332 m s^-1

C
320 m s^-1

D
293 m s^-1

Solution

Here, $v_{H_{2}}=1328 \mathrm{m} \mathrm{s}^{-1} ; \rho_{H_{2}}=\frac{\rho_{\text {air }}}{16} ; v=\sqrt{\frac{\gamma P}{8}}$
Let $\mathrm{v}_{\text {air }}$ be the velocity of sound in air at $\mathrm{N} .$ T.P
$N o w, \frac{v_{a i r}}{v_{H_{2}}}=\sqrt{\frac{\rho_{H_{2}}}{\rho_{\text {air }}}} \quad\left\{\begin{array}{c}\text { since } \\ \gamma_{H_{2}}=\gamma_{\operatorname{air}}\end{array}\right\}$
or $\quad v_{\text {air}}=v_{H_{2}} \sqrt{\frac{\rho_{H_{2}}}{\rho_{\text {art }}}}$
$=332 \mathrm{m} \mathrm{s}^{-1}$
Hence, the correct option is (B)