Physics Test

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Physics Test - 40

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Question 1
1 / -0

A balloon that is initially flat, is inflated by filling it from a tank of compressed air. The final volume of the balloon is $5 m ^{3} .$ The barometer reads $95 kPa$. The work done in this process is :

A

$4.75 J$

B

$4.75 \times 10^{7} J$

C

$4.75 \times 10^{3} J$

D

$4.75 \times 10^{5} J$

Solution

$W=P \Delta V=95 \times 10^{3} \times 5.0 $
$ =4.75 \times 10^{5} J$
Hence, the correct option is (D)
Question 2
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A 0.5 kg block slides from the point A, on a horizontal track, with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2N/m. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (g = 10 m/s²)

A
3.23 m

B
4.14 m

C
4.24 m

D
4.67 m

Solution

As the track AB is frictionless, the block moves this distance without loss in its initial $K E=$
$\frac{1}{2} m v^{2}=\frac{1}{2} \times 0.5 \times 3^{2}=2.25 J .$ In the path BD as friction is present, so work done against friction
$=\mu_{k} m g s=0.2 \times 0.5 \times 10 \times 2.14=2.14 J$
So, at D the KE of the block is $=2.25-2.14=0.11 \mathrm{J}$
Now, if the spring is compressed by $\mathrm{x}$ $0.11=\frac{1}{2} \times k \times x^{2}+\mu_{k} m g x$
i.e., $0.11=\frac{1}{2} \times 2 \times x^{2}+0.2 \times 0.5 \times 10 x$
or $x^{2} x-0.11=0$
which on solving gives positive value of $x=0.1 \mathrm{m}$ After moving the distance $\mathrm{x}=0.1 \mathrm{m}$ the block comes to rest. Now the compressed spring exerts a force:
$F=k x=2 \times 0.1=0.2 N$
On the block while limiting frictional force between block and track is $f_{L}=\mu_{s} m g$
$=0.22 \times 0.5 \times 10=1.1 \mathrm{N} .$ Since, $Fmoved by the block \(=A B+B D+0.1$
$=2+2.14+0.1$
$=4.24 \mathrm{m}$
Hence, the correct option is (C)
Question 3
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A resistance of 2 Ω is connected across one gap of a meter-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 Ω , is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any end corrections, the unknown resistance is :

A
4 $Ω$

B
1.33 $Ω$

C
6 $Ω$

D
3 $Ω$

Solution

$R>2 \Omega \therefore 100-l>l$

Applying $\frac{P}{Q}=\frac{R}{S}$
We have $\frac{2}{R}=\frac{l}{100-1} \quad \ldots \ldots \ldots .(i)$
On intercharging balance point must shift to right as $\mathrm{R}>2 \Omega$
$$
\frac{R}{2}=\frac{l+20}{100-l-20} \quad \ldots \ldots \ldots
$$
Solving Equation. $(i)$ and $(i i),$ we get $, R=3 \Omega$
$l^{2}+20 l=(100-l)(80-l)$
$=8000-100 l=80 l+l^{2}$
$=8000+l^{2}-180 l$
$200 l=8000$
$l=40$
Hence, the correct option is (D)
Question 4
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Two trains, which are moving along different tracks in opposite directions, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when they are 300 m apart. Graphs given below show their velocities as a function of time as they slow down. The separation between the trains, when both have stopped, is :

A
120 m

B
280 m

C
60 m

D
20 m

Solution

Initial distance between trains is 300 m. Displacement of 1st train is calculated by area under V-t.
curve of train $1 = \frac{1/}{2} \times 10 \times 40 = 200 m$ .

Displacement of train $2 = \frac{1/}{2} \times 8 \times (- 20) = - 80 m$
Which means it moves towards left.
$∴$ Distance between the two is = 300 - 280 = 20 m
Hence, the correct option is (D)
Question 5
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A horizontally insulated cylindrical vessel of length 2ℓ is separated by a thin insulating piston into two equal parts each of which contains n moles of an ideal monoatomic gas at temperature T. The piston is connected to the end faces of the vessel by undeformed springs of force constant k each.

The left part is in contact with a thermostat (a device which maintains a constant temperature). When an amount of heat Q is supplied to the gas in the right part, the piston is displaced to the left by a distance x = ℓ/2. Determine the heat Q' given away to the thermostat by the left part of the piston.

A
Q'=Q−(5/2)Kl²−3nRT

B
$\mathrm{Q}^{\prime}=\mathrm{Q}-(5 / 2) \mathrm{K} \mathrm{l}^{2}-4 \mathrm{nRT}$

C
$\mathrm{Q}^{\prime}=\mathrm{Q}-(3 / 2) \mathrm{K} \mathrm{l}^{2}-4 \mathrm{nRT}$

D
$\mathrm{Q}^{\prime}=\mathrm{Q}-(4 / 2) \mathrm{K} \mathrm{l}^{2}-8 \mathrm{nRT}$

Solution

Total heat given to the system is $\mathrm{Q}-\mathrm{Q}^{\prime}$. So from first law of thermodynamics, $\mathrm{Q}-\mathrm{Q}^{\prime}=$ total work done by the gas in both the chambers $(\mathrm{W})+$ change in internal energies of both the gases $(\Delta U)$
There $W=$ sum of potential energies stored in the springs
$\mathbf{W}=2\left[\frac{1}{2} k\left(\frac{\ell}{2}\right)^{2}\right]=\frac{k \ell^{2}}{4} \quad \ldots \ldots \ldots \ldots(2)$
Since the temperature of left part remains constant (piston does not conduct heat), internal energy of left part does not change. $\Delta U$ of right part can be given as
$\Delta \mathrm{U}=n \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}=\frac{3}{2} n \mathrm{R} \Delta \mathrm{T} \quad \ldots \ldots \ldots \ldots$
$\Delta T$ can be found from the condition of equilibrium at the end of the process. pressure on right side = pressure on left side
or $\frac{n R(T+\Delta T)}{A(\ell+\ell / 2)}=\frac{n R T}{A(\ell-\ell / 2)}+\frac{2 K(\ell / 2)}{A}$
Simplifying this we get
or $\Delta \mathbf{T}=\frac{3 k e^{2}}{2 n \mathbb{R}}+2 \mathbf{T} \quad \ldots \ldots \ldots \ldots(4)$
From equations (1),(2),(3) and $(4),$ we get
$Q-Q^{\prime}=\frac{+K l^{2}}{4}+\frac{3}{2} n R\left(\frac{3 K l^{2}}{2 n R}+2 T\right) \quad$ or $\quad Q^{\prime}=Q-\frac{5}{2} k l^{2}-3 n R T$
Hence, the correct option is (A)
Question 6
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Which element has a hydrogen-like spectrum and whose lines have wavelengths four times shorter than those of atomic hydrogen?

A
Helium ion

B
Hydrogen

C
Lithum ion

D
None of these

Solution

For Hydrogen like atom, $\frac{1}{\lambda^{\prime}}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
For Hydrogen atom, $\quad \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
Dividing Eqn (ii) by Eqn (i), we get
or
$$
\begin{array}{r}
\frac{\lambda^{\prime}}{\lambda}=\frac{R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)}{R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)} \\
\frac{\lambda^{\prime}}{\lambda}=\frac{1}{Z^{2}}
\end{array}
$$
According to problem, $\lambda^{\prime}=\frac{\lambda}{4} \quad \therefore \frac{\lambda^{\prime}}{4 \lambda^{\prime}}=\frac{1}{Z^{2}}$
$z=2$
The atomic number of Helium atom is 2 . Hence, required element is helium.
Hence, the correct option is (A)
Question 7
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Two graphs of the same projectile motion (in the xy plane) projected from origin are shown. X axis is along horizontal direction & Y axis is vertically upwards. Take g = 10 m/s².

Find initial velocity of projectile.

A
$\mathbf{u}_{y}=5 \mathbf{m} / \mathbf{s}$, u_x = 4 m/s

B
$\mathbf{u}_{y}=3 \mathbf{m} / \mathbf{s}$,ux=4 m/s

C
$\mathbf{u}_{y}=3 \mathbf{m} / \mathbf{s}$,u_x=9m/s

D
None of these

Solution

From graph (1)$: V_{y}=0$ at $t=\frac{1}{2} s e c$
i.e., time taken to reach maximum height H is,
$t=\frac{u_{y}}{g}=\frac{1}{2} \Rightarrow u_{y}=5 m / s \cdots \cdots \cdots \cdots$ Ans. $(i)$
From graph(2): $\mathbf{v}_{y}=0$ at $x=2 m$
i.e., when the particle is at maximum height,
its displacement along horizontal $\mathrm{x}=2 \mathrm{m}$
$x=u_{x} \times t \Rightarrow 2=u_{x} \times \frac{1}{2} \Rightarrow u_{x}=4 \mathrm{m} / \mathrm{s} \quad \cdots \cdots \cdots$ Ans. (ii)
Hence, the correct option is (A)
Question 8
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A network of Four capacitors of capacity equal to C₁ = C, C₂ = 2C, C₃ = 3C and C₄ = 4C, are connected to a battery as shown in the figure. The ratio of the charges on C₂ and C₄ is :

A
4/7

B
3/22

C
7/4

D
22/3

Solution
Question 9
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An infinite wire placed along z-axis, has current $I_{i}$ in positive z-direction. A conducting rod placed in xy plane parallel to y-axis has current $I_{2}$ in positive y-direction. The ends of the rod subtend angles of +30^o and -60^o at the origin with positive x-direction. The rod is at a distance 'a' from the origin. Find net force on the rod.

A
$\mathrm{F}=\frac{\mathrm{R}_{9}}{4 \pi} \mathrm{I}_{1} \mathrm{I}_{2} \ln 3(-\hat{\mathrm{k}})$

B
$\mathrm{F}=\frac{\mathrm{N}_{9}}{4 \pi} 2 \mathrm{I}_{1} \mathrm{I}_{2} \ln 3(-\hat{\mathrm{k}})$

C

$\mathrm{F}=\frac{\mathrm{H}_{9}}{4 \mathrm{x}} \mathrm{I}_{1} \mathrm{I}_{2} \ln 3(\hat{\mathrm{k}})$

D
$\mathrm{F}=\frac{\mathrm{k}_{9}}{4 \pi} 2 \mathrm{I}_{1} \mathrm{I}_{2} \ln 3(\hat{\mathrm{k}})$

Solution
Question 10
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The electric potential at a point (x, y, z) is given by V = -x²y - xz³ + 4
The electric field $\underset{E}{\to}$ at that point is :

A
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 2 \mathrm{xy}+\hat{\mathrm{j}}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\hat{\mathrm{k}}\left(3 \mathrm{xz}-\mathrm{y}^{2}\right)$

B
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} \mathrm{z}^{3}+\hat{\mathrm{j}} \mathrm{xyz}+\hat{\mathrm{k}} \mathrm{z}^{2}$

C
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 \mathrm{xy}-\mathrm{z}^{3}\right)+\hat{\mathrm{j}} \mathrm{xy}^{2}+\hat{\mathrm{k}} 3 \mathrm{z}^{2} \mathrm{x}$

D
$\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\hat{\mathrm{j}} \mathrm{x}^{2}+\hat{\mathrm{k}} 3 \mathrm{xz}^{2}$

Solution

The electric potential at a point,
$$
\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{xz}^{3}+4
$$
The field $\overrightarrow{\mathrm{E}}=-\vec{\nabla} \mathrm{V}=-\left(\frac{\partial \mathrm{v}}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial \mathrm{v}}{\partial \mathrm{z}} \widehat{\mathrm{k}}\right)$
$\therefore \overrightarrow{\mathbf{E}}=\overrightarrow{\mathbf{i}}\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\hat{\mathrm{j}} \mathrm{x}^{2}+\widehat{\mathrm{k}}\left(3 \mathrm{xz}^{2}\right)$
Hence, the correct option is (D)