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Physics Test - 16

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Weekly Quiz Competition

Question 1
4 / -1

When a system is taken from state i to state f along the path iaf, it is found that Q = 50 cal and W = 20 cal . Along the path ibf Q = 36 cal. W along the path ibf is:

A
6 cal

B
16 cal

C
66 cal

D
14 cal

Solution

According to the First Law of the thermodynamics,
$\Delta Q=\Delta U+W$
For the path iaf, $\Delta Q=50$ cal $, W=20$ cal
Change in the internal energy $\Delta U=\Delta Q-W=50-20=30$ cal
For the path ibf', change in the internal energy will be same as for the path iaf' as the initial and the final states are same.
Hence work done for the path ibf is
$W=\Delta Q-\Delta U=36-30=6 \mathrm{cal}$
Question 2
4 / -1

An engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62oC, its efficiency is doubled. The temperature of the source is

A
$124 ° C$

B
$37 ° C$

C
$62 ° C$

D
$99 ° C$

Solution

Maximum efficiency of an engine working between temperatures $T_{2}$ and $T_{1}$ is given by the
fraction of the heat absorbed by the engine, which can be converted into work. Mathematically, Efficiency, $\eta=\left(T_{2}-T_{1}\right) / T_{2}=1 / 6$
$6 \mathrm{T}_{2}-6 \mathrm{T}_{1}=\mathrm{T}_{2}$
Therefore, $T_{2}=1.2 T_{1} \ldots \ldots \ldots \ldots \ldots \ldots . .(1)$
Where $T_{2}$ is the source temperature and $T_{1}$ is the sink temperature
If the sink temperature $\left(\mathrm{T}_{1}\right)$ is reduced by $62^{\circ} \mathrm{C}$, then the efficiency gets doubled, i.e.
$\eta=T_{2}-\left(T_{1}-62\right) / T_{2}=2 \times 1 / 6=2 / 6$
$6 \mathrm{T}_{2}-6 \mathrm{T}_{1}+372=2 \mathrm{T}_{2}$
$6 \mathrm{T}_{2}-2 \mathrm{T}_{2}-6 \mathrm{T}_{1}+372=0$
$4 T_{2}-6 T_{1}+372=0$
Substituting the value of $T_{2}$ from equation $(1),$ we get
$4\left(1.2 T_{1}\right)-6 T_{1}+372=0$
$4.8 \mathrm{T}_{1}-6 \mathrm{T}_{1}+372=0$
$372=1.2 \mathrm{T}_{1}$
$\mathrm{T}_{1}=310 \mathrm{K}$
Therefore, $T_{2}=1.2 \times 310=372 \mathrm{K}$
Hence, the temperatures of the source and the sink are $372 \mathrm{K}$ and $310 \mathrm{K}$, respectively. Source temperature $=372-273=99^{\circ} \mathrm{C}$
Question 3
4 / -1

A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was R, then the resistance of the bundle will be

A
$\frac{R}{4}$

B
$\frac{R}{8}$

C
$\frac{R}{16}$

D
$\frac{R}{32}$

Solution

Resistance of the wire is proportional to the length of the wire, $R \propto l$.
If the length of the wire is reduced by four times, then the resistance will also reduce by four
times.

Hence, the resistance of each part will be $\frac{R}{4}$. When these four parts are connected in parallel, the equivalent resistance will be $\mathrm{R}^{\prime}$.
$\frac{1}{R^{\prime}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}=\frac{16}{R}$
$R^{\prime}=\frac{R}{16}$
Question 4
4 / -1

Equation of a transverse wave travelling in a rope is given by $y=5 \sin (4 t-$ $0.02 \mathrm{x})$ where $\mathrm{y}$ and $\mathrm{x}$ are expressed in $\mathrm{cm}$ and time in seconds. Calculatevelocity of the wave.

A
200 cm/s

B
400 cm/s

C
600 cm/s

D
None of these

Solution

Given:
$y=5 \sin (4 t-$ $0.02 \mathrm{x})$
Comparing this with the standard equation of wave motion:
$y=A \sin \left(2 \pi f t-\frac{2 \pi}{\lambda} \mathrm x\right)$
Where $\mathrm{A}, \mathrm{f}$ and $\lambda$ are amplitude, frequency and wavelength respectively.
Thus, amplitude:
$ A =5 \mathrm{~cm} $
$ 2 \pi f =4 $
$ \text { Frequency } f =\frac{4}{2 \pi}=0.637 \mathrm{~Hz}$
$ \text { Again }: \frac{2 \pi}{\lambda} =0.02$
$ \text { Wavelength } \lambda =\frac{2 \pi}{0.02}=100 \pi \mathrm{cm} $
Velocity of the wave:
$v=f \lambda=\frac{4}{2 \pi} \frac{2 \pi}{0.02}=200 \mathrm{~cm} / \mathrm{s}$
Question 5
4 / -1

In an arrangement of resistances shown in the diagram, the potential difference between B and D will be zero, when the unknown resistance X is

A
4 ohm

B
2 ohm

C
3 ohm

D
None of these

Solution

For the potential difference across $\mathrm{B}$ and $\mathrm{D}$ to be zero
$\frac{x}{16}=\frac{0.5}{4}$
$x=2 \Omega$
Question 6
4 / -1

A uniform ladder rest in limiting equilibrium with its lower end on a rough horizontal plane and its upper end against a smooth vertical wall, if $\theta$ is an angle of inclination of the ladder to the vertical wall and $\mu$ is the coefficient of friction, then tan $\theta$ is equal to

A
$\mu$

B
1/2 $μ$

C
$\frac{3 μ}{2}$

D
$μ$ + 1

Solution

$\mathrm{mg}=\mathrm{N}_{1}-\mathrm{ci}$
$\mathrm{f}=\mu \mathrm{N}_{1}=\mathrm{N}_{2}=\ldots-(\mathrm{ii})$
$\mu \mathrm{mg}=\mathrm{N}_{2}$
$\tan \theta=\mathrm{y} / \mathrm{x}$
$\mathrm{N}_{2} \mathrm{y}=\mathrm{mg}(\mathrm{x} / 2)$
$\Rightarrow \mu \mathrm{mg} \mathrm{y}=\mathrm{mg}(\mathrm{x} / 2)$
$\Rightarrow \mathrm{y} / \mathrm{x}=1 / 2 \mu$
$\tan \theta=1 / 2 \mu$
Question 7
4 / -1

The pressure and density of a diatomic gas $\left(Y=\frac{7}{5}\right)$ change adiabatically from $\left(P_{1}, \rho_{1}\right)_{\text {to }}\left(P_{2}, \rho_{2}\right),$ if $\frac{\rho_{2}}{\rho_{1}}=32,$ then what should be the value of $\frac{P_{2}}{P_{1}} ?$

A
16

B
32

C
64

D
128

Solution
Question 8
4 / -1

The engine of a train crosses an electric pole with velocity 'u', and the last compartment of the train crosses the same pole with a velocity 'v'. The velocity with which the middle part of the train crosses the pole is

A
u

B
v

C
$\frac{(u+v)}{2}$

D
$\sqrt{\frac{\left(u^{2}+v^{2}\right)}{2}}$

Solution

Let $2 L$ be the length of the train. The distance of middle part of the train from engine is $L$. Let 'a' be the acceleration of the train. If $\mathrm{v}^{\prime}$ is the velocity with which the middle part of the train crosses the pole, then for the first half, we have
$v^{\prime 2}-u^{2}=2 a L \quad-\ldots-A$
and for the second half, we have
$v^{2}-v^{\prime 2}=2 a L \quad-\ldots$
From $\mathrm{A}$ and $\mathrm{B}$ :
$v^{\prime 2}-u^{2}=v^{2}-v^{\prime 2}$
$2 v^{\prime 2}=u^{2}+v^{2} \Rightarrow v^{\prime}=\sqrt{\frac{\left(u^{2}+v^{2}\right)}{2}}$
Question 9
4 / -1

Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and a negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc, perpendicular to its plane is

A
1 kg m²

B
0.1 kg m²

C
2 kg m²

D
0.2 kg m²

Solution

sum of moment of inertia=5 $^{*} \mathrm{m}^{*}(\mathrm{r})^{2}$
$\begin{array}{ll}\Rightarrow & =5^{*} 2^{*}(0.1)^{2} \\ =>& =0.1 \mathrm{kgm}^{2}\end{array}$
Question 10
4 / -1

Two masses m1and m2are suspended together by a mass-less spring of spring constant K. When the masses are in equilibrium, m1 is removed without disturbing the system.The angular frequency and amplitude of oscillation of m2 are

A
$\sqrt{\frac{K}{m_{2}}} \cdot \frac{m_{1} g}{K}$

B
$\sqrt{\frac{K}{m_{1}}}, \frac{m_{1} g}{K}$

C
$\sqrt{\frac{K}{m_{1}}}, \frac{m_{2} g}{K}$

D
$\sqrt{\frac{K}{m_{2}}}, \frac{m_{2} g}{K}$

Solution

When only the mass $\mathrm{m}_{2}$ is suspended let the elongation of the spring be $\mathrm{x}_{2}$.
So we have $m_{2} g=K x_{1}$ -1
When both the masses $\left(\mathrm{m}_{2}+\mathrm{m}_{2}\right)$ together are suspended, the elongation of the spring is $\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)$.
Thus, we have
$\left(m_{1}+m_{2}\right) g=K\left(x_{1}+x_{2}\right)$
Where $\mathrm{K}$ is the spring constant.
From 1 and 2 we have
$m_{1} g=K x_{2}$
Thus, $x_{2}$ is the elongation of the spring due to the mass $m_{1}$ only. When the mass $m_{1}$ is removed the mass $m_{2}$
executes SHM with the amplitude $x_{2}$
Amplitude of vibration $x_{2}=\frac{m_{1} g}{K}$

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Re-Attempt Weekly Quiz Competition

Physics Test - 16

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Physics Test - 16

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SHARING IS CARING

Question 1

6 cal

16 cal

66 cal

14 cal

Solution

Question 2

124°C

37°C

62°C

99°C

Solution

Question 3

R4

R8

R16

R32

Solution

Question 4

200 cm/s

400 cm/s

600 cm/s

None of these

Solution

Question 5

4 ohm

2 ohm

3 ohm

None of these

Solution

Question 6

\(\mu\)

1/2μ

3μ2

μ+ 1

Solution

Question 7

16

32

64

128

Solution

Question 8

u

v

\(\frac{(u+v)}{2}\)

\(\sqrt{\frac{\left(u^{2}+v^{2}\right)}{2}}\)

Solution

Question 9

1 kg m2

0.1 kg m2

2 kg m2

0.2 kg m2

Solution

Question 10

\(\sqrt{\frac{K}{m_{2}}} \cdot \frac{m_{1} g}{K}\)

\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{1} g}{K}\)

\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{2} g}{K}\)

\(\sqrt{\frac{K}{m_{2}}}, \frac{m_{2} g}{K}\)

Solution

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$124 ° C$

$37 ° C$

$62 ° C$

$99 ° C$

$\frac{R}{4}$

$\frac{R}{8}$

$\frac{R}{16}$

$\frac{R}{32}$

1/2 $μ$

$\frac{3 μ}{2}$

$μ$ + 1

1 kg m²

0.1 kg m²

2 kg m²

0.2 kg m²