Physics Test

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Physics Test - 38

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Question 1
1 / -0

A particle executes S.H.M. given by

$x = 0 \cdot 24 cos (400 t - 0.5)$

in SI units. Find amplitude ?

A
0.24 m

B
0.25 m

C
0.26 m

D
-0.25 m

Solution
Question 2
1 / -0

In the circuit shown in diagram, the equivalent resistance between point A and B is

A
$\frac{3 R}{4}$

B
$\frac{R}{2}$

C
$\frac{5 R}{8}$

D
2 R

Solution

The circuit shown in diagram (1) can be redrawn as shown in diagram (2).

Hence, the correct option is (C)
Question 3
1 / -0

Two heavy metallic plates are joined together at 90° to each other. A laminar sheet of mass 30 kg is hinged at the line AB joining the two heavy metallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its centre of mass is 1.2 kg - m². Two rubber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5 m from the line AB. This distance is chosen, so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacle is 6N-s,

find the location of the centre of mass of the laminar sheet from AB.

A
0.1 m

B
1.2 m

C
2.1 m

D
0.4 m

Solution

Let the perpendicular distance of CM from the line AB be r and $\omega$ the angular velocity of the sheet just after colliding with rubber obstacle for the first time.
Obviously the linear velocity of CM before and after collision will be $\nu_{1}=(r)(1 \mathrm{rad} / \mathrm{s})=\mathrm{r}$ and $\nu_{\mathrm{f}}=\mathrm{r} \omega$
$\overrightarrow{\mathbf{v}_{i}}$ and $\overrightarrow{\mathbf{v}_{f}}$ will be in opposite directions.
Now, linear impulse on $\mathrm{CM}=$ change in linear momentum of $\mathrm{CM}$
or $\quad 6=\mathrm{m}\left(\nu_{\mathrm{f}}+\nu_{\mathrm{i}}\right)=30(\mathrm{r}+\mathrm{r} \omega)$
or $\quad \mathrm{r}(1+\omega)=\frac{1}{5}$
$\ldots$
Similarly, angular impulse about $\mathrm{AB}=$ change in angular momentum about $\mathrm{AB}$
Angular impulse = Linear impulse $\mathrm{x}$ perpendicular distance of impulse from $\mathrm{AB}$
Hence, $6(0.5 \mathrm{m})=\mathrm{I}_{\mathrm{AB}}(\omega+1)$
(Initial angular velocity $=1$ rad/s)
or $3=\left[\mathrm{I}_{\mathrm{CM}}+\mathrm{Mr}^{2}\right](1+\omega)$
or $3=\left[1.2+30 \mathrm{r}^{2}\right](1+\omega)$
Solving Eqs. (i) and (ii) for $r$, we get
$r=0.4 \mathrm{m}$ and $r=0.1 \mathrm{m}$
But at $r=0.4 \mathrm{m}, \omega$ comes out to be negative $(-0.5 \mathrm{rad} / \mathrm{s})$ which is not acceptable. Therefore,
$r=d i$ stance of $C M$ from $A B=0.1 \mathrm{m}$
Hence, the correct option is (A)
Question 4
1 / -0

The energy gap of pure $\mathrm{Si}$ is $1.1 \mathrm{eV}$. The mobilities of electrons and holes are respectively $0.135 \mathrm{m}^{2} \mathrm{V}^{-1} \mathrm{s}^{-1}$ and $0.048 \mathrm{m}^{2} \mathrm{V}^{-1} \mathrm{s}^{-1}$ and can be taken as independent of temperature. The intrinsic carrier concentration is given by

$\mathbf{n}_{1}=\mathbf{n}_{0} \mathbf{e}^{-\mathbf{E}_{\mathbf{q}} / 2 \mathbf{k} \mathbf{T}}$

where $n_{0}$ is a constant, $E_{g}$ the gap width and $k$ the Boltzman's constant whose value is $1.3810^{-23} \mathrm{JK}^{-1}$. Find the ratio of the electrical conductivities of Si at $600 \mathrm{K}$ and $300 \mathrm{K}$

A
$\mathrm{e}^{10.63} \approx 4 \times 10^{4}$

B

C

D

Solution

The total electrical conductivity of a semiconductor is given by
$$
\sigma=e\left(n_{\theta} \mu_{e}+n_{n} \mu_{n}\right)
$$
For an intrinsic semiconductor,
$$
\mathrm{n}_{\mathrm{e}}=n_{\mathrm{h}}=n_{\mathrm{i}}
$$
We can thus write for the conductivity
$$
\sigma=e\left(\mu_{e}+\mu_{h}\right) n_{i}
$$
or $\sigma=\mathrm{e}\left(\mu_{\mathrm{e}}+\mu_{\mathrm{h}}\right) \mathrm{n}_{0} \mathrm{e}^{-\mathrm{E}_{\mathrm{k}} / 2 \mathrm{k} \mathrm{T}}$
As the mobilities $\mu_{e}$ and $\mu_{n}$ are independent of temperature, they can be regarded as constant. The ratio of the conductivities at $600 \mathrm{K}$ and $300 \mathrm{K}$ is then
As per given data, $E g=1.1 \mathrm{eV}$
and
$$
\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1} \text {or }\left(\frac{1.38 \times 10^{-23}}{1.6 \times 10^{-19}}\right) \mathrm{eVK}^{-1}
$$
$\therefore \quad \mathrm{k}=8.625 \times 10^{-5} \mathrm{eVK}^{-1}$
We thus have,
$$
\begin{aligned}
\frac{\sigma_{600}}{\sigma_{300}} &=\mathrm{e}^{1.1 /\left(1200 \times 8.625 \times 10^{-5}\right)} \\
& \approx \mathrm{e}^{10.63} \approx 4 \times 10^{4}
\end{aligned}
$$
Hence, the correct option is (A)
Question 5
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A vessel has the shape shown in figure. Water, which has density 10³kg/m³, is filled in the vessel. The pressure at the point A, ignoring the atmospheric pressure, is:

(Take g=10m/s²)

A
1.2 × 10⁴N/m²

B
1.0 × 10⁴N/m²

C
2.2 × 10⁴N/m²

D
2.4 × 10⁴N/m²

Solution

Pressure at A will be due to water height (1.0 + 1.0 + 0.2) m

∴ PA = ρgh = 10³ × 10 × 2.2 = 2.2 × 10⁴Nm⁻²
Hence, the correct option is (C)
Question 6
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Equal volume of two immiscible liquids of densities ρ and 2ρ are filled in a vessel as shown in figure. Two small holes are made at depth h/2 and 3h/2 from the surface of lighter liquid. If v₁ and v₂ are the velocities of efflux at these two holes, then v₁/v₂ will be:

A
1/√2

B
1/2√2

C
1/2

D
1/4

Solution
Question 7
1 / -0

When an air bubble moves up from the bottom of a lake ?

a) velocity decreases and becomes zero

b) acceleration increases and becomes zero

c) velocity increases and becomes constant

d) acceleration decreases and becomes zero

A
a, d are correct

B
a, b are correct

C
c, d are correct

D
c is correct

Solution

As the bubble rises up, the net forces acting on it are given by$m a=V \rho g-m g-f_{r}$
$\Longrightarrow m a=V \rho g-V d g-6 \pi \eta r v$
$\Longrightarrow a=\frac{d v}{d t}=\frac{1}{m}(V(\rho-d) g-6 \pi \eta r v)$
Hence the speed of bubble continuously rises. However as the speed increases, the frictional force increases and hence the
acceleration decreases.
Hence, the correct option is (C)
Question 8
1 / -0

If a body floats with (m/n)^th of its volume above the surface of water, then the relative density of the material of the body is:

A
(n - m)/n

B
m/n

C
n/m

D
(n - m)/m

Solution

As body floats with $\left(\frac{m}{n}\right)^{t h}$ volume above. Volume of liquid displaced $=\left[1-\left(\frac{m}{n}\right)\right] V$
$\therefore$ weight of cube $=$ weight of displaced liquid $\Rightarrow V \rho_{s}=\left[1-\left(\frac{m}{n}\right)\right] V \rho$
$\frac{\rho_{s}}{\rho}=\left[1-\left(\frac{m}{n}\right)\right]=\left[\frac{n-m}{n}\right]$
Hence, the correct option is (A)
Question 9
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Calculate the velocity of efflux of kerosene oil from an orifice of a tank in which pressure is 4atm. The density of kerosene oil = 720kgm⁻³and 1 atmospheric pressure = 1.013 × 10⁵Nm⁻².

A
3.355m/s

B
33.55m/s

C
335.5m/s

D
33.55cm/s

Solution

From Bernoulli's Theorem, $P_{1}=\frac{1}{2} \rho v^{2}$
$\Rightarrow 4 a t m=\frac{1}{2} \times \rho V^{2}$
$\Rightarrow \frac{4 \times 1.013 \times 10^{5} \times 2}{720}=V^{2}$
$\Rightarrow V=33.549 \mathrm{m} / \mathrm{s}$
$\Rightarrow V \approx 33.55 m / s$
Hence, the correct option is (B)
Question 10
1 / -0

Two sinusoidal plane waves of same frequencies having intensities I₀ and 4I₀ are travelling in the same direction . The resultant intensity at a point at which waves meet with a phase difference of zero radian is :

A
I₀

B
5I₀

C
9I₀

D
3I₀

Solution

When two waves of same frequencies meet at a point with no phase difference, then the resultant amplitude is the sum of the individual ones. Intensity is proportional to the square of the amplitude. Hence $I_{0}=k A_{1}^{2}$ and
$4 I_{0}=k A_{2}^{2}$ where $\mathrm{k}$ is the constant of proportionality. Hence $A_{1}=\sqrt{k^{-1} I_{0}}$ and
$A_{2}=2 \sqrt{k^{-1} I_{0}}$
Resultant amplitude is $A_{R}=A_{1}+A_{2}=3 \sqrt{k^{-1} I_{0}}$
Resultant intensity $9 I_{0}$
Hence, the correct option is (C)