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Physics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Half-life of a radioactive substance A is 4 days. The probability of a nucleus that, from the given sample that it will decay in two half-lives is

A
1/4

B
3/4

C
1/2

D
1

Solution

After two half-lives 1/4 th fraction of nuclei will remain undecayed. Or, 3/4 th fraction will decay. Hence, the probability that a nucleus decays in two half lives is 3/4
Question 2
1 / -0

A block of mass m is connected rigidly with a smooth wedge (plank) by a light spring of stiffness k. If the wedge is moved with constant velocity v0, find the work done by the external agent till the maximum compression of the spring.

A
m v²₀

B
2m v²₀

C
3m v²₀

D
2m v³₀

Solution

Let us take wedge $+$ spring $+$ block as a system. The forces responsible for performing work are spring force $\mathrm{kx}(\leftarrow)$ and the external force $\mathrm{F}(\rightarrow)$.

Work Energy theorem for block + plank relative to ground :
Applying work-energy theorem, we have $\mathrm{W}_{\mathrm{ext}}+\mathrm{W}_{\mathrm{sp}}=\Delta \mathrm{K}$
where $\mathrm{W}_{\mathrm{sp}}$ the total work done by the spring on wedge and block $-\frac{1}{2} \mathrm{k} x^{2}$ and $\Delta \mathrm{K}=$ change in $\mathrm{KE}$ of the block (because the plank does not change its kinetic energy)
Then, $\quad \mathrm{W}_{\text {ext }}=\frac{1}{2} \mathrm{k} x^{2}+\Delta \mathrm{K}$
As the block was initially stationary and it will acquire a velocity $v_{0}$ equal to that of the plank at the time of maximum compression of the spring, the change in kinetic energy of the block relative to ground is
$$
\Delta \mathrm{K}=\frac{1}{2} m v_{0}^{2}
$$
Substituting $\Delta \mathrm{K}$ in the above equation, we have
$$
\mathbf{W}_{\text {ext }}=\frac{1}{2} \mathrm{K} x^{2}+\frac{1}{2} m v_{0}^{2}
$$
W-E theorem for block $+$ plank relative to the plank $W_{e x t}+W_{s p}=D K$ The plank moves with constant velocity, there is no pseudo-force acting on the block. $W_{e x t}=O$ Then the net work done on the system (block + plank), due to the spring, can be given as
$$
\mathbf{W}_{\mathrm{SP}}=-\frac{1}{2} \mathrm{k} x^{2}
$$
As the relative velocity between the observer (plank) and block decreases from $\mathrm{v}_{0}$ to zero at the time of maximum compression of the spring, the change in kinetic energy of the block is $\Delta \mathrm{K}=-\frac{1}{2} m v_{0}^{2}$
Substituting $\mathrm{W}_{\text {sp }}$ and $\stackrel{\Delta K}{\longleftrightarrow}$ in above equation
$-\frac{1}{2} K X^{2}=-\frac{1}{2} m v_{o}^{2} \ldots \ldots$
From $(i) \&(i i)$
$W e x t=\frac{1}{2} m v_{o}^{2}+\frac{1}{2} m v_{o}^{2}=m v_{o}^{2}$
Question 3
1 / -0

Three capacitors each of capacity 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by

A
Connecting all of them in series

B
Connecting all of them in parallel

C
Connecting two in series and one in parallel with the series combination

D
Connecting two in parallel and one in series with the parallel combination

Solution

To get equivalent capacitance $6 \mu \mathrm{F}$. Out of the $4 \mu \mathrm{F}$ capacitance, two are connected in series and third one is connected in parallel.

$\mathrm{C}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}+4=2+4=6 \mu \mathrm{F}$
Question 4
1 / -0

An elevator is going upward with an acceleration a = g/4 and a ball is released from rest relative to the elevator at a distance h₁ above the floor. The speed of the elevator at the time of ball release is v₀. Then the bounce height h₂ of the ball with respect to the elevator is (the coefficient of restitution for the impact is e)

A
eh₁

B
e² h₁

C
e ³ h¹

D
None of these

Solution

$u_{\mathrm{rel}}=0$ and $a_{\mathrm{rel}}=g+\frac{g}{4}=\frac{5 g}{4}$
$\left(\mathrm{v}_{\mathrm{b} / \mathrm{e}}\right)^{2}=2\left(\frac{5 g}{4}\right) h_{1}$
(just before collision)
$\mathbf{v}_{b / e}^{\prime}=\mathbf{e v}_{b / e}$
(just after collision)
$\Rightarrow \quad \mathrm{e}^{2}\left(\mathrm{v}_{b / e}\right)^{2}=2\left(\frac{5 g}{4}\right) h_{2} \Rightarrow \quad h_{2}=\mathrm{e}^{2} h_{1}$
$\therefore \quad e^{2} h_{1}$
Question 5
1 / -0

Three point charges Q, +q and +q are placed at the vertices of a right angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero, if Q is equal to

A
$\frac{-q}{1+\sqrt{2}}$

B

$\frac{-\sqrt{2} q}{1+\sqrt{2}}$

C
$-2q$

D
$q$

Solution

Potential energy of the system $U=k \frac{Q_{1} Q_{2}}{r}$

Where, $k=\frac{1}{4 \pi \varepsilon_{0}}$
Potential energy of the configuration $U=k \frac{Q q}{a}+k \frac{q^{2}}{a}+k \frac{q Q}{a \sqrt{2}}=0$
$Q=\frac{-\sqrt{2} q}{1+\sqrt{2}}$
Hence, $Q$ is equal to $\frac{-\sqrt{2} q}{1+\sqrt{2}}$
Question 6
1 / -0

A particle is projected vertically upwards with a velocity u, from a point O. When it returns to the point of projection, which of the following is incorrect ?

A
Its average velocity is zero

B
Its displacement is zero

C
Its average speed is u

D
Its average speed is u/2

Solution

Total displacement is 0 as the particle returns to the original position, thus the average velocity is zero.

Total distance travelled is 2 s and total time taken is 2t.
Question 7
1 / -0

An inverted bell, lying at the bottom of lake 47.6 m deep, has 50 cm³ of air trapped in it. The bell is brought to the surface of lake. The volume of the trapped air will become (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm³)

A
350 cm³

B
300 cm³

C
250 cm³

D
22 cm³

Solution

According to Boyle's law, pressure and volume are inversely proportional to each other i.e. $\mathrm{p} \propto \frac{1}{\mathrm{v}}$

$\Rightarrow \mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$
$\Rightarrow\left(\mathrm{P}_{0}+\mathrm{h} \rho_{\mathrm{w}} \mathrm{g}\right) \mathrm{V}_{1}=\mathrm{P}_{0} \mathrm{V}_{2}$
$\Rightarrow \mathrm{V}_{2}=\left(1+\frac{\mathrm{h} \rho_{\mathrm{w}} \mathrm{g}}{\mathrm{P}_{0}}\right) \mathrm{V}_{1}$
$V_{2}=\left(1+\frac{47.6^{×}+1000^{×}+10}{70^{×} 10^{-2}+13.6^{×} 1000^{×} 10}\right)$
$\left[A s P_{2}=P_{0}=70 \mathrm{cm}\right.$ of $\left.\mathrm{Hg}=70^{×} 10^{-2} × 13.6^{×} 1000^{*} 10\right]$
$\Rightarrow \mathrm{V}_{2}=(1+5) 50 \mathrm{cm}^{3}=300 \mathrm{cm}^{3}$
Question 8
1 / -0

An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV. The kinetic energy is E and its de Broglie wavelength is λλ. Then

A
E=6.8 eV, λ=6.6×10⁻¹⁰m

B
E=3.4 eV, λ=6.6×10⁻¹⁰m

C
E=3.4 eV, λ=6.6×10⁻¹¹m

D
E=6.8 eV, λ=6.6×10⁻¹¹m

Solution

$K . E .=\frac{K Z e^{2}}{2 r}$
$P . E .=\frac{-K Z e^{2}}{r}$
$T . E .=P . E .+K . E .=\frac{-K Z e^{2}}{2 R}$
Therefore, $T E=-K E=\frac{P E}{2}=-3.4 \mathrm{ev}$
So, $K E=3.14 \mathrm{ev}$
Let $p=$ momentum and $m=$ mass of the electron.
$\therefore \mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$ or $\mathrm{p}=\sqrt{2 \mathrm{m} \mathrm{E}}$
de Broglie wavelength,
$\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
On substituting the values, we get
$\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}}$
$=6.6 \times 10^{-10} \mathrm{m}$
Question 9
1 / -0

A clock, which keeps correct time at 25^oC, has a pendulum made of brass. The coefficient of linear expansion for brass is 0.000019 ^oC-1. How many seconds a day will it gain if the ambient temperature falls to 0^oC ?

A
20.52 sec

B
15.00 sec

C
52.10 sec

D
63.10 sec

Solution

Let $L_{0}$ and $L_{25}$ be the length of pendulum at $0^{\circ} \mathrm{C}$ and $25^{\circ} \mathrm{C}$ respectivel
We know that
$$
\begin{array}{l}
L_{25}=L_{0}(1+a T)=L_{0}(1+0.000019 \times 25)=1.000475 L_{0} \\
\Rightarrow \Delta L=L_{25}-L_{0}=0.000475 L_{0}
\end{array}
$$
Also, $T=2 \pi \sqrt{\frac{L}{g}}$
From method of errors
$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta L}{L}$
here, $\Delta T=$ gain in time
$$
\begin{array}{l}
T=\text {Total time of day} \\
=24 \times 60 \times 60 \mathrm{sec}
\end{array}
$$
$\Rightarrow \Delta T=\frac{1}{2}\left(\frac{0.000475 L_{0}}{L_{0}}\right) \times 24 \times 60 \times 60 \mathrm{sec}$
$\Rightarrow \Delta T=20.52 \mathrm{sec}$
Question 10
1 / -0

Two blocks each of mass m, lie on a smooth table. They are attached to two other masses as shown in figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces with respect to each other is

A
5g/6

B
5g/3

C
17g/12

D
17g/6

Solution

$\vec{a}_{A B}=\frac{3 m g(-i)}{m+3 m}=\frac{3 g}{4}(-\hat{i})$
$\vec{a}_{C D}=\frac{2 m g(-i)}{m+2 m}=\frac{2 g}{3}(\hat{i})$
In a plane mirror the image moves twice as faster as mirror, thus
acceleration of image in mirror $\mathrm{AB}=\mathrm{a} \rightarrow \mathrm{AB}$
acceleration of image in mirror $\mathrm{CD}=\mathrm{a} \rightarrow \mathrm{cd}$
acceleration of images w.r.t. each other is
$\vec{a}_{x l}=2 \vec{a}_{C D}-2 \vec{a}_{A B}$
$=2\left(\frac{2 g}{3}\right) \hat{i}-2\left(\frac{3 g}{4}\right) \quad(-\hat{i})$
$=\frac{17 g}{6} \hat{i}$

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Physics Test - 21

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Physics Test - 21

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SHARING IS CARING

Question 1

1/4

3/4

1/2

1

Solution

Question 2

m v20

2m v20

3m v20

2m v30

Solution

Question 3

Connecting all of them in series

Connecting all of them in parallel

Connecting two in series and one in parallel with the series combination

Connecting two in parallel and one in series with the parallel combination

Solution

Question 4

eh1

e2 h1

e 3 h1

None of these

Solution

Question 5

\(\frac{-q}{1+\sqrt{2}}\)

\(\frac{-\sqrt{2} q}{1+\sqrt{2}}\)

\(-2q\)

\(q\)

Solution

Question 6

Its average velocity is zero

Its displacement is zero

Its average speed is u

Its average speed is u/2

Solution

Question 7

350 cm3

300 cm3

250 cm3

22 cm3

Solution

Question 8

E=6.8 eV, λ=6.6×10−10m

E=3.4 eV, λ=6.6×10−10m

E=3.4 eV, λ=6.6×10−11m

E=6.8 eV, λ=6.6×10−11m

Solution

Question 9

20.52 sec

15.00 sec

52.10 sec

63.10 sec

Solution

Question 10

5g/6

5g/3

17g/12

17g/6

Solution

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m v²₀

2m v²₀

3m v²₀

2m v³₀

eh₁

e² h₁

e ³ h¹

350 cm³

300 cm³

250 cm³

22 cm³

E=6.8 eV, λ=6.6×10⁻¹⁰m

E=3.4 eV, λ=6.6×10⁻¹⁰m

E=3.4 eV, λ=6.6×10⁻¹¹m

E=6.8 eV, λ=6.6×10⁻¹¹m