Physics Test

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Physics Test - 31

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Question 1
1 / -0

The electron in a hydrogen atom makes a transition $\mathrm{n}_{1} \rightarrow \mathrm{n}_{2},$ where $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ are the principal quantum numbers of the two energy states. Assume Bohr's model to be valid. The time period of the electron in the initial state is eight times that in the final state. What are the possible values of $\mathrm{n}_{1}$ and $\mathrm{n}_{2} ?$

A
$n_{1}=8, n_{2}=1$

B
$n_{1}=4, n_{2}=2$

C
$n_{1}=2, n_{2}=4$

D
$n_{1}=1, n_{2}=8$

Solution

In a hydrogen atom the time period is given by
$T \propto n^{3}$
Here, $\mathrm{Z}=1$ (Hydrogen atom)
We can derive the expression for the time period of electron in $\mathrm{n}^{\mathrm{th}}$ orbit as
$\mathrm{T}_{\mathrm{n}}=\frac{1}{\mathrm{f}_{\mathrm{n}}} \propto \mathrm{n}^{3}$
Thus,
$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{n}_{1}^{3}}{\mathrm{n}_{2}^{3}}$.
As $\mathrm{T}_{1}=8 \mathrm{~T}_{2},$ therefore we get
$8=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3} \Rightarrow \mathrm{n}_{1}=2 \mathrm{n}_{2}$
Thus, the possible values of $n_{1}$ and $n_{2}$ are $\mathrm{n}_{1}=2, \mathrm{n}_{2}=1, \mathrm{n}_{1}=4, \mathrm{n}_{2}=2, \mathrm{n}_{1}=6, \mathrm{n}_{2}=3$ and so on.
Question 2
1 / -0

The correct Boolean operation represented by the circuit diagram drawn is :

A
AND

B
OR

C
NAND

D
NOR

Solution

From the given logic circuit LED will glow, when voltage across LED is high. This is out put of NAND gate.
Question 3
1 / -0

Two different metals are joined end to end. One end is kept at constant temperature and the other end is heated to a very high temperature. The high depicting the thermo emf is:

A

B

C

D

Solution

$E=\alpha t+\frac{1}{2} \beta t^{2},$

The above equation is in the standard form of the parabola. According to this, the graph will be a parabola, such that first emf increases and then decreases.

The graph between $E$ and $t$:
Question 4
1 / -0

A toroidal solenoid has $3000$ turns and a mean radius of $10 c m$ . It has a soft iron core of relative permeability $2000$ . What is the magnitude of the magnetic field in the core, when a current of $1 A$ is passed through the solenoid?

A
$0.012 T$

B
$0.12 T$

C
$1.2 T$

D
$12 T$

Solution

Given,
$N = 3000$
$r = 10$ cm $= 0.1$ m
$n = \frac{N}{2 πr}$
$μ_{r} = 2000$
$I = 1 A$
For a solenoid magnetic field, $B = μ n I = μ_{r} n_{0} n I$
$B = 2000 \times 4 π \times 10^{- 7} \times \frac{N}{2 πr} \times 1$
$= 2000 \times 4 \times 10^{- 7} \times \frac{3000}{2 π \times 0.1} \times 1 = 12 T$
Question 5
1 / -0

A board is balanced on a rough horizontal semi-circular log. Equilibrium is obtained with the help of addition of a weight to one of the ends of the board when the board makes an angle θ with the horizontal. Coefficient of friction between the log and the board is:

A
$\tan \theta$

B
$\cos \theta$

C
$\cot \theta$

D
$\sin \theta$

Solution

Free body diagram of the system when the weight is at one of the end of board,

Now, as the net force on board is zero.

Net horizontal force $= 0$

$f \cos θ – N sin θ = 0$

$f = N tan θ...(1)$

As we know, frictional force is

$f = μN...(2)$

From equation (1) and (2), we get

$\mu=\tan \theta$
Question 6
1 / -0

A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$. Four small particles each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

A
$\frac{M \omega }{(M+4 m)}$

B
$\frac{{M} \omega }{ 4 {m}}$

C
$\frac{(M+4 m)\omega} { M}$

D
$\frac{M \omega }{(M-4 m)}$

Solution

According to the relation between inertia moment and angular velocity,
$I \omega=$ constant
$I_{1} \omega_{1}=I_{2} \omega_{2}$
$\Rightarrow M R^{2} \omega=(M+4 m) R^{2} \omega_{2}$
$\Rightarrow \omega_{2} = \frac{M \omega }{(M+4 m)}$
Question 7
1 / -0

If momentum $ (p) $, area $ (A) $ and time $ (T) $ are taken to be fundamental quantities, then energy has the dimensional formula:

A
$\left[p A^{-1} T^{1}\right]$

B
$\left[p^2 A T\right]$

C
$\left[p A^{-1} T\right]$

D
$\left[p A^{\frac 12} T^{-1}\right]$

Solution

According to the problem, fundamental quantities are momentum $ (p) $, area $ (A) $ and time $ (T) $ and we have to express energy in these fundamental quantities.
Let energy $E$,
$E \propto p^{a} A^{b} T^{c}$
$\Rightarrow E=k p^{a} A^{b} T^{c}$.....(i)
Where, $k$ is dimensionless constant of proportionality.
Dimensional formula of energy,$[E]=\left[M L^{2} T^{-2}\right]$
$[p]=\left[M L T^{-1}\right]$
$[A]=\left[L^{2}\right]$
$[T]=[T]$
Putting all the dimensions in equation (i), we get
$M L^{2} T^{-2}=\left[M L T^{-1}\right]^{a}\left[L^{2}\right]^{b}[T]^{c}$
$=M^{a} L^{a+2 b} T^{a+c}$
According to the principle of homogeneity of dimensions, we get
$a=1$....(ii)
$a+2 b=2$...(iii)
$-a+c=-2$...(iv)
By solving these equations, we get
$a=1, b=\frac{1}{2}, c=-1$
Dimensional formula for $E$ $=\left[p A^{\frac 12} T^{-1}\right]$
Question 8
1 / -0

If a box of mass 25 kg is pushed 15 m by a force of ‘F’ N and work done in the process is 480 J. Find F:

A
50

B
32

C
16

D
25

Solution

Given that:
Mass (m) = 25 kg
Distance (s) = 15 m
Work (W) = 480 J
Force = F
Using equation:
Work $=$ Force $\times$ Distance
$\therefore 480=\mathrm{F} \times 15$
$\mathrm{F}=\frac{480}{15}=32 \mathrm{~N}$
F $=3 2 \mathrm{~N}$
Question 9
1 / -0

Diagram shows a bar magnet and two infinite long wires W₁ and W₂ carrying equal currents in opposite directions. The magnet is free to move and rotate. P is the mid-point of magnet. For this situation mark out the correct statement(s).

A
Magnet experiences a net torque in clockwise direction and zero net force.

B
Magnet experiences a net force towards left and a net torque in anti-clockwise direction.

C
Magnet experiences a net force towards right and a net torque in anti-clockwise direction.

D
Magnet experiences zero net force and a net torque in anti-clockwise direction.

Solution

Both the poles of the magnet experience a force due to magnetic field produced by wires W₁ and W₂

North pole experiences a force $\mathrm{F}_{1}$ due to $\mathrm{W}_{1}$, and $\mathrm{F}_{2}$ due to $\mathrm{W}_{2}$ shown in diagram. Similarly, south pole experience a force

$\mathrm{F}_{1}$ due to $\mathrm{W}_{2}$ and $\mathrm{F}_{2}$ due to $\mathrm{W}_{1}$.

From free body diagram of magnet, it is clear that magnet experiences a net force towards right and a torque in anti- clockwise direction.
Question 10
1 / -0

If number of neutrons become more than the number of electrons in the element then it will become:

A
Positively charged

B
Negatively charged

C
Neutral

D
Can't say

Solution

Positive charge: A body having a deficiency of electrons and the number of protons becomes more than the number of electrons.
Negative charge: A body having an excess of electrons and the number of electrons becomes more than the number of protons.
Neutron is not responsible for the charge on the element.
The number of neutrons becomes more than the number of electrons in the element, by this statement, we can't predict whether the number of protons is more than, less than, or equal to the number of electrons in the element.
So, we can't predict the type of charge on the element.