Tag: physics

Questions Related to physics

Assertion : In damped oscillations, the energy of the system is dissipated continuously.
Reason : For the small damping, the oscillations remain approximately periodic.

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.

Show answer
Correct Option: B
Explanation:

In damped oscillation like the motion of simple pendulum swinging in air, motion dies out eventually. This is because of the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually.

Assertion : In forced oscillations, the steady state motion of the particle is simple harmonic.
Reason : Then the frequency of particle after the free oscillations die out, is the natural frequency of the particle.

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.

Show answer
Correct Option: C
Explanation:

In forced oscillations, the frequency of particle after free oscillations die out, is the frequency of the driving force, not the natural frequency of the particle.

A force F= -4x-8 is acting on a block where x is position of block in meter. The energy of oscillation is 32 J, the block oscillate between two points Position of extreme position is:

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. 6

  2. 0

  3. 4

  4. 3

Show answer
Correct Option: A
Explanation:

$F=-4x-8\F=-4(x+2)\ \Rightarrow F=-4[x-(-2)]\F=-kx\rightarrow$ Distance mean position. So, this will be SHM

with $x=-2$ as mean position 
as $K _SHM=4$
Energy of SHM$=\cfrac{1}{2}KA^2=32\ \Rightarrow A^2=\cfrac{2\times32}{K}=\cfrac{2\times32}{4}=16\Rightarrow A=4$
and mean position $=-2$
Extereme position $=(-2+4)\quad and (-2-4)\=2\quad and\quad -6$

A block of $0.5\ kg$ is placed on a horizontal platform. The system is making vertical oscillations about a fixed point with a frequency of $0.5\ Hz.$ Find the maximum amplitude of oscillation if the block is not to lose contact with the horizontal platform?

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $0.6542\ m$

  2. $0.9927\ m$

  3. $0.7428\ m$

  4. $0.852\ m$

Show answer
Correct Option: D

A highly rigid cubical block A of small mass M and side L is fixed rigidly on another cubical block B of the same dimensions and of low modulus of rigidity $\eta $ such that the lower face of A completely covers the upper face of B.  The lower face of B is rigidly held on horizontal surface.  A small force is applied perpendicular to the side faces of A.  After the force is withdrawn, block A executes small oscillations the time period of which is given by 

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $2\pi \sqrt{M\eta L}$

  2. $2\pi \sqrt{\frac{M-\eta }{L}}$

  3. $2\pi \sqrt{\frac{M-L}{\eta }}$

  4. $2\pi \sqrt{\frac{M-N}{\eta L}}$

Show answer
Correct Option: A

In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is $2 : 3$, the balance point from left is?

physics current electricity meter bridge meter bridge and problems on it galvanometer

  1. $60$cm

  2. $50$cm

  3. $40$cm

  4. $20$cm

Show answer
Correct Option: C
Explanation:

Let $X$ is the left gap resistance and $R$ is the right gap resistance

$l _1$ be the balance point from left

From meter Bridge principle:-
$\implies \dfrac XR= \dfrac{l _1}{100-l _1}=\dfrac23$

$\implies 200-2l _1= 3l _1 \implies 200=5l _1$

$l _1= 40\ cm $

Hence option $(C)$ is correct

In a meter bridge, a standard resistor of R ohm is connected in the left gap and two wires A and B are connected one after the other in the right gap. The balancing length measured from the left is 50 cm for either of them. If the two wires are connected is series and put in the right gap, the balancing length measured from the left would be (in cm)

physics current electricity meter bridge meter bridge and problems on it galvanometer

  1. 25

  2. 33.3

  3. 66.7

  4. 75

Show answer
Correct Option: B
Explanation:

let the resistance of wire A be a and that of  wire B and  b,then
$\dfrac{R}{50}=\dfrac{a}{100-50}$$\therefore$  a$=R$
also,$\dfrac{R}{50}=\dfrac{b}{100-50}$ $\therefore$  b$=R$
when both are connected in series$\dfrac{R}{i}=\dfrac{2a}{100-i}$
or $100-i=  \ 2i$
or $i=\dfrac{100}{3}$

In the metre bridge experiment of resistances, the known and unknown resistances are inter-changed. The error so removed is:

physics current electricity meter bridge meter bridge and problems on it galvanometer

  1. end correction

  2. index error

  3. due to temperature effect

  4. random error

Show answer
Correct Option: A
Explanation:

$ \alpha, \beta$ are the end correction on left and right side.


case 1:- Without interchanging.

$ \dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{X+\alpha+l _{1}P}{(Y+\beta+(100-l _{1})P)}$       ..........( 1 )

case 2:- After interchanging.

$\dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{Y+\propto + l _{2}P}{X + \beta+(100-l _{2})P}$    ............( 2 )

on simplification of eq.  (1) and (2) we get
$X = Y +(l _{2}-l _{1})P$
$\therefore$ By interchanging the end correction is removed.

A lamp of 6 V and 30 W is used in a laboratory but the supply is of 120 V. what will be done to make use of the lamp?
(1) A resistance may be used
(2) A resistance may be used in series with lamp.
(3) The resistance should be of 18 $\Omega$. 

physics current electricity meter bridge meter bridge and problems on it galvanometer

  1. 1, 2 and 3 are correct

  2. 1 and 2 are correct

  3. 1 and 3 are correct

  4. 2 and 3 are correct

Show answer
Correct Option: B
Explanation:

Here, the resistance should be used in series with lamp in order to divide the voltage or to reduce potential drop across the lamp.

Lamp can withstand a maximum of 6 V , and a current of $\dfrac { 6 }{ { R } _{ lamp } } $ where ${R} _{lamp}$ is $\dfrac { { 6 }^{ 2 } }{ 30 } =1.2\Omega $
hence maximum current that lamp can with stand is  $\dfrac { 6 }{ { 1.2} }= 5A $
so the resistance which is to be connected in series would also get same current of 5 A hence if it will be of $18 \Omega$ then voltage drop accros it will be 90 V and maximum of 6 V across lamp but both voltage drop does not make 120 V . hence the resistance value should be more than $18 \Omega$.  So 1 and 2 is correct but 3 is incorrect.

Why is the Wheatstone bridge better than the other methods of measuring resistances?

physics current electricity meter bridge meter bridge and problems on it galvanometer

  1. It does not involve Ohm's law

  2. It is based on Kirchoff's law

  3. It has four resistor arms

  4. It is a null method

Show answer
Correct Option: D
Explanation:

The Wheatstone bridge is used to measure the unknown resistance by using null method. i.e, when the bridge is balanced, no current through the galvanometer. Using this null method, we can easily measure the unknown resistance if the other three arm's resistor are given.