Tag: galvanometer

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Questions Related to galvanometer

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

In the measurement of resistance by a metre bridge, the known and unknown resistance are interchanged to eliminate 

  1. end error

  2. index error

  3. random error

  4. error due to thermoelectric effect
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

End errors in a meter bridge arise due to end resistances at the copper strips and resistance of the connecting wires at the zero and hundred centimeter marks. Interchanging the known and unknown resistances helps cancel out these systematic end errors by taking the average of the two balancing lengths.

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

Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by $10\%$, the balancing point shift :

  1. $10\%$ to right
  2. $10\%$ to left
  3. $9.6\%$ to right
  4. $4.8\%$ to right
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the Resistance be R,

Initially
$\dfrac{R}{R}  = \dfrac{l}{100-l}$
     $ \Rightarrow  l = 50 cm$
After 10% increase it is,

$ \Rightarrow  \dfrac{1.1R}{R} = \dfrac{l}{100-l}$

$ \Rightarrow  110  = 2.1 l$

$ \Rightarrow  l = \dfrac{110}{2.1} = 52.38$

$ \Rightarrow  \dfrac{\Delta l}{l} \times 100 = \dfrac{2.38}{50}\times 100  \approx 4.8$%
Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

The balancing point in a meter bridge is 44 cm. If the resistances in the are  gaps are  inchanged   the new balance point is 

  1. 44 cm

  2. 56 cm

  3. 50 cm

  4. 22 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At balance position of meter bridge $=\dfrac{I}{{\left( {100 - I} \right)}}$ $=\frac{R}{S}$

Where $I$ is the balancing length$.$
So$,$ if value of $R$ and $S$ are inter changed the new $I$ limit will be $100-44=56$
Hence,
option $(B)$ is correct answer.

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

In a meter bridge setup, which of the following should be the properties of the one meter long wire?

  1. High resistivity and low temperature coefficient

  2. Low resistivity and low temperature coefficient

  3. Low resistivity and high temperature coefficient

  4. High resistivity and high temperature coefficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The meter bridge wire should have high resistivity to ensure a measurable resistance and a low temperature coefficient of resistance so that its resistance does not change significantly with temperature fluctuations during the experiment.

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

A $6\Omega$ resistance is connected in the left gap of a meter bridge. In the second gap $3\Omega$ and $6\Omega$ are joined in parallel. The balance point of the bridge is at __

  1. $75cm$
  2. $60cm$
  3. $30cm$
  4. $25cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the right gap, a 3 ohm and 6 ohm resistor are in parallel, giving an equivalent resistance of (3 * 6) / (3 + 6) = 18 / 9 = 2 ohms. Let the balancing length from the left be l. For a meter bridge, R1 / R2 = l / (100 - l), so 6 / 2 = l / (100 - l), which gives 3 = l / (100 - l). Solving this yields 300 - 3l = l, so 4l = 300, giving l = 75 cm.

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

The length of a potentiometer wire is $l$ A cell of emf E is balanced at a length $\dfrac{l }{ 3}$ from the positive end of the wire. If the length of the wire is increased by $\dfrac{l} { 2}$.
At what distance will the same cell give a balance point?

  1. $

    \dfrac{2l}{3}

    $
  2. $

    \dfrac{l}{2}

    $
  3. $

    \dfrac{l}{6}

    $
  4. $

    \dfrac{4l}{3}

    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The potential gradient V/L = E/l1. Initially, E = (V/l) * (l/3) = V/3. If the length becomes l + l/2 = 3l/2, the new potential gradient is V/(3l/2) = 2V/3l. The balance length l2 for the same cell E is E = (2V/3l) * l2. Substituting E = V/3: V/3 = (2V/3l) * l2. Thus, l2 = l/2.

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

In Wheatstone's bridge $  P=9  $ ohm, $  Q=11  $ ohm, $  R=4  $ ohm and $  S=6  $ ohm. How much resistance must be put in parallel to the resistance $  S  $ to balance the bridge

  1. $24 ohm$
  2. $ \frac{44}{9} ohm$
  3. $26.4 \mathrm{ohm} $
  4. $18.7 ohm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a balanced Wheatstone bridge, P/Q = R/S'. Here P/Q = 9/11. R/S' = 4/S'. So 9/11 = 4/S', S' = 44/9. The original S is 6. To get 44/9, we add a parallel resistance x: (6*x)/(6+x) = 44/9. 54x = 264 + 44x, 10x = 264, x = 26.4 ohms.

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

In metre bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is 'l'. If the temperature of water is increased, the new balancing length is

  1. >l

  2. none

  3. =0

  4. =l
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \frac { { { R _{ unknown } } } }{ { { R _{ s\tan  dard } } } } =\frac { l }{ { \left( { 1-l } \right)  } } . \ If\, \, temperature\, \, increases,\, \, resis\tan  ce\, \, increases. \end{array}$

Hence, Option $A$ is correct.

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

An electron is released from the origin at a place where a uniform electric field $\overrightarrow{E}$ and uniform magnetic field $\overrightarrow{B}$ exit along the negative y-axis and the negative z-axis respectively.

  1. At time $t$ the y-component of velocity of the electron becomes $u _y = \dfrac{E}{B}\sin \omega t$ where $\omega = \dfrac{eB}{m}$.
  2. At $t = \pi m / eB$ the electron will have only x-component of velocity.
  3. At $t = \dfrac{2\pi m}{3eB}$, the y-component of velocity becomes zero.
  4. The displacement along y-axis is $\dfrac{2Em}{eB^2}$ when the velocity of electron becomes perpendicular to the y-axis
Reveal answer Fill a bubble to check yourself
A,B Correct answer