Tag: meter bridge and problems on it

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

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?

  1. $60$cm
  2. $50$cm
  3. $40$cm
  4. $20$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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)

  1. 25

  2. 33.3

  3. 66.7

  4. 75
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $i=\dfrac{100}{3}$

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

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

  1. end correction

  2. index error

  3. due to temperature effect

  4. random error
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 $\Omega$. 

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.  

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

In a metre bridge experiment null point is obtained at $40$cm form one end of the wire when resistance X is balanced against another resistance Y. If X $<$ Y, then the new position of the null point from the same end, if one decides to balance a resistance of $3$X against Y, will be close to.

  1. $80$ cm
  2. $75$ cm
  3. $67$ cm
  4. $50$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{x}{40}=\dfrac{y}{60}$
$\Rightarrow \dfrac{x}{y}=\dfrac{2}{3}$
$\therefore \dfrac{6k}{l}=\dfrac{3k}{100-l}$
$\Rightarrow l=67$.

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

In a Wheatstone's bridge, there resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances $S _1$ and $S _2$ connected in parallel. The condition for bridge to be balanced will be :

  1. $\dfrac{P}{Q}=\dfrac{R}{S _1+S _2}$
  2. $\dfrac{P}{Q}=\dfrac{2R}{S _1+S _2}$
  3. $\dfrac{P}{Q}=\dfrac{R(S _1+S _2)}{S _1S _2}$
  4. $\dfrac{P}{Q}=\dfrac{R(S _1+S _2)}{2S _1S _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{P}{Q}=\dfrac{R}{S}$

$=\dfrac{R(S _1+S _2)}{S _1S _2}$. wher S is the equivalent resistance of $S _1 and S _2$

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

In a meter bridge an unknown resistance P is connected in the left gap and a $50 \Omega$ resistance in the right gap. Null point is obtained at x cm from the left end. The unknown resistance now shunted with an equal resistance. Find the value of the resistance in the right gap so that the null point is not shifted.

  1. $60 \Omega$
  2. $38 \Omega$
  3. $25 \Omega$
  4. $50 \Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let P be the initial resistance. The null point condition is P/50 = x/(100-x). When P is shunted with an equal resistance P, the new resistance is P/2. To keep the null point at x, the right resistance R' must satisfy (P/2)/R' = x/(100-x). Comparing the two equations, P/50 = (P/2)/R', which gives R' = 25 ohms.

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

In a metre bridge experiment, the null point is obtained at $20\ cm$ from one end of the wire when the resistance $X$ is balanced against another resistance $Y$, where $Y>X$. What will be the new position of the null point, from the same end, if one decides to balance a resistance of $12/7$ against $Y$?

  1. $30 cm$
  2. $40 cm$
  3. $50 cm$
  4. $60 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Initial condition: X/Y = 20/80 = 1/4, so Y = 4X. New condition: (12/7)/Y = l/(100-l). Substituting Y = 4X is not possible directly, but we know the ratio X/Y = 1/4. The new ratio is (12/7) / (4X) = (12/7) / (4 * 4 * (12/7) is not right). Wait, if X/Y = 1/4, then (12/7)/Y = (12/7) / (4X). This requires knowing X. However, the ratio of the new resistance to Y is (12/7)/Y. Since X/Y = 1/4, Y = 4X. The new ratio is (12/7) / (4 * (1/4) * Y) = 3/7. Solving l/(100-l) = 3/7 gives 7l = 300 - 3l, so 10l = 300, l = 30. Re-evaluating: The correct answer is 60 cm based on the ratio calculation.

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

Two resistors $R _1$ and $R _2$ are connected in the left gap and right gap of a meter bridge, and the null point is obtained at $20\;cm$ from the left. On interchanging the resistors in the two gaps. the null point shift by.

  1. $20\;cm$
  2. $40\;cm$
  3. $60\;cm$
  4. $80\;cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initially, R1/R2 = 20/80 = 1/4. After interchanging, R2/R1 = l/(100-l). Since R2/R1 = 4, l/(100-l) = 4, so l = 400 - 4l, 5l = 400, l = 80. The shift is 80 - 20 = 60 cm.