- 1(current)
Q. No.In the metre bridge arrangement, it is observed that the balance is obtained with a length of \(l_b=25~\text{cm},\) on a \(100~\text{cm}\)-wire. If the left gap has both \(P\) & \(Q\) connected (in parallel!), while the right gap has only \(Q\) connected in it, then the new balance length will be:
1. \(20~\text{cm}\)
2. \(30~\text{cm}\)
3. \(40~\text{cm}\)
4. \(80~\text{cm}\)
1. \(20~\text{cm}\)
2. \(30~\text{cm}\)
3. \(40~\text{cm}\)
4. \(80~\text{cm}\)
Subtopic: Â Meter Bridge |
 78%
Level 2: 60%+
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In a metre bridge, there is an error due to excess resistance at the ends \(A\) & \(B.\) Due to this, the correct expression to be used is:
\(\dfrac PQ=\dfrac{\theta_A+l}{(100-l)+\theta_B},\)
where \(l\) is the measured balance length & \(\theta_A,\theta_B\) are the respective "end corrections". In the metre bridge shown in the figure, the following data was collected.
| \(P\) | \(Q\) | \(l\) |
| \(100~\Omega\) | \(100~\Omega\) | \(50~\text{cm}\) |
| \(42~\Omega\) | \(60~\Omega\) | \(41~\text{cm}\) |
| \(240~\Omega\) | \(x~\Omega\) | \(79~\text{cm}\) |
From the data, it can be concluded that:
| 1. | \(\theta_A=\theta_B\) | 2. | \(\theta_A=2\theta_B\) |
| 3. | \(\theta_B=2\theta_A\) | 4. | \(\theta_A=-\theta_B\) |
Subtopic: Â Meter Bridge |
 80%
Level 1: 80%+
To unlock all the explanations of this course, you need to be enrolled.
Hints
To unlock all the explanations of this course, you need to be enrolled.
In a metre bridge, there is an error due to excess resistance at the ends \(A\) & \(B.\) Due to this, the correct expression to be used is:
\(\dfrac PQ=\dfrac{\theta_A+l}{(100-l)+\theta_B},\)
where \(l\) is the measured balance length & \(\theta_A,\theta_B\) are the respective "end corrections". In the metre bridge shown in the figure, the following data was collected.
| \(P\) | \(Q\) | \(l\) |
| \(100~\Omega\) | \(100~\Omega\) | \(50~\text{cm}\) |
| \(42~\Omega\) | \(60~\Omega\) | \(41~\text{cm}\) |
| \(240~\Omega\) | \(x~\Omega\) | \(79~\text{cm}\) |
The value of \(\theta_A,\) from the data, is:
| 1. | \(1~\text{cm}\) | 2. | \(0.5~\text{cm}\) |
| 3. | \(2~\text{cm}\) | 4. | \(-1~\text{cm}\) |
Subtopic: Â Meter Bridge |
 70%
Level 2: 60%+
To unlock all the explanations of this course, you need to be enrolled.
Hints
To unlock all the explanations of this course, you need to be enrolled.
In a metre bridge, there is an error due to excess resistance at the ends \(A\) & \(B.\) Due to this, the correct expression to be used is:
\(\dfrac PQ=\dfrac{\theta_A+l}{(100-l)+\theta_B},\)
where \(l\) is the measured balance length & \(\theta_A,\theta_B\) are the respective "end corrections". In the metre bridge shown in the figure, the following data was collected.
| \(P\) | \(Q\) | \(l\) |
| \(100~\Omega\) | \(100~\Omega\) | \(50~\text{cm}\) |
| \(42~\Omega\) | \(60~\Omega\) | \(41~\text{cm}\) |
| \(240~\Omega\) | \(x~\Omega\) | \(79~\text{cm}\) |
The value of the unknown resistance \((x)\) is:
| 1. | \(63~\Omega\) | 2. | \(66~\Omega\) |
| 3. | \(60~\Omega\) | 4. | \(64~\Omega\) |
Subtopic: Â Meter Bridge |
 62%
Level 2: 60%+
To unlock all the explanations of this course, you need to be enrolled.
Hints
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- 1(current)
Q. No.
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