The resistances in the two arms of the meter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In the first case,at the balance point:$\frac{5}{R} = \frac{l_1}{100 - l_1}$  $....(i)$In the second case,the resistance $R$ is shunted with an eq

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) In the first case,at the balance point:$\frac{5}{R} = \frac{l_1}{100 - l_1}$  $....(i)$In the second case,the resistance $R$ is shunted with an equal resistance $R$,so the equivalent resistance becomes $R' = \frac{R 	imes R}{R + R} = \frac{R}{2}$.The new balance point is at $1.6\,l_1$. Thus,at the balance point:$\frac{5}{R/2} = \frac{1.6\,l_1}{100 - 1.6\,l_1}$  $....(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{5/R}{5/(R/2)} = \frac{l_1 / (100 - l_1)}{1.6\,l_1 / (100 - 1.6\,l_1)}$$\frac{1}{2} = \frac{l_1}{100 - l_1} 	imes \frac{100 - 1.6\,l_1}{1.6\,l_1}$$\frac{1}{2} = \frac{100 - 1.6\,l_1}{1.6(100 - l_1)}$$0.8(100 - l_1) = 100 - 1.6\,l_1$$80 - 0.8\,l_1 = 100 - 1.6\,l_1$$0.8\,l_1 = 20$$l_1 = 25\,cm$Substituting $l_1 = 25\,cm$ into equation $(i)$:$\frac{5}{R} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$$R = 15\,\Omega$.

$Sl. No.$	$R \, (\Omega)$	$l \, (cm)$
$1.$	$1000$	$60$
$2.$	$100$	$13$
$3.$	$10$	$1.5$
$4.$	$1$	$1.0$

The resistances in the two arms of the meter bridge are $5 \,\Omega$ and $R \,\Omega$ respectively. When the resistance $R$ is shunted with an equal resistance,the new balance point is at $1.6\,l_1$. The resistance $R$ is .................. $\Omega$.

Similar Questions

In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in the figure.

$Sl. No.$ $R \, (\Omega)$ $l \, (cm)$

$1.$ $1000$ $60$

$2.$ $100$ $13$

$3.$ $10$ $1.5$

$4.$ $1$ $1.0$

Which of the readings is inconsistent?

In the given arrangement of the experiment on a meter bridge,if $AD$ corresponding to the null deflection of the galvanometer is $X$,what would be its value if the radius of the wire $AB$ is doubled?

When a metal conductor connected to the left gap of a meter bridge is heated,the balancing point

The experimental setup of a meter bridge is shown in the figure. If the null deflection point of the galvanometer is at distance $x$ from end $A$,what will be the new null point if the radius of the wire $AB$ is doubled?

The figure shows a meter-bridge circuit,with $AB = 100\, cm$,$X = 12\,\Omega$ and $R = 18\,\Omega$,and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$,through what distance will $J$ have to be moved to obtain balance? .............. $cm$

Vedclass Test Series

Exam Paper Generator

Online Exam Module