In a meter bridge circuit as shown in the figure, the bridge is balanced when AJ=20 cm . On interchanging P and Q the balance length shifts by

Question

In a meter bridge circuit as shown in the figure, the bridge is balanced when AJ=20 cm . On interchanging P and Q the balance length shifts by <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ubimcn2bwgducrup.jpg" /> (A) zero (B) 80 cm (C) 40 cm (D) 60 cm

Tardigrade Admin · Accepted Answer

Initially balancing point is J , Hence (AJ/BJ)=(P/Q)=(20/80) Now if P & Q are interchanged, Let balancing point be J1 (AJ1/BJ1)=(Q/P)=(80/20) The balanced point shifts by AJ1-AJ=80-20=60 cm .

Q. In a meter bridge circuit as shown in the figure, the bridge is balanced when $A J = 20 c m$ . On interchanging $P$ and $Q$ the balance length shifts by

zero

$80 c m$

$40 c m$

$60 c m$

Initially balancing point is $J$ ,
Hence $\frac{A J}{B J} = \frac{P}{Q} = \frac{20}{80}$
Now if $P$ & $Q$ are interchanged,
Let balancing point be $J_{1}$
$\frac{A J _{1}}{B J _{1}} = \frac{Q}{P} = \frac{80}{20}$
The balanced point shifts by $A J_{1} - A J = 80 - 20 = 60 c m$ .

Q. In a meter bridge circuit as shown in the figure, the bridge is balanced when AJ=20cm . On interchanging P and Q the balance length shifts by

zero

80cm

40cm

60cm

Initially balancing point is J , Hence BJAJ​=QP​=8020​ Now if P & Q are interchanged, Let balancing point be J1​ BJ1​AJ1​​=PQ​=2080​ The balanced point shifts by AJ1​−AJ=80−20=60cm .

Q. In a meter bridge circuit as shown in the figure, the bridge is balanced when $A J = 20 c m$ . On interchanging $P$ and $Q$ the balance length shifts by

$80 c m$

$40 c m$

$60 c m$

Initially balancing point is $J$ ,
Hence $\frac{A J}{B J} = \frac{P}{Q} = \frac{20}{80}$
Now if $P$ & $Q$ are interchanged,
Let balancing point be $J_{1}$
$\frac{A J _{1}}{B J _{1}} = \frac{Q}{P} = \frac{80}{20}$
The balanced point shifts by $A J_{1} - A J = 80 - 20 = 60 c m$ .