“If I have seen a little further it is by standing on the shoulders of
Giants.”

– Sir Isaac Newton

The metre bridge is basically a Wheatstone
bridge, the condition of balance of which is conveniently used in measuring
unknown resistances. Today we shall discuss a few questions (MCQ) involving
metre bridge:

(1) Resistances R

_{1}and R_{2}are connected respectively in the left gap and right gap of a metre bridge. If the balance point is located at 55 cm, the ratio R_{2}/R_{1}is
(a) 4/5

(b) 5/4

(c) 9/11

(d) 11/9

(e) 6/7

The distance of the balance point from the

*left*end of the wire in the metre bridge is 55 cm. Therefore the distance of the balance point from the*right*end of the wire is 45 cm. Thus we have
R

_{1}/R_{2}= 55/45
Or, R

_{2}/R_{1 }= 45/55 = 9/11
(2) A 15 Ω resistance is connected in the left
gap and an unknown resistance less than 15 Ω is connected in the right gap of a
metre bridge. When the resistances are interchanged, the balance point is found
to shift by 20 cm. The unknown resistance is

(a) 5 Ω

(b) 6 Ω

(c) 8 Ω

(d) 10 Ω

(e) 12 Ω

Initially the balance point will be at J

_{1}as indicated in the figure. On interchanging the resistances the balance point will shift to J_{2}so that the length of the bridge wire between J_{1}and J_{2}is 20 cm. The balance points J_{1}and J_{2}must be*equidistant*from the mid point (50 cm mark) of the wire so that J_{1}is at 60 cm and J_{2}is at 40 cm.
Therefore we have

R/X
= 60/40 = 6/4

Or, 15/X = 6/4

This gives X = 10 Ω

(3) Resistances 4 Ω and 6 Ω are connected across
the left gap and right gap respectively of a metre bridge. When a 2 Ω
resistance is connected in series with the 4 Ω resistance in the leftt gap, the
shift in the balance point is

(a) 10 cm

(b) 15 cm

(c) 20 cm

(d) 25 cm

(e) 30 cm

Initially when the left gap and right gap contain
4 Ω and 6 Ω respectively, the condition of balance is

4/6
=

*L*/(100 –*L*) where*L*is the balancing length
This gives

*L*= 40 cm.
When a 2 Ω resistance is connected in series with
the 4 Ω resistance in the leftt gap, the balancing length becomes 50 cm (since
the gaps contain equal resistances).

Therefore the shift in the balance point is (50
cm – 40 cm) = 10 cm.

sir question are simple and direct but sometimes it tricks us...

ReplyDelete