1
MHT CET 2021 23rd September Evening Shift
+1
-0

To determine the internal resistance of a cell by using a potentiometer, the null point is at $$1 \mathrm{~m}$$ when shunted by $$3 \Omega$$ resistance and at a length $$1.5 \mathrm{~m}$$, when cell is shunted by $$6 \Omega$$ resistance The internal resistance of the cell is

A
$$1 \Omega$$
B
$$4 \Omega$$
C
$$2 \Omega$$
D
$$6 \Omega$$
2
MHT CET 2021 23rd September Evening Shift
+1
-0

In the given circuit, the current in 8$$\Omega$$ resistance is 1.5 A. The total current (I) flowing in the circuit is

A
5 A
B
4.5 A
C
3 A
D
5.5 A
3
MHT CET 2021 23th September Morning Shift
+1
-0

A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $$1 \Omega \mathrm{m}^{-1}$$. The current drawn from the battery is (Internal resistance of battery is negligible)

A
0.44 A
B
0.66 A
C
0.88 A
D
0.22 A
4
MHT CET 2021 23th September Morning Shift
+1
-0

In potentiometer experiment, cells of e.m.f. '$$E_1$$' and '$$E_2$$' are connected in series $$\left(E_1>E_2\right)$$ the balancing length is $$64 \mathrm{~cm}$$ of the wire. If the polarity of $$E_2$$ is reversed, the balancing length becomes $$32 \mathrm{~cm}$$. The ratio $$\frac{E_1}{E_2}$$ is

A
$$1: 1$$
B
$$6: 1$$
C
$$3: 1$$
D
$$2: 1$$
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