1
JEE Main 2021 (Online) 25th July Evening Shift
+4
-1
Out of Syllabus
In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, $${{{\varepsilon _1}} \over {{\varepsilon _2}}}$$ is :

A
$${5 \over 3}$$
B
$${8 \over 5}$$
C
$${4 \over 3}$$
D
$${3 \over 2}$$
2
JEE Main 2021 (Online) 25th July Evening Shift
+4
-1
Out of Syllabus
The given potentiometer has its wire of resistance 10$$\Omega$$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2$$\Omega$$ resistor is :

A
10 V
B
5 V
C
$${{40} \over 9}$$ V
D
$${{40} \over 11}$$ V
3
JEE Main 2021 (Online) 25th July Morning Shift
+4
-1
Out of Syllabus
In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 $$\Omega$$ per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :

A
5 V
B
2.25 V
C
6 V
D
2.75 V
4
JEE Main 2021 (Online) 22th July Evening Shift
+4
-1
A Copper (Cu) rod of length 25 cm and cross-sectional area 3 mm2 is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.

(Take Resistivity of Copper = 1.7 $$\times$$ 10$$-$$8 $$\Omega$$m and Resistivity of Aluminium = 2.6 $$\times$$ 10$$-$$8 $$\Omega$$m)

A
0.0858 m$$\Omega$$
B
1.420 m$$\Omega$$
C
0.858 m$$\Omega$$
D
2.170 m$$\Omega$$
EXAM MAP
Medical
NEET