A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure. The current in the 10 Ω resistor is 

• 0.27 A P₂ to P₁

• 0.03 A P₁ to P₂

• 0.03 A P₂ to P₁

• 0.27 A P₁ to P₂

C.

0.03 A P₂ to P₁

A long straight wire of radius ‘a’ carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2 a is 

• 1/4

• 4

• 1

• 1/2

C.

1

A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio

• 2

• 1

• 1/4

• 1/2

A.

2

A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio AA /AB of their respective lengths must be

• 2

• 1

• 1/4

• 3

A.

2

A rectangular loop has a sliding connector PQ of length

• 
• 
• 
• 

C.

A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I₁ + I₂

Applying Kirchhoff's law
I₁R + IR -vBl = 0 .... (i)
I₂R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I₁ = I₂ = VBI/3R