Q: A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:

$ R = l A $ Now, $1=1+ 1 10 = 111 10 $ and therefore, $ A = 10 ~A 11 $ So $ R ^ = p ( 111 10 ) 10 ~A 11 = l A (11)^ 2 (10)^ 2 =1.21 R $ Now resistance becomes $1.21$ times of initial and specific resistance is the intrinsic property so remains same.

Question 1

In the given circuit, E_1=E_2=E_3=2 ~V and R_1=R_2=4 , then current flowing through the branch A B is

Accepted Answer

The given circuit can be redrawn as, where, equivalent emf of the combination, aligned E_ eq & = E_1 R_1+E_2 R_2 R_1+R_2 \ & = 2 4+2 4 4+4 =2 ~V aligned and R_ e q = R_1 R_2 R_1+R_2 = 4 4 4+4 = 16 8 =2 Net emf of the circuit, aligned E^ & =E_2+E_ eq \ & =2+2 \ & =4 ~V aligned Current flowing through branch A B, I= E^ R_ e q = 4 2 =2 ~A Clearly from figure, it is clear that direction of current is from point A to B clockwise is loop.

Question 2

In the following circuit diagram, when 3 resistor is removed, then equivalent resistance of the network

Accepted Answer

The equivalent Wheatstone's bridge network of the given circuit is shown in figure. Here, The points B and D are at the same potential as the bridge is balanced. So, the 3 resistance in B D arm is ineffective and can be omitted from the circuit. Hence, equivalent resistance remains same when 3 resistor is removed.

Question 3

A conducting wire is stretched by applying a deforming force, so that its diameter decreases to 40 \% of original value. The percentage change in its resistance will be

Accepted Answer

On stretching, volume of wire remains constant. So, V=A l or l= V A array ll & R= l A = V A^2 = V ^2 D^4 16 = 16 V ^2 D^4 \ & R R =-4 D D =-4(-0.4)=1.6 \% array

Question 4

A wire of resistance 160 is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

Accepted Answer

Let Initial length =l_1 Final length =l_2 Initial area = A _1 Final area =A_1 Volume remains same A _1 l _1= A _2 l _2 ~A _1 l _1= A _2 l _1 4 4 ~A _1= A _2 Initial resistance, R _1= l_1 A_1 =160 (given) Final resistance, R _2= l _2 ~A _2 R _2 R _1 = l _2 ~A _1 ~A _2 l _1 = l_1 4 ~A _1 4 ~A _1 l_1 R_2= 1 16 R_1= 1 16 160=10

Question 5

Five cells each of emf E and internal resistance r send the same amount of current through an external resistance R whether the cells are connected in parallel or in series. Then the ratio ( R r ) is

Accepted Answer

Given : Number of cells, n=5, emf of each cell =E Internal resistance of each cell =r In series, current through resistance R I= n E n r+R = 5 E 5 r+R In parallel, current through resistance R I^ = E r n +R = n E r+n R = 5 E r+5 R According to question, I=I^ 5 E 5 r+5 R = 5 E r+5 R 5 r+R=r+5 R or R=r R r =1

Question 6

Ratio of thermal energy released in two resistor R and 3 R connected in parallel in an electric circuit is :

Accepted Answer

Thermal energy is given by H = P t = V ^2 R t Here, voltage V is same. H _1 H _2 = V ^2 t R V ^2 t 3 R =3: 1

Question 7

The resistance of a wire is 5 . It's new resistance in ohm if stretched to 5 times of its original length will be :

Accepted Answer

Let resistance of a wire R and length 1 . R = A =5 Volume of wire is constant in stretching V_i=V_f A_i _i=A_f _f A = A ^ (5 ) A ^ = A 5 R_f= _f A_f = (5 ) ( A 5 ) =25 ( A )=25 5=125

Question 8

In the arrangement shown in figure, when the switch S_2 is open, the galvanometer, shows no deflection for l=50 ~cm when the switch S_2 is closed, the galvanometer shows no deflection for l=0.416 ~m .

Accepted Answer

When S_2 open. Assume resistance of X Y=R. Resistance of wire per unit length, x= R L =R m ^ -1 I=E_0 / R Now, the potential drop across 50 ~cm length is 6 ~V , So array cc & E_0 R R 50 100 =6 \ & E_0=12 ~V array When S_2 closed, potential drop across 0.416 ~cm length, V_1= E_0 R R 0.416=12 0.416 5 ~V Hence, E-I r=5 ~V aligned & & 6-I r & =5 \ & & I & = 5 10 \ & & 6-5 & = 5 10 r \ & & r & =2 aligned

Question 9

An electric current I enters and leaves a uniform circular wire of radius r through diametrically opposite points. A charged particle q moves along the axis of circular wire passes through its centre

Accepted Answer

Force on a moving charged particle in uniform magnetic field F=B q v Since, charge particle moves along the axis of circular current carrying loop, therefore, =0^ or 180^ When =0^ ; F=B q v 0^ F=0 When =180^ , F=B q u 180^ F=0

Question 10

A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:

Accepted Answer

$ R = l A $ Now, $1=1+ 1 10 = 111 10 $ and therefore, $ A = 10 ~A 11 $ So $ R ^ = p ( 111 10 ) 10 ~A 11 = l A (11)^ 2 (10)^ 2 =1.21 R $ Now resistance becomes $1.21$ times of initial and specific resistance is the intrinsic property so remains same.

Question 11

A moving coil galvanometer has resistance of 10 and full scale deflection of 0.01 ~A . It can be converted into voltmeter of 10 ~V full scale by connecting into resistance of

Accepted Answer

R = V I _ g - G = 10 0.01 -10=990 in series.

Current Electricity — Previous Year Questions & Solutions