Answer: Option B. -> 12 cells in series 4 such groups in parallel
:
B
Let m cells be connected in series and n such groups are connected in parallel
If the emf of each cell is E and internal resistance r then the total emf of m in series is mE and the total internal resistance is mr. When n such groups are in parallel the effective internal resistance is mr/n. Then the current through an external resistance R is
$l=\frac{mE}{R+\frac{mr}{n}}=\frac{mnE}{nR+mr}=\frac{mnE}{(\sqrt{nR}-{\sqrt{mr)}}^2+2\sqrt{mnRr}}$
Now i will be maximum if the denominator is the minimum i.e. if nR=mr
Using R= 3ohm and r=1ohm we have 3n = m
But mn = 48
therefore $\frac{m\times m}{3}=48$
which gives m= 12
thus n=4

Answer: Option B. -> x  
:
B
The condition for no delflection of the galvanometer is
$\frac{R_1}{R_2}=\frac{R_{AC}}{R_{CB}}$
where $R_{AC}$ and $R_{CB}$ are the resistances of the bridge wire of length ACand CBrespectively. If the radius of the wire AB is doubled the ratio
$\frac{R_{AC}}{R_{CB}}$ will remain unchanged.Hence the balance length will remain the same

Answer: Option D. -> copper decreases and of germanium increases
:
D
copper is a conductor whereas Germanium is a semi conductor

Answer: Option B. -> both the length and the radius of the wire are doubled
:
B
$Q=\frac{V^2}{R}$
But $R=\frac{pl}{\pi r^2}$
Therefore $Q=\left(\frac{\pi V^2}{p}\right)\frac{r^2}{1}$
Q is doubled if both l and r are doubled

Answer: Option B. -> 7 ohm
:
B

The circuit can be redrawn as shown in figure. Let a bettery of emf E and of negligible resistance be connected as shown. The current in various branches are shown.Applying Kirchoff's second law to loops ACDA,CBDC and ADBGFA we have
$5i_2+5i_3-10i_1=0$
$10(i_2-i_3)-5(i_1+i_3)-i_3=0$
$10i_1+5(i_1+i_3)-E=0$
equation (i) and (ii) give $i_2=\frac{3i_1}{2}$ and $i_3=\frac{i_1}{2}.$ If R is the effective resistance between A and B
then $R=\frac{E}{i_1+i_2}=\frac{2E}{5i_1}$
from equation (iii) we have
$E=15i_1+5i_3=\frac{35i_1}{2}$
$R=\frac{2E}{5i_1}=\frac{2\ast 35i_1}{5i_1\ast 2}=7$ Ohm

Answer: Option C. -> If a voltmeter is connected across the 6 V battery it will read 7 V
:
C
Total resistance = 4 + 1 + 0.5 + 0.5 = 6ohm
Net voltage in the circuit is 6 V
Current $l=\frac{6}{6}=1A$ in the anticlockwise direction
$V_{PR}=1\times 4=4V$
Since R is connected to earth $V_R=0$ hence $V_p=4V$
$V_{SQ}=0.5\times 1=0.5V$
S is at a higher potential than Q
$V_Q=-0.5V$
Current is being forced into the 6 V battery in the opposite direction.Hence $V_6=E+l_r=6+1X$
1 = 7 V

Answer: Option A. -> 44+π ohm
:
A
Circumference of the circle = $2\pi r$
Therefore the resistance per unit length of the wire = $\frac{R}{2\pi r}$
where r= 4 ohm is the resistance of wire
Now the length of the specimen connected along the diameter = 2r
Therefore the resistance of this specimen is
$R_1=\frac{R}{2\pi r}\times 2r=\frac{R}{\pi }$
also the resistance of each semicircle is
$R_2=\frac{R}{2}$
equivalent resistance R' across the specimen is given by
$\frac{1}{R^{\prime }}=\frac{2}{R}+\frac{2}{R}+\frac{\pi }{R}=\frac{4+\pi }{R}$
$R^{\prime }=\frac{R}{4+\pi }=\frac{4}{4+\pi }$ ohm

Answer: Option C. -> 48 ohm
:
C
If k is the current sensitivity of the galvanometer the current in the galvanometer is
$l_g=50$ K
when a shunt S is connected across it the current through the galavnometer becomes
$l_g^{\prime }=\frac{l_{g}S}{G+S}=10$ K
where G is the resistance of the galvanometer dividing the equations we get
$\frac{G+S}{S}=5$
which gives $G=4\times S=4\times 12=48$ ohm

Answer: Option B. -> switch
:
B
A switch is a device used to make or break the circuit. When it is switched on, it completes the circuit, letting the current to flow. When it is switched off, it breaks the path, restricting the flow of electric current.

Answer: Option A. -> electric circuit
:
A
An electric circuit is a complete pathway for electricity to flowbetween the two terminals of the electric cell. A simpleelectric circuit has components like cell, wire, bulb or fan, resistoretc.