For zero galvanometer deflection, the bridge must be balanced. This is a Wheatstone bridge arrangement.
The condition for a balanced Wheatstone bridge is:
$\frac{R_1}{R_2} = \frac{R_3}{R_4}$
Where R1, R2, R3, and R4 represent the resistance in each arm of the Wheatstone bridge. Here, R1 and R2 are the resistances of the wire segments AJ and JB respectively, and R3 = 8 Ω and R4 = 12 Ω.
The length of AB is 40 cm, thus, $\frac{R_1}{R_2} = \frac{8}{12} = \frac{2}{3}$
$\frac{R_1}{R_2} = \frac{x}{40 - x} = \frac{2}{3}$
3x = 80 − 2x 5x = 80 x = 16 cm
Distance of J from B = 40 cm − 16 cm = 24 cm
List I | List II | ||
A | Down’s syndrome | I | 11th chormosome |
B | α-Thalassemia | II | ‘X’ chromosome |
C | β-Thalassemia | III | 21st chromosome |
D | Klinefelter’s syndrome | IV | 16th chromosome |
The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :