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
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :