Solution:
Key concept: Meter bridge: In case of meter bridge, the resistance wire AC is 100cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 - l) so that $\frac{\text{Q}}{\text{P}}=\frac{(100-\text{I})}{\text{I}}$. Also $\frac{\text{P}}{\text{Q}}=\frac{\text{R}}{\text{S}}>\text{S}=\frac{(100-\text{I})}{\text{I R}}$.
When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to the right of D, the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D in the galvanometer wire.
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The frequency of a light ray is 6 
. Its frequency when it propagates in a medium of refractive index 1.5, will be
|
(a) 1.67 |
(b) 9.10 |
(c) 6 |
(d) 4 |
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of this cube will be
|
(a) |
(b) |
(c) |
(d) Zero |
The coils of a step down transformer have 500 and 5000 turns. In the primary coil an ac of 4 ampere at 2200 volts is sent. The value of the current and potential difference in the secondary coil will be
|
(a) 20 A, 220 V |
(b) 0.4 A, 22000 V |
(c) 40 A, 220 V |
(d) 40 A, 22000 V |
The value of stopping potential in the following diagram
|
(a) – 4V |
(b) – 3 V |
(c) – 2V |
(d) – 1 V |
The given truth table is of
|
(a) OR gate |
(b) AND gate |
(c) NOT gate |
(d) None of above |
Two charges + 5μC and +10 μC are placed 20 cm apart. The net electric field at the mid-Point between the two charges is
|
(a) 4.5 |
|
(b) 4.5 |
|
(c) 13.5 |
|
(d) 13.5 |
