Take note that the question is looking for specific latent heat of vaporisation during boiling. Hence information of mass m1 before heating is not relevant. There is no change of state of the water, it is merely the heating of the water (which involves specific heat capacity of water). Solutions: Option D Refer to the worked solution below .

Solutions: Option B As lamp X is thicker and shorter , these two factors make the resistance of X lower than Y . Recall: Length increases, R increases & cross-sectional area increases, R decreases. [R = pL/A] As both bulbs are connected to the same mains, (assume 240V), the potential difference across the bulbs are 240 V. Brightness of the bulb depends on power , P = IV or V2/R, since V is constant, I across X is higher due to lower R,hence power of X is greater than power of Y, hence X is brighter .

Sensitivity of the ammeter means that how small a current the ammeter can measure. Example, if the original set up is 0.2 A per division, and the modified set up can measure up to 0.1 A per division, so we say the latter one is more sensitive as it is able to measure smaller change in the currrent.

Based on individual calculations, the highest effective resistance will only be achieved if the junctions are connected at Q and S.

Solutions: Option B Whatever that goes in has to come out. With this basic concept, you will be able to solve this question easily. Ignoring the unknown branch, Total current going into O = 8A + 6A = 14A Total current going out of O = 12A + 4A = 16A Hence in order to be equal, 2A must be flowing in direction PO. Hence answer is B.

Double-insulated appliances are common around us. The air purifiers in our classroom are one good example. In these appliances, only two wires are needed, namely Live and Neutral wires . Earth wire is not required as the casing is made of insulator like plastic for this case. The classic air purifier in our classroom The “double squares” symbol means that this appliance is double-insulated. As you see, the earth wire is not required.

The concepts needed to solve such questions Length increases, resistance increases and vice versa. Finding resistors in parallel and in series.