Some of you may understand the principles in physics very well but your capacity for numerical manipulations may be poor. Practice can make you strong in solving questions involving numerical manipulations so that you will not waste your precious time on such questions. Here are two questions which appeared in AIPMT 2008 question paper:
(1) An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20º C? The temperature of boiling water is 100º C.
(1) 8.4 min
(2) 12.6 min
(3) 4.2 min
(4) 6.3 min
We have VIt =mcθ where V is the voltage, I is the current, t is the time of flow of the current, m c is the specific heat of water (which is approximately 4200 Jkg–1 K–1) and θ is the temperature rise.
Therefore, 220×4×t = 1×4200×(100 – 20)
This will give t = 381 sec. (nearly) which is approximately 6.3 min.
(2) a galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
(1) 5550 Ω
(2) 6050 Ω
(3) 4450 Ω
(4) 5050 Ω
Since the current through an ordinary galvanometer is directly proportional to the deflection (remember that in a tangent galvanometer this is not the case) we have
3/(50+2950) = k×30 when the deflection is 30 divisions.
Here k is the proportionality constant (figure of merit of the galvanometer).
If the resistance in series for reducing the deflection to 20 divisions is X we have
3/(50+X) = k×20
On dividing the first equation by the second,
(50+X)/3000 = 3/2 from which X = 4450 Ω.
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