“Do not worry about your problems with
mathematics, I assure you mine are far greater."

– Albert Einstein

The following questions appeared in JEE Advanced 2014 question paper. Question No. 1 is a multiple choice question in which one or more than one options are correct. The second question also is multiple choice type but it has just one correct option.

(1) Heater of an electric kettle is made of a wire of length

*L*and diameter*d*. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length*L*and diameter 2*d*. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K ?
(A) 4 if wires are
in parallel

(B) 2 if wires are
in series

(C) 1 if wires are
in series

(D) 0.5 if wires
are in parallel

The resistance

*R*of the heater wire in the first kettle is given by*R =*

*ρL/A*where

*ρ*is the resistivity (specific resistance) of the wire and

*A*is the cross section area of the wire.

[Note that

*A =*π(d/2)^{2}]
The second kettle has

*two*heater wires, each of the same length*L*but cross section area 4*A*.
[Since the diameter of each wire in the second kettle is twice that of
the wire in the first kettle, the cross section area is 4 times].

The resistance of each wire in the second kettle is

*R*/4.
When the wires are connected in parallel, the
effective resistance is

*R/*8.
The power of the first kettle is

*V*^{2}/*R*where as the power of the second kettle is*V*^{2}/(*R/*8) which is equal to 8*V*^{2}/*R*.
Since the power of the second kettle is 8 times that
of the first one, the time required for heating the water is reduced by a
factor of 8 and is equal to 4/8 minutes which is 0.5 minute. Therefore (D) is a
correct option.

When the wires are connected in series, the
effective resistance of the combination is 2×(

*R/*4) =*R/*2. The power of the second kettle is then*V*^{2}/(*R/*2) which is equal to 2*V*^{2}/*R*.
Since the power of the second kettle in this case twice
that of the first one, the time required for heating the water is half the time
taken by the first kettle and is equal to 2 minutes. Therefore (B) too is a
correct option.

(2) During an experiment with a metre bridge, the
galvanometer shows a null point when the jockey is pressed at 40.0 cm using a
standard resistance of 90 Ω, as shown in the figure. The least count of the
scale used in the metre bridge is 1 mm. The unknown resistance is

(A) 60 ± 0.15 Ω

(B) 135 ± 0.56 Ω

(C) 60 ± 0.25 Ω

(D) 135 ± 0.23 Ω

We have

*R/*90*=*40/(100 – 40) from which*R =*60 Ω.
Since the
least count of the scale used in the metre bridge is 1 mm only, there is an
error in the value of

*R*calculated above. In order to take this inaccuracy into account, we write the balance condition of the metre bridge as*R/*90 =

*x/*(100 –

*x*) where

*x*is the balancing length (on the side of

*R*as shown in the figure).

Taking logarithms,

ln

*R =*ln 90 + ln*x*– ln(100 –*x*)
Therefore ∆

*R/R =*∆*x/x*– ∆(100 –*x*)/(100 –*x*)
Or, ∆

*R/R =*∆*x/x*+ ∆*x*/(100 –*x*)
Since ∆

*x =*0.1 cm,*x =*40 cm and*R*= 60 Ω, we have
∆

*R/*60*=*(0.1/40) + (0.1/60)
Therefore,
∆

*R =*60[(0.1/40) + (0.1/60)] = 0.25 Ω
The unknown resistance is

*R*± ∆*R =*60 ± 0.25 Ω
How is delta x =0.1cm ? 0.1cm is the least count right?

ReplyDeleteIts the least count of the scale.

DeleteYes, indeed.

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