Manipal Entrance Test 2020 Physics Paper

Let m be the mass of the given wire, d its density and p its resistivity. These quantities remain unchanged when the wire is drawn out. Let l1 be the length of the smaller wire and r1 its radius of cross section and If I2 and r2 represent the length and radius of cross section of the elongated wire we have m=πr12l2d=πr22I2d r12I1=r22I2
It is given that I2=3l1
Therefore r22=‌

r12

3

If R1 is the resistance of the smaller wire and R2 that of the elongated wire we have R1=‌

pl

πr12

R2=‌

pl‌l2

πr22

‌

R2

R1

=‌

l2

l1

×‌

r12

r22

=9
R1=5‌ohm
R2=45‌ohm