Question

the coil of the heater is cut into two equal parts and one of them is used as heating element. What is the ratio of heat produced by this half coil to that of the original coil​

Answer 1

Answer:

We know that R=p L / A

Hence, on cutting half R = R by 2 .

And for same end potential difference as in house hold circuits-

H= V square / R.

⟹H∝ 1/2

⟹ H / H = R / R = 2

Hence the answer is 2:1.

Answer 2

$\sf\huge\bold{Answer:}$

Given :

the coil of the heater is cut into two equal parts and one of them is used as heating element.

To find :

the ratio of heat produced by this half coil to that of the original coil

H':H

Solution :

$\fbox{Resistance}$

$\bf R=\rho× \frac{L}{A}$

where,

R→Resistance of coil

ρ→Resistivity of material of coil

L→Length of coil

A→Area of coil

Resistance of new coil

$\rm R'= \frac{R}{2}$

where,

R'→Resistance of new coil

R→ Resistance of original coil

Heat produced by original coil

$\bf H= \frac{V²}{R}$

where,

H→Head produced by original coil

V→potential difference between two ends of coil

R→Resistance of original coil

Head produced by original coil(H) is directly proportional to potential difference between two ends of coil(V) and inversely proportional to Resistance of original coil (R)

$\bf H∝ \frac{1}{R}$

Now,

H→Head produced by original coil

H'→Head produced by new coil

R→Resistance of original coil

R'→Resistance of new coil

$:⟶\tt\ \frac{ \: \: H'}{H} = \frac{ \frac{1}{ \: \: R'} }{ \frac{1}{R} }$

$\tt\ \frac{ \: \: H'}{H} = \frac{R}{ \: \: R'}$

$\tt\ \frac{ \: \: H'}{H} = \frac{R}{ \frac{R}{2} }$

$\tt\ \frac{ \: \: H'}{H} = \frac{2R}{ R }$

$\tt{\frac{H'}{H} =}$$\displaystyle{\tt { \cancel{ \frac{2R}{R} }}}$

$\tt{ \frac{ \: \: H'}{H} = \frac{2}{1} }$

$\huge\mathrm{Ratio\:Is}$

$\huge\tt\purple{ \frac{ \: \: H'}{H} = \frac{2}{1} }$