JEE Main 16 March 2021 Shift 2 Solved Paper

Given,
Heat energy, H1=500J
Initial current, I1=1.5A, final current, I2=3A
and time, t=20s
According to Joule's law of heating,
H=I2Rt
⇒‌‌H1=l12Rt.....(i)
and H2=I22Rt....(ii)
[ Resistance and time being the same in both cases]
Dividing Eq. (i) by Eq. (ii), we get
⇒‌‌‌

H1

H2

=‌

I12Rt

I22Rt

⇒‌

H1

H2

=(‌

I1

I2

)2
⇒‌‌‌

H1

H2

=(‌

1.5

3

)2=(‌

15

30

)2⇒‌

H1

H2

=(‌

1

2

)2
⇒‌‌‌

H1

H2

=‌

1

4

⇒‌

500

H2

=‌

1

4

⇒H2=500×4
⇒‌‌H2=2000J