CGPET 2021 Solved Paper

Conduction of heat in rod is given by relation

Q=KA.∆θ.t

l

where, K = coefficient of thermal conductivity of heat,
A = cross - sectional area,
∆θ = temperature difference,
t = time takn by heat flow
and l = length of rod.
If the radius of rod be r, then
A=πr2⇒Q=

K.πr2.∆θ.t

l

Rate of flow of heat through conductor,

Q

t

=

Kπr2∆θ

l

Here, K,π‌and∆θ are constants.
∴ Q∝

r2

l

For more value of Q,
(i) r should be maximum.
(ii) l should be minimum.
The value of (

r2

r

) for each observation,
(a) r=1‌cm=10−2m,l=1m
Q1∝

r2

l

=

(10−2)2

1

=10−4
(b) r=2‌cm=2×10−2m,l=2m
Q2∝

r2

l

=

(2×10−2)2

2

=2×10−4
(c) r=1‌cm=10−2m,l=

1

2

‌cm=

1

2

×10−2m
Q3∝

r2

l

=

(10−2)2

( 1 2 ×10−2)

=2×10−2
(d) r=2‌cm=2×10−2m,l=

1

2

m
Q4∝

r2

l

=

(2×10−2)2

1 2

=8×10−4
The value of

r2

l

is maximum for the dimensions given in option (c), hence it will conduct maximum heat.