Coefficient of thermal conductivity is the product of heat, distance and reciprocal of (area × difference in temperature × time). The new value of unit of coefficient of thermal conductivity, if fundamental units are 21.6 kg, 1 decimetre, 4 K and 1 minute will be × 10-6 new units.

Question

Coefficient of thermal conductivity is the product of heat, distance and reciprocal of (area × difference in temperature × time). The new value of unit of coefficient of thermal conductivity, if fundamental units are 21.6 kg, 1 decimetre, 4 K and 1 minute will be × 10-6 new units. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Coefficient of thermal conductivity, K =(Q × l/A × Δ θ × t) K =((m × l2/t2) × l/l2 × Δ θ × t) K =(m × l/Δ θ × t3)=(21.6 × 0.1/4 ×(60)3) =2.5 × 10-6 new units

Q. Coefficient of thermal conductivity is the product of heat, distance and reciprocal of (area × difference in temperature × time). The new value of unit of coefficient of thermal conductivity, if fundamental units are 21.6kg,1 decimetre, 4K and 1 minute will be __________×10−6 new units.

Coefficient of thermal conductivity, K=A×Δθ×tQ×l​ K=l2×Δθ×tt2m×l2​×l​ K=Δθ×t3m×l​=4×(60)321.6×0.1​ =2.5×10−6 new units

Coefficient of thermal conductivity,
$K = \frac{Q \times l}{A \times Δ θ \times t}$
$K = \frac{\frac{m \times l ^{2}}{t ^{2}} \times l}{l ^{2} \times Δ θ \times t}$
$K = \frac{m \times l}{Δ θ \times t ^{3}} = \frac{21.6 \times 0.1}{4 \times ( 60 ) ^{3}}$
$= 2.5 \times 1 0^{- 6}$ new units