Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
415 views
in General by (95.6k points)
closed by

The properties of mercury at 300 K are: density = 13529 kg/m3, specific heat at constant pressure = 0.1393 kJ/kg-K, dynamic viscosity = 0.1523 × 10-2 Ns/m2 and thermal conductivity = 8.540 W/m-K. the Prandtl number of mercury at 300 K is:


1. 0.0248
2. 2.48
3. 24.8
4. 248

1 Answer

0 votes
by (98.5k points)
selected by
 
Best answer
Correct Answer - Option 1 : 0.0248

Concept:

Prandtl number is the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{\nu }{\alpha } = \frac{μ }{{\frac{{ρ k}}{{ρ {C_p}}}}} = \frac{{μ {C_p}}}{k}\)

Calculation:

Given:

ρ = 13529 kg/m3, Cp = 0.1393 kJ/kg-K, μ = 0.1523 × 10-2 Ns/m2, k = 8.540 W/m-K

\(Pr = \frac{{μ {C_p}}}{k}\)

\(Pr = \frac{{0.1523\;\times\;10^{-2}\;\times\; {0.1393\;\times\;10^3}}}{8.540}=0.0248\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...