Exams › NEET › Physics

If dimensions of critical velocity v₀ of a liquid flowing through a tube are expressed as [ηⁿρʳrᵐ], where η, ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by:

1. -1, -1, 1
2. -1, -1, -1
3. 1, 1, 1
4. 1, -1, -1

Correct answer: -1, -1, 1

Solution

The critical velocity depends on viscosity (η), density (ρ), and radius (r). Using dimensional analysis, equating the dimensions of velocity [L¹T⁻¹] with the given expression [ηⁿρʳrᵐ], we find n = -1, r = -1, and m = 1.

Related NEET Physics questions

• \( \mathbf{1}^{o} \boldsymbol{A} \) is
• Which of the following is different from the rest?
• Assertion nm is not same as mN Reason \( \ln m=10^{-9} m \operatorname{and} 1 m N=10^{-3} N \)
• If the dimension of a physical quantity is given by \( \left[M^{a} L^{b} T^{c}\right] \) then the physical quantity will be
• In a new system, unit of mass is \( 10 \mathrm{kg} \) unit of length is \( 5 \mathrm{m} \) and unit of time is 10s. Then in the new system 5N is equal to :
• A cube has a side of length \( 2.342 m \). The surface area and volume in correct significant figures are respectively:

⚔️ Practice NEET Physics free + battle 1v1 →