# Tutor profile: Jishnu T.

## Questions

### Subject: Physics (Thermodynamics)

A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi. Determine the absolute pressure in the chamber.

Solution:The gage pressure of a vacuum chamber is given. The absolute pressure in the chamber is to be determined. The absolute pressure is easily determined from equation Pabs = Patm - Pvac =14.5-5.8 =8.7 psi Note that the local value of the atmospheric pressure is used when determining the absolute pressure.

### Subject: Physics (Fluid Mechanics)

A 0.6-mm-diameter glass tube is inserted into water at 20°C in a cup. Determine the capillary rise of water in the tube

Solution: The rise of water in a slender tube as a result of the capillary effect is to be determined. Assumptions: 1 There are no impurities in the water and no contamination on the surfaces of the glass tube. 2 The experiment is conducted in atmospheric air. Properties: The surface tension of water at 20°C is 0.073 N/m (Table 2–3). The contact angle of water with glass is 0° (from preceding text). We take the density of liquid water to be 1000 kg/m^3. Analysis: Capillary rise h=$$\frac{2*\sigma_{s} *cos\phi }{\rho gR}$$ =$$\frac{2*.073*cos0^{0}}{1000*9.81*0.3*10^{-3}}$$ =0.050m Therefore, water rises in the tube 5 cm above the liquid level in the cup. Discussion :Note that if the tube diameter were 1 cm, the capillary rise would be 0.3 mm, which is hardly noticeable to the eye. Actually, the capillary rise in a large-diameter tube occurs only at the rim. The center does not rise at all. Therefore, the capillary effect can be ignored for large-diameter tubes.

### Subject: Mechanical Engineering

Basic Strength of Material Question: A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in Newton) required to maintain equilibrium of the ladder is

Exp:Taking moments about B ∑ MB = 0 ⇒ W × 2 = R A × 3 ⇒ R A = (600 × 2)/3 = 400 ∑ H = 0 ⇒ R A = P = 400 N

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