Abrasive Jet Machining (AJM) MCQ Quiz in தமிழ் - Objective Question with Answer for Abrasive Jet Machining (AJM) - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Mar 20, 2025

பெறு Abrasive Jet Machining (AJM) பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Abrasive Jet Machining (AJM) MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Abrasive Jet Machining (AJM) MCQ Objective Questions

Top Abrasive Jet Machining (AJM) MCQ Objective Questions

Abrasive Jet Machining (AJM) Question 1:

During the abrasive jet machining mixing ratio used is 0.2. Calculate mass ratio, if ratio of density of abrasive & density of carrier gas is equal to 20.

Answer (Detailed Solution Below) 0.8

Abrasive Jet Machining (AJM) Question 1 Detailed Solution

\(\begin{array}{l} Mixing\;ratio = \frac{{{V_a}}}{{{V_{gas}}}} = 0.2\\ Mass\;ratio\;\left( \alpha \right) = \frac{{{M_a}}}{{{M_{a + g}}}} = \frac{{{\rho _a}{V_a}}}{{{\rho _a}{V_a} + {\rho _g}{V_g}}} \end{array}\)

\(\begin{array}{l} \frac{1}{\alpha } = \frac{{{\rho _a}{V_a} + {\rho _g}{V_g}}}{{{\rho _a}{V_a}}}\\ = 1 + \frac{{{\rho _g}}}{{{\rho _a}}}\frac{{{V_g}}}{{{V_a}}} \end{array}\)

\(\begin{array}{l} = 1 + \frac{1}{{20}} \times \frac{1}{{0.2}}\\ \frac{1}{\alpha } = 1 + 0.25\\ \alpha = \frac{1}{{1.25}} = 0.8 \end{array}\)

Abrasive Jet Machining (AJM) Question 2:

In an abrasive jet machining (AJM) process, the metal removal rate (MRR) for Q flow rate of abrasives and d mean diameter of the abrasive is 10 mm3/s. If the flow rate of abrasives is doubled and the mean diameter of abrasives is halved, then the MRR will be _______ mm3/s.

Answer (Detailed Solution Below) 2.5

Abrasive Jet Machining (AJM) Question 2 Detailed Solution

In AJM given data,

Metal removal rate (MRR) = 10 mm3/sec

Flow rate = Q

Mean diameter of abrasive = d

We know,

\(MRR = KQ{d^3}{V^{\frac{3}{2}}}{\left( {\frac{l}{{2H}}} \right)^{\frac{3}{4}}}\)

∴ MRR α (Qd3)

If flow rate (Q) is double and ‘d’ is made half, then

\(\begin{array}{l} MRR' = 2Q\frac{{{d^3}}}{8} = \frac{{MRR}}{4}\\ \therefore MRR' = \frac{{10}}{4} = 2.5\;m{m^3}/sec \end{array}\)

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