Abrasive Jet Machining (AJM) MCQ Quiz in বাংলা - Objective Question with Answer for Abrasive Jet Machining (AJM) - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Apr 4, 2025

পাওয়া Abrasive Jet Machining (AJM) उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Abrasive Jet Machining (AJM) MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

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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