Abrasive Jet Machining (AJM) MCQ Quiz in मराठी - Objective Question with Answer for Abrasive Jet Machining (AJM) - मोफत PDF डाउनलोड करा

Last updated on Mar 20, 2025

पाईये Abrasive Jet Machining (AJM) उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Abrasive Jet Machining (AJM) एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

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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