Economics of Machining MCQ Quiz in मल्याळम - Objective Question with Answer for Economics of Machining - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Given:

V₁ = 50 m/min, T₁ = 50 min, V₂ = 80 m/min, T₂ = 15 min, T_c = 10 min, C_t = Rs. 80, C_m = Rs. 5/min

Calculation:

\({V_1}T_1^n = {V_2}T_2^n\)

50(50)ⁿ = 80(15)ⁿ

∴ n = 0.39

For maximum productivity,

\({T_{opt}} = {T_c}\left( {\frac{{1 - n}}{n}} \right)\)

\({T_{opt}} = 10\left( {\frac{{1 - 0.39}}{{0.39}}} \right) = 15.64\;min\)

Now,

For maximum profit,

\({T_{opt}} = \left( {{T_c} + \frac{{{C_t}}}{{{C_m}}}} \right)\left( {\frac{{1 - n}}{n}} \right)\)

\({T_{opt}} = \left( {10 + \frac{{80}}{5}} \right)\left( {\frac{{1 - 0.39}}{{0.39}}} \right)\)

∴ T_opt = 40.66 minutes.

The correct option is 1. In the given assertion and reason, both statements are true and the reason correctly explains the assertion. Manufacturing cells typically reduce material handling costs because machines are arranged in sequences that follow the process flow, minimizing the distance materials travel. The reason statement supports this by indicating that when machines are not located in sequence, it leads to increased material handling costs. Therefore, proper sequencing of machines in a manufacturing cell is crucial for cost efficiency.

The concept involved in the question pertains to the classification of data types within an industrial or production context. The focus is on identifying the type of data that encompasses metrics related to the functioning and performance of a production system. This includes monitoring variables such as the rate at which products are manufactured, the efficiency of production processes, and the consumption of resources necessary for production. Understanding these data types is crucial for optimizing operations, improving productivity, and reducing costs.

Concept:

For maximum productivity,

\({\rm{Tool\;life\;}}\left( {{T_0}} \right) = {T_c}\left( {\frac{{1 - n}}{n}} \right)\)

Calculation:

Given:

Tc = 6 min, n = 0.3, c = 200

\({\rm{Tool\;life\;}}\left( {{T_0}} \right) = {T_c}\left( {\frac{{1 - n}}{n}} \right)\)

\( {{T_0}} = {6}\left( {\frac{{1 -0.4}}{0.4}} \right)\)

To = 9 minutes

Explanation:

• Expanses of production refers to the total money needed for the production of a particular product.
• These expanses  are usually classified as direct or indirect costs.
• Direct costs tend to be variable costs, while indirect costs are more likely to be either fixed costs or period costs.
• Examples of direct costs are direct labor, direct materials, commissions, piece rate wages, and manufacturing supplies.
• Examples of indirect costs are production supervision salaries, quality control costs, insurance, and depreciation.

For Maximum production rate or Minimum production time,

\(Tool\;Life\; = \;{T_c}\left( {\frac{{1 - n}}{n}} \right)\; = \;2\left( {\frac{{1 - 0.25}}{{0.25}}} \right)\; = \;6min\; = \;360sec\) 

Minimum cost implies for minimum tool wear. Maximum production rate means higher speed irrespective of tool wear.

∴ V_{r max} > V_{c min}

From graph

V₁ = 120 m/min

T₁ = 10 min

V₂ = 80 m/min

T₂ = 50 min

\(\eqalign{ & \therefore V{T^n} = C \cr &\therefore {{\rm{V}}_1}{\rm{T}}_1^{\rm{n}} = {{\rm{V}}_2}{\rm{T}}_2^{\rm{n}} \cr & \therefore \ln {V_1} + n\ln {T_1} = \ln {V_2} + n \ln {T_2} \cr &\therefore {\rm{n}} = {{ \ln \left( {{{{{\rm{V}}_2}} \over {{{\rm{V}}_1}}}} \right)} \over { \ln \left( {{{{{\rm{T}}_1}} \over {{{\rm{T}}_2}}}} \right)}} \cr & \therefore {\rm{n}} = {{ \ln \left( {{{80} \over {120}}} \right)} \over {\ln \left( {{{10} \over {50}}} \right)}} = 0.252 \cr & C = {{\rm{V}}_1}{\rm{T}}_1^{\rm{n}} = {{\rm{V}}_2}{\rm{T}}_2^{\rm{n}} = 214.37{\rm{m}} \cr}\)

Economical speed / optimum speed base on cost criteria.

\({V_{opt}} = C{\left\{ {{n \over {1 - n}} \times {{{L_m}} \over {{C_g}}}} \right\}^n}\)

L_m = Labour charge per min.

C_g = cost of regrind.

\({V_{opt}} = 214.37{\left\{ {{{0.252} \over {1 - 0.252}} \times {{{{500} \over {60}}} \over {45}}} \right\}^{0.252}} = 106.54\ m/\min \)