100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 573 InterviewSolution

28601.	If the camber on the two front wheels is not equal, the vehicle will try to pull towards the side where.a. The camber is higher. b. The camber is lower c. Both (a) & (b) d. None of the above
Answer» Correct option: a. The camber is higher

Discussion

28602.	Generally, when the temperature is increased, the viscosity of oil will _______. a. remains same. b. increase. c. decrease. d. None of above.
Answer» Correct option: c. decrease

Discussion

28603.	Name the component of the cooling system which helps to increase the pressure inside the radiator and raise the boiling point of coolant.
Answer» Radiator’s Pressure cap or Pressure cap helps to increase the pressure inside the radiator and raise the boiling point of coolant.

Discussion

28604.	A 1.25 cm diameter steel bar is subjected to a load of 2500 kg. The stress induced. in the bar will be (a) 200 MPa (b) 210 MPa (c) 220 MPa (d) 230 MPa
Answer» Correct option is (a) 200 MPa Axial stress, σ = \(\cfrac PA\) = \(\cfrac{2500\times9.81}{\cfrac{\pi}4(12.5)^2}\) = 200 MPa

28604.

A 1.25 cm diameter steel bar is subjected to a load of 2500 kg. The stress induced. in the bar will be (a) 200 MPa (b) 210 MPa (c) 220 MPa (d) 230 MPa

Answer»

Correct option is (a) 200 MPa

Axial stress, σ = \(\cfrac PA\) = \(\cfrac{2500\times9.81}{\cfrac{\pi}4(12.5)^2}\)

= 200 MPa

Discussion

28605.	Which of the following are the sources of motivation and inspiration____ a. Music b. Books c. Both of the above
Answer» Which of the following are the sources of motivation and inspiration Both of the above.

Discussion

28606.	The maximum energy which can be stored in a body up to the elastic limit is called. (a) Proof resilience (b) Modulus of resilience (c) Impact toughness (d) Endurance strength
Answer» (a) Proof resilience The maximum energy which can be stored in a body upto elastic limit, is called proof resilience.

28606.

The maximum energy which can be stored in a body up to the elastic limit is called. (a) Proof resilience (b) Modulus of resilience (c) Impact toughness (d) Endurance strength

Answer»

(a) Proof resilience

The maximum energy which can be stored in a body upto elastic limit, is called proof resilience.

Discussion

28607.	Listening is _______ Process. a. Active b. Passive c. Bothe the above
Answer» Correct option: a. Active

Discussion

28608.	A rubber band extends 50 mm when a force of 300 N is applied to it. Assuming the band is within the elastic limit, determine the extension produced by a force of 60 N.
Answer» Hooke's law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x α F or x = kF where k is a constant. When x = 50 mm, F = 300 N, thus 50 = k(300), from which, constant k = \(\cfrac{50}{300}\) = \(\cfrac{1}6\) When F = 60 N, then x = k(60) i.e. x = \(\left(\cfrac{1}6\right)\) (60) = 10 mm Thus, a force of 60 N stretches the wire 10 mm.

28608.

A rubber band extends 50 mm when a force of 300 N is applied to it. Assuming the band is within the elastic limit, determine the extension produced by a force of 60 N.

Answer»

Hooke's law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x α F or x = kF where k is a constant.

When x = 50 mm, F = 300 N, thus 50 = k(300), from which, constant k = \(\cfrac{50}{300}\) = \(\cfrac{1}6\)

When F = 60 N, then x = k(60) i.e. x = \(\left(\cfrac{1}6\right)\) (60) = 10 mm

Thus, a force of 60 N stretches the wire 10 mm.

Discussion

28609.	A wire is stretched 1.5 mm by a force of 300 N. Determine the force that would stretch the wire 4 mm, assuming the elastic limit of the wire is not exceeded.
Answer» Hooke's law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x α F or x = kF where k is a constant. When x = 1.5 mm, F = 300 N, thus 1.5 = k(300), from which, constant k = \(\cfrac{1.5}{300}\) = 0..005 When x = 4 mm, then 4 = kF i.e. 4 = 0.005 F from which, force F = \(\cfrac{4}{0.005}\) = 800 N Thus to stretch the wire 4 mm, a force of 800 N is required.

28609.

A wire is stretched 1.5 mm by a force of 300 N. Determine the force that would stretch the wire 4 mm, assuming the elastic limit of the wire is not exceeded.

Answer»

Hooke's law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x α F or x = kF where k is a constant.

When x = 1.5 mm, F = 300 N, thus 1.5 = k(300), from which, constant k = \(\cfrac{1.5}{300}\) = 0..005

When x = 4 mm, then 4 = kF i.e. 4 = 0.005 F

from which, force F = \(\cfrac{4}{0.005}\) = 800 N

Thus to stretch the wire 4 mm, a force of 800 N is required.

Discussion

28610.	Write the classification of spark plug according to the heat range.
Answer» The spark plugs are classified according to the temperature range, which are of two types and they are designated as hot plug and cold plug. It will be seen that the hot plug has a longer heat path giving delayed cooling than the cold plug. In general, it may be remembered that the hot plugs have a much longer insulator nose than the cold plugs.

28610.

Write the classification of spark plug according to the heat range.

Answer»

The spark plugs are classified according to the temperature range, which are of two types and they are designated as hot plug and cold plug. It will be seen that the hot plug has a longer heat path giving delayed cooling than the cold plug.

In general, it may be remembered that the hot plugs have a much longer insulator nose than the cold plugs.

Discussion

28611.	A cast iron bed plate for a pump has a crack length of 100. µm. If the Young’s modulus of cast iron is 210 GN/m2 and the specific surface energy is 10 J/m2, the fracture strength required will be nearly (a) 1.0 × 108 N/m2 (b) 1.2 × 108 N/m2 (c) 1.4 × 108 N/m2 (d) 1.6 × 108 N/m2
Answer» Correct option is (d) 1.6 × 10⁸ N/m² Crack length = 2c = 100 µm E = 210 GN/m² = 210 × 109 N/m² Surface energy, γ = 10 J/m² According to Griffith’s theory of brittle fracture, Fracture strength, σ = \(\left(\cfrac{2Eγ}{\pi c}\right)^{1/2}\) = \(\left[\cfrac{2\times(210\times10^9)\times10}{\pi\times50\times10^{-6}m}\right]^{1/2}\) = 1.6 × 10⁸ N/m²

28611.

A cast iron bed plate for a pump has a crack length of 100. µm. If the Young’s modulus of cast iron is 210 GN/m2 and the specific surface energy is 10 J/m2, the fracture strength required will be nearly (a) 1.0 × 108 N/m2 (b) 1.2 × 108 N/m2 (c) 1.4 × 108 N/m2 (d) 1.6 × 108 N/m2

Answer»

Correct option is (d) 1.6 × 10⁸ N/m²

Crack length = 2c = 100 µm

E = 210 GN/m² = 210 × 109 N/m²

Surface energy, γ = 10 J/m²

According to Griffith’s theory of brittle fracture,

Fracture strength, σ = \(\left(\cfrac{2Eγ}{\pi c}\right)^{1/2}\)

= \(\left[\cfrac{2\times(210\times10^9)\times10}{\pi\times50\times10^{-6}m}\right]^{1/2}\)

= 1.6 × 10⁸ N/m²

Discussion

28612.	A rectangular block of plastic material 400 mm long by 15 mm wide by 300 mm high has its lower face fixed to a bench and a force of 150 N is applied to the upper face and in line with it. The upper face moves 12 mm relative to the lower face. Determine (a) the shear stress, and (b) the shear strain in the upper face, assuming the deformation is uniform.
Answer» (a) Shear stress, \(\tau\) = \(\cfrac{force}{area\, parallel \,to \,the\,force}\) Area of any face parallel to the force = 400 mm x 15 mm = (0.4 x 0.015) m² = 0.006 m² Hence, shear stress \(\tau\) = \(\cfrac{150\,N}{0.006\,m^2}\) = 25000 Pa or 25 kPa (b) Shear strain, \(\gamma\) = \(\cfrac XL\) = \(\cfrac{12}{300}\) = 0.04 (or 4%)

28612.

A rectangular block of plastic material 400 mm long by 15 mm wide by 300 mm high has its lower face fixed to a bench and a force of 150 N is applied to the upper face and in line with it. The upper face moves 12 mm relative to the lower face. Determine (a) the shear stress, and (b) the shear strain in the upper face, assuming the deformation is uniform.

Answer»

(a) Shear stress, \(\tau\) = \(\cfrac{force}{area\, parallel \,to \,the\,force}\)

Area of any face parallel to the force = 400 mm x 15 mm

= (0.4 x 0.015) m² = 0.006 m²

Hence, shear stress \(\tau\) = \(\cfrac{150\,N}{0.006\,m^2}\) = 25000 Pa or 25 kPa

(b) Shear strain, \(\gamma\) = \(\cfrac XL\)

= \(\cfrac{12}{300}\) = 0.04 (or 4%)

Discussion

28613.	A force of 25 kN applied to a piece of steel produces an extension of 2 mm. Assuming the elastic limit is not exceeded, determine (a) the force required to produce an extension of 3.5 mm, (b) the extension when the applied force is 15 kN.
Answer» From Hooke’s law, extension x is proportional to force F within the limit of proportionality, i.e. x α F or x = kF, where k is a constant. If a force of 25 kN produces an extension of 2 mm, then 2 = k(25), from which, constant k = \(\cfrac{2}{25}\) = 0.08 (a) When an extension x = 3.5 mm, then 3.5 = k(F), i.e. 3.5 = 0.08 F, from which, force F = \(\cfrac{3.5}{0.08}\) = 43.75 kN (b) When force F = 15 kN, then extension x = k(15) = (0.08)(15) = 1.2 mm

28613.

A force of 25 kN applied to a piece of steel produces an extension of 2 mm. Assuming the elastic limit is not exceeded, determine (a) the force required to produce an extension of 3.5 mm, (b) the extension when the applied force is 15 kN.

Answer»

From Hooke’s law, extension x is proportional to force F within the limit of proportionality, i.e.

x α F or x = kF, where k is a constant. If a force of 25 kN produces an extension of 2 mm, then

2 = k(25), from which, constant k = \(\cfrac{2}{25}\) = 0.08

(a) When an extension x = 3.5 mm, then 3.5 = k(F), i.e. 3.5 = 0.08 F,

from which, force F = \(\cfrac{3.5}{0.08}\) = 43.75 kN

(b) When force F = 15 kN, then extension x = k(15) = (0.08)(15) = 1.2 mm

Discussion

28614.	A 13 mm diameter tensile specimen has 50 mm gauge length. If the load corresponding. to the 0.2% offset is 6800 kg, the yield stress will be nearly (a) 31 kg/mm2 (b) 43 kg/mm2 (c) 51 kg/mm2 (d) 63 kg/mm2
Answer» Correct option is (c) 51 kg/mm² Yield stress, σ = \(\cfrac{6800}{\cfrac{\pi}{4}\times(13)^2}\) = 51.25 kg/mm²

28614.

A 13 mm diameter tensile specimen has 50 mm gauge length. If the load corresponding. to the 0.2% offset is 6800 kg, the yield stress will be nearly (a) 31 kg/mm2 (b) 43 kg/mm2 (c) 51 kg/mm2 (d) 63 kg/mm2

Answer»

Correct option is (c) 51 kg/mm²

Yield stress, σ = \(\cfrac{6800}{\cfrac{\pi}{4}\times(13)^2}\)

= 51.25 kg/mm²

Discussion

28615.	The graph shown in the represents the emission of a pollutant from an SI engine for different fuel/air ratios. The pollutant in question is (a) CO (b) CO2 (c) hydrocarbon (d) NOx.
Answer» Correct option is (d) NO_x.

Discussion

28616.	A pipe has an outside diameter of 20 mm, an inside diameter of 10 mm and length 0.30 m and it supports a compressive load of 50 kN. The pipe shortens by 0.6 mm when the load is applied. Determine (a) the compressive stress, (b) the compressive strain in the pipe when supporting this load.
Answer» Compressive force F = 50 kN = 50000 N, and cross-sectional area A = \(\cfrac{\pi}4\)(D²- d²) where D = outside diameter = 20 mm and d = inside diameter = 10 mm. Hence, A = \(\cfrac{\pi}4\) (20² - 10²) mm² = \(\cfrac{\pi}4\) (20² - 10² ) x 10^-6 m² = 2.3562 x 10^-4 m² (a) Compressive stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{50000\,N}{2.3562\times10^{-4}m^2}\) = 212.2 x10⁶ Pa = 212.2 MPa (b) Contraction of pipe when loaded, x = 0.6 mm = 0.0006 m, and original length L = 0.30 m. Hence, compressive strain \(\varepsilon\) = \(\cfrac XL\) = \(\cfrac{0.0006}{0.3}\) = 0.002 (or 0.20%)

28616.

A pipe has an outside diameter of 20 mm, an inside diameter of 10 mm and length 0.30 m and it supports a compressive load of 50 kN. The pipe shortens by 0.6 mm when the load is applied. Determine (a) the compressive stress, (b) the compressive strain in the pipe when supporting this load.

Answer»

Compressive force F = 50 kN = 50000 N, and cross-sectional area A = \(\cfrac{\pi}4\)(D²- d²)

where D = outside diameter = 20 mm and d = inside diameter = 10 mm.

Hence, A = \(\cfrac{\pi}4\) (20² - 10²) mm² = \(\cfrac{\pi}4\) (20² - 10² ) x 10^-6 m² = 2.3562 x 10^-4 m²

(a) Compressive stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{50000\,N}{2.3562\times10^{-4}m^2}\) = 212.2 x10⁶ Pa = 212.2 MPa

(b) Contraction of pipe when loaded, x = 0.6 mm = 0.0006 m, and original length L = 0.30 m.

Hence, compressive strain \(\varepsilon\) = \(\cfrac XL\) = \(\cfrac{0.0006}{0.3}\) = 0.002 (or 0.20%)

Discussion

28617.	To get an illness from someone is to ______. A) pick it up B) truck it in C) take it away
Answer» Correct option is A) pick it up

Discussion

28618.	Consider the following statements : The maximum temperature produced by the combustion of an unit mass of fuel depends upon 1. LCV 2. ash content 3. mass of air supplied 4. pressure in the furnace. Of these statements : (a) 1 alone is correct (b) 1 and 3 are correct (c) 2 and 4 are correct (d) 3 and 4 are correct.
Answer» (b) 1 and 3 are correct

Discussion

28619.	An 80 cm long bar contracts axially by 0.2 mm when a compressive load is applied to it. Determine the strain and the percentage strain.
Answer» Strain \(\varepsilon\) = \(\cfrac{extension}{original\, lengh}\) = \(\cfrac{0.2\,mm}{800\,mm}\) = 0.00025 Percentage strain = 0.00025 x 100 = 0.025%

28619.

An 80 cm long bar contracts axially by 0.2 mm when a compressive load is applied to it. Determine the strain and the percentage strain.

Answer»

Strain \(\varepsilon\) = \(\cfrac{extension}{original\, lengh}\) = \(\cfrac{0.2\,mm}{800\,mm}\) = 0.00025

Percentage strain = 0.00025 x 100 = 0.025%

Discussion

28620.	To heat food again that has already been cooked is to ______. A) ruffle it up B) warm it up C) pick it up
Answer» Correct option is B) warm it up

Discussion

28621.	Consider the following statements : The difference between higher and lower heating values of the fuels is due to 1. heat carried by steam from the moisture content of fuel. 2. sensible heat carried away by the flue gases. 3. heat carried away by steam from the combustion of hydrogen in the fuel. 4. heat lost by radiation. Of these statements : (a) 2, 3 and 4 are correct (b) 1 and 2 are correct (c) 3 alone is correct (d) 1, 2, 3 and 4 are correct.
Answer» (c) 3 alone is correct

Discussion

28622.	An ideal cycle is shown in the . Its thermal efficiency is given by(a) 1 - \(\cfrac{\left(\cfrac{v_2}{v_1}-1\right)}{\left(\cfrac{p_2}{p_1}-1\right)}\)(b) 1 - \(\cfrac 1γ\) \(\cfrac{\left(\cfrac{v_3}{v_1}-1\right)}{\left(\cfrac{p_2}{p_1}-1\right)}\)(c) 1 - γ \(\cfrac{(v_3-v_1)}{(p_2-p_1)}\) \(\cfrac{p_1}{v_1}\)(d) 1 - \(\cfrac 1γ\) \(\cfrac{(p_3-p_1)}{(v_3-v_1)}\) \(\cfrac{v_1}{p_1}\)
Answer» (c) 1 - γ \(\cfrac{(v_3-v_1)}{(p_2-p_1)}\) \(\cfrac{p_1}{v_1}\)

Discussion

28623.	A single cylinder, four-stroke cycle oil engine is fitted with a rope brake. The diameter of the brake wheel is 600 mm and the rope diameter is 26 mm. The dead load on the brake is 200 N and the spring balance reads 30 N. If the engine runs at 600 rpm, what will be the nearest magnitude of the brake power of the engine?(a) 3.3 kW (b) 5.2 kW(c) 7.3 kW (d) 7.2 kW
Answer» (a) 3.3 kW r = \(\cfrac{600+26}2\) = 313 mm bp = \(\cfrac{2\pi N(W-S)r}{60\times1000}\) = \(\cfrac{2\pi \times 600\times(170)\times313}{60\times1000}\) = 3.343 kW

28623.

A single cylinder, four-stroke cycle oil engine is fitted with a rope brake. The diameter of the brake wheel is 600 mm and the rope diameter is 26 mm. The dead load on the brake is 200 N and the spring balance reads 30 N. If the engine runs at 600 rpm, what will be the nearest magnitude of the brake power of the engine?(a) 3.3 kW (b) 5.2 kW(c) 7.3 kW (d) 7.2 kW

Answer»

(a) 3.3 kW

r = \(\cfrac{600+26}2\) = 313 mm

bp = \(\cfrac{2\pi N(W-S)r}{60\times1000}\)

= \(\cfrac{2\pi \times 600\times(170)\times313}{60\times1000}\)

= 3.343 kW

Discussion

28624.	A circular section bar is 2.5 m long and has a diameter of 60 mm. When subjected to a compressive load of 30 kN it shortens by 0.20 mm. Determine Young's modulus of elasticity for the material of the bar.
Answer» Force, F = 30 kN = 30000 N and cross-sectional area A \(\pi\)r² = \(\left(\cfrac{60\times10^{-3}}2\right)^2\) = 2.8274 x 10^-3 m² Stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{30000}{2.8274\times10^{-3}}\) = 10.61 MPa Bar shortens by 0.20 mm = 0.00020 m Stress \(\sigma\) = \(\cfrac XL\) = \(\cfrac{0.00020}{2.5}\) = 0.00008a Modulus of elasticity, E = \(\cfrac{stress}{strain}\) = \(\cfrac{10.61\times10^6}{0.00008}\) = 132.6 x 10⁹ = 132 .6 Gpa

28624.

A circular section bar is 2.5 m long and has a diameter of 60 mm. When subjected to a compressive load of 30 kN it shortens by 0.20 mm. Determine Young's modulus of elasticity for the material of the bar.

Answer»

Force, F = 30 kN = 30000 N and cross-sectional area A \(\pi\)r² = \(\left(\cfrac{60\times10^{-3}}2\right)^2\)

= 2.8274 x 10^-3 m²

Stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{30000}{2.8274\times10^{-3}}\) = 10.61 MPa

Bar shortens by 0.20 mm = 0.00020 m

Stress \(\sigma\) = \(\cfrac XL\) = \(\cfrac{0.00020}{2.5}\) = 0.00008a

Modulus of elasticity, E = \(\cfrac{stress}{strain}\) = \(\cfrac{10.61\times10^6}{0.00008}\)

= 132.6 x 10⁹ = 132 .6 Gpa

Discussion

28625.	A metal bar 2.5 m long extends by 0.05 mm when a tensile load is applied to it. Determine (a) the strain, (b) the percentage strain.
Answer» (a) Strain \(\varepsilon\) = \(\cfrac{extension}{original\, lengh}\) = \(\cfrac{0.05\,mm}{2.5\times10^3\,mm}\) = \(\cfrac{0.05}{2500}\) = 0.00002 (b) Percentage strain = 0.00002  100 = 0.002%

28625.

A metal bar 2.5 m long extends by 0.05 mm when a tensile load is applied to it. Determine (a) the strain, (b) the percentage strain.

Answer»

(a) Strain \(\varepsilon\) = \(\cfrac{extension}{original\, lengh}\) = \(\cfrac{0.05\,mm}{2.5\times10^3\,mm}\)

= \(\cfrac{0.05}{2500}\) = 0.00002

(b) Percentage strain = 0.00002  100 = 0.002%

Discussion

28626.	To drink a lot of alcohol is to ______. A) knock it over B) knock it in C) knock it back
Answer» Correct option is C) knock it back

Discussion

28627.	Consider the following statements regarding Otto cycle : 1. It is not a reversible cycle. 2. Its efficiency can be improved by using a working fluid of higher value of ratio of specific heats. 3. The practical way of increasing its efficiency is to increase the compression ratio. 4. Carburetted gasoline engines working on Otto cycle can work with compression ratios more than 12. Of these statements : (a) 1, 3 and 4 are correct (b) 1, 2 and 3 are correct (c) 1, 2 and 4 are correct (d) 2, 3 and 4 are correct.
Answer» (b) 1, 2 and 3 are correct

Discussion

28628.	A bar of thickness 20 mm and having a rectangular cross-section carries a load of 82.5 kN. Determine (a) the minimum width of the bar to limit the maximum stress to 150 MPa, (b) the modulus of elasticity of the material of the bar if the 150 mm long bar extends by 0.8 mm when carrying a load of 200 kN.
Answer» (a) Force, F = 82.5 kN = 82500 N and cross-sectional area A = (20x)10^-6 m² , where x is the width of the rectangular bar in millimetres. Stress \(\sigma\) = \(\cfrac FA\), from which, A = \(\cfrac F{\sigma}\) = \(\cfrac{82500\,N}{150\times10^6\,pa}\) = 5.5 x10^-4 m² = 5.5 x 10^-4 x 10⁶ mm² = 5.5 x10² mm²= 550 mm² Hence, 550 = 20x, from which, width of bar, x = \(\cfrac{550}{20}\) = 27.5 mm (b) Stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{200000}{550\times10^{-6}}\) = 363.64 MPa Extension of bar = 0.8 mm Strain \(\varepsilon\) = \(\cfrac XL\) = \(\cfrac{0.8}{150}\) = 0.005333 Modulus of elasticity, E = \(\cfrac{stress}{strain}\) = \(\cfrac{363.64\times10^6}{0.005333}\) = 68.2 x 10⁹ = 68.2 GPa

28628.

A bar of thickness 20 mm and having a rectangular cross-section carries a load of 82.5 kN. Determine (a) the minimum width of the bar to limit the maximum stress to 150 MPa, (b) the modulus of elasticity of the material of the bar if the 150 mm long bar extends by 0.8 mm when carrying a load of 200 kN.

Answer»

(a) Force, F = 82.5 kN = 82500 N and cross-sectional area A = (20x)10^-6 m² ,

where x is the width of the rectangular bar in millimetres.

Stress \(\sigma\) = \(\cfrac FA\), from which, A = \(\cfrac F{\sigma}\) = \(\cfrac{82500\,N}{150\times10^6\,pa}\) = 5.5 x10^-4 m² = 5.5 x 10^-4 x 10⁶ mm²

= 5.5 x10² mm²= 550 mm²

Hence, 550 = 20x, from which, width of bar, x = \(\cfrac{550}{20}\) = 27.5 mm

(b) Stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{200000}{550\times10^{-6}}\) = 363.64 MPa

Extension of bar = 0.8 mm

Strain \(\varepsilon\) = \(\cfrac XL\) = \(\cfrac{0.8}{150}\) = 0.005333

Modulus of elasticity, E = \(\cfrac{stress}{strain}\) = \(\cfrac{363.64\times10^6}{0.005333}\) = 68.2 x 10⁹ = 68.2 GPa

Discussion

28629.	They were looking for a good _____ car. A) use B) used C) to use D) using
Answer» Correct option is B) used

Discussion

28630.	What “H word” is a tool used to knock nails in with?
Answer» Correct option is hammer

Discussion

28631.	In an Otto cycle, air is compressed from 2.2 l to 2.26 l from an initial pressure of 1.2 kg/cm2. The net output/cycle is 440 kJ. What is the mean effective pressure of the cycle? (a) 227 kPa (b) 207 kPa (c) 192 kPa (d) 185 kPa
Answer» (a) 227 kPa Mean effective pressure = \(\cfrac{Net \,work \,output}{Swept \,volume}\) = \(\cfrac{440}{(2.2-0.26)\times10^{-3}}\) = 226.8 kpa = 227 kpa

28631.

In an Otto cycle, air is compressed from 2.2 l to 2.26 l from an initial pressure of 1.2 kg/cm2. The net output/cycle is 440 kJ. What is the mean effective pressure of the cycle? (a) 227 kPa (b) 207 kPa (c) 192 kPa (d) 185 kPa

Answer»

(a) 227 kPa

Mean effective pressure

= \(\cfrac{Net \,work \,output}{Swept \,volume}\)

= \(\cfrac{440}{(2.2-0.26)\times10^{-3}}\)

= 226.8 kpa

= 227 kpa

Discussion

28632.	Aduct of rectangular cross-section 600 mm x 400 mm carries 90 m3/min of air having density of 1.2 kg/m3. When the quantity of air in both cases is same, the equivalent diameter of a circular duct will be nearly (a) 0.86 m (b) 0.76 m (c) 0.64 m (d) 0.54 m
Answer» (d) 0.54 m A₁ = A₂ 0.6 × 0.4 = \(\cfrac{\pi}4\) x D² = 0.55 m

28632.

Aduct of rectangular cross-section 600 mm x 400 mm carries 90 m3/min of air having density of 1.2 kg/m3. When the quantity of air in both cases is same, the equivalent diameter of a circular duct will be nearly (a) 0.86 m (b) 0.76 m (c) 0.64 m (d) 0.54 m

Answer»

(d) 0.54 m

A₁ = A₂

0.6 × 0.4 = \(\cfrac{\pi}4\) x D²

= 0.55 m

Discussion

28633.	Another word to describe a good-looking man is ______
Answer» Correct option is handsome

Discussion

28634.	Membrane ion channels A. Consist mainly of carbohydrate and lipid. B. Have a specific structure for each ion species. C. For sodium may be blocked by tetrodotoxin. D. May be opened by a given change in transmembrane potential.E. Remain open as long as the activating signal is present.
Answer» A. False They are usually proteins. B. False There is usually a family of channels for each ion species. C. True This toxin from the puffer fish specifically blocks sodium channels. D. True These channels are ‘voltage-gated’ channels; ‘ligand-gated’ channels are opened by chemical substances such as acetylcholine. E. False They close automatically after a certain interval.

28634.

Membrane ion channels A. Consist mainly of carbohydrate and lipid. B. Have a specific structure for each ion species. C. For sodium may be blocked by tetrodotoxin. D. May be opened by a given change in transmembrane potential.E. Remain open as long as the activating signal is present.

Answer»

A. False They are usually proteins.

B. False There is usually a family of channels for each ion species.

C. True This toxin from the puffer fish specifically blocks sodium channels.

D. True These channels are ‘voltage-gated’ channels; ‘ligand-gated’ channels are opened by chemical substances such as acetylcholine.

E. False They close automatically after a certain interval.

Discussion

28635.	The vehicles which are generally running in hilly-road, are given. a. Positive camber b. Zero camber c. Negative camber d. None of the above
Answer» Correct option: c. Negative camber

Discussion

28636.	In a two-stroke Petrol engine, fuel loss is maximum after :(a) Opening the exhaust port(b) Closing the exhaust port(c) Opening the inlet port(d) Closing the inlet port
Answer» (a) Opening the exhaust port With engines working on Otto cycle, a part of fresh mixture is lost as it escapes through the exhaust port during scavenging. This increase the fuel comsumption and reduces the thermal efficiency. If the fuel is supplied to the cylinder after the exhaust ports are closed, there will be no loss of fuel.

28636.

In a two-stroke Petrol engine, fuel loss is maximum after :(a) Opening the exhaust port(b) Closing the exhaust port(c) Opening the inlet port(d) Closing the inlet port

Answer»

(a) Opening the exhaust port

With engines working on Otto cycle, a part of fresh mixture is lost as it escapes through the exhaust port during scavenging. This increase the fuel comsumption and reduces the thermal efficiency.

If the fuel is supplied to the cylinder after the exhaust ports are closed, there will be no loss of fuel.

Discussion

28637.	For a private new vehicle the initial fitness certificate is valid for a) 20 years b) 15 years c) 10 years d) 5 years
Answer» Correct option: b) 15 years

Discussion

28638.	For a commercial new vehicle the initial fitness certificate is valid for ________ a) 15 years b) 10 years c) 5 yearsd) 2 years
Answer» d) 2 years For a commercial new vehicle the initial fitness certificate is valid for 2 years.

28638.

For a commercial new vehicle the initial fitness certificate is valid for ________ a) 15 years b) 10 years c) 5 yearsd) 2 years

Answer»

d) 2 years

For a commercial new vehicle the initial fitness certificate is valid for 2 years.

Discussion

28639.	Write any four hair accessories are used to make a good hair style.
Answer» 1. Hair pins, 2. False hair, 3.Tiara, 4. Flowers.

28639.

Write any four hair accessories are used to make a good hair style.

Answer»

1. Hair pins,

2. False hair,

3.Tiara,

4. Flowers.

Discussion

28640.	It’s worn on your head for either fashion or warmth. It’s a ______
Answer» Correct option is hat

Discussion

28641.	Aroom having dimensions of 5 m x 5 m x 3 m contains air at 25°C and 100 kPa at a relative humidity of 75%. The corresponding value of ps is 3.169 kPa. The partial pressure of dry air will be nearly (a) 106 kPa (b) 98 kPa (c) 86 kPa (d) 78 kPa
Answer» (b) 98 kPa φ = \(\cfrac{p_v}{p_s}\) 0.75 = \(\cfrac{p_v}{3.169}\) P_v = 2.376 kPa P_t = P_a + P_v 101.3 = P_a + 2.376 P_a = 98.92 kPa

28641.

Aroom having dimensions of 5 m x 5 m x 3 m contains air at 25°C and 100 kPa at a relative humidity of 75%. The corresponding value of ps is 3.169 kPa. The partial pressure of dry air will be nearly (a) 106 kPa (b) 98 kPa (c) 86 kPa (d) 78 kPa

Answer»

(b) 98 kPa

φ = \(\cfrac{p_v}{p_s}\)

0.75 = \(\cfrac{p_v}{3.169}\)

P_v = 2.376 kPa

P_t = P_a + P_v

101.3 = P_a + 2.376

P_a = 98.92 kPa

Discussion

28642.	The octet rule is not valid for (a) CO2 (b) H2O (c) O2(d) CO
Answer» The octet rule is not valid for CO

Discussion

28643.	The distributor serves the following purposes in the ignition system.a. It operates break and make mechanism b. It distributes high tension current to spark plug at correct time c. both (a) and (b)d. None of above
Answer» Correct option: c. both (a) and (b)

Discussion

28644.	Emission includes_______ a) NOx b) SOx c) Hydrocarbons d) All of the above
Answer» Correct answer is d) All of the above

28644.

Emission includes_______ a) NOx b) SOx c) Hydrocarbons d) All of the above

Answer»

Correct answer is

d) All of the above

Discussion

28645.	The hydrogen ion concentration in a solution A. Equals the hydroxyl ion concentration at pH 7. B. Rises with increasing acidity. C. Of pH 5 is 100 times greater than that in a solution of pH 7. D. Is inversely related to the hydroxyl ion concentration. E. Is approximately doubled if the pH falls by 0.3 units.
Answer» A. True This is the neutral pH of water. B. True The pH falls. C. True pH is minus log₁₀ hydrogen ion concentration. D. True The product of these ionic concentrations is constant. E. True Log = 0.301.

28645.

The hydrogen ion concentration in a solution A. Equals the hydroxyl ion concentration at pH 7. B. Rises with increasing acidity. C. Of pH 5 is 100 times greater than that in a solution of pH 7. D. Is inversely related to the hydroxyl ion concentration. E. Is approximately doubled if the pH falls by 0.3 units.

Answer»

A. True This is the neutral pH of water.

B. True The pH falls.

C. True pH is minus log₁₀ hydrogen ion concentration.

D. True The product of these ionic concentrations is constant.

E. True Log = 0.301.

Discussion

28646.	Ameasure of feeling warmth or coolness by the human body in response to the air temperature, moisture content and air motion is called (a) Dry bulb temperature (b) Effective temperature (c) Wet bulb temperature (d) Dew point temperature
Answer» (b) Effective temperature

Discussion

28647.	The historigram is the graphical presentation of data which are classified: (a) Geographically (b) Numerically (c) Qualitatively (d) According to time
Answer» (d) According to time

Discussion

28648.	For graphic presentation of a frequency distribution, the paper to be used is: (a) Carbon paper (b) Ordinary paper (c) Graph paper (d) Butter paper
Answer» (c) Graph paper

Discussion

28649.	The human cell nucleus A. Has a membrane which is permeable to nucleic acid. B. In somatic cells contains 44 chromosomes. C. Stores its genetic material in the nucleolus. D. Has a skeleton of fine filaments. E. Is essential for cell division.
Answer» A. True Messenger RNA crosses the nuclear membrane. B. False The total is 46; i.e. 44 + XX or XY. C. False The nucleolus is a condensation of RNA; the genetic material is in chromosomes. D. True These are attached to the nuclear membrane. E. True For example, red blood cells cannot divide.

28649.

The human cell nucleus A. Has a membrane which is permeable to nucleic acid. B. In somatic cells contains 44 chromosomes. C. Stores its genetic material in the nucleolus. D. Has a skeleton of fine filaments. E. Is essential for cell division.

Answer»

A. True Messenger RNA crosses the nuclear membrane.

B. False The total is 46; i.e. 44 + XX or XY.

C. False The nucleolus is a condensation of RNA; the genetic material is in chromosomes.

D. True These are attached to the nuclear membrane.

E. True For example, red blood cells cannot divide.

Discussion

28650.	Historigram and histogram are: (a) Always same (b) Not same (c) Off and on same (d) Randomly same
Answer» (b) Not same

Discussion

Explore topic-wise InterviewSolutions in .

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In a two-stroke Petrol engine, fuel loss is maximum after :(a) Opening the exhaust port(b) Closing the exhaust port(c) Opening the inlet port(d) Closing the inlet port

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For a commercial new vehicle the initial fitness certificate is valid for ________ a) 15 years b) 10 years c) 5 yearsd) 2 years

Write any four hair accessories are used to make a good hair style.

It’s worn on your head for either fashion or warmth. It’s a ______

Aroom having dimensions of 5 m x 5 m x 3 m contains air at 25°C and 100 kPa at a relative humidity of 75%. The corresponding value of ps is 3.169 kPa. The partial pressure of dry air will be nearly (a) 106 kPa (b) 98 kPa (c) 86 kPa (d) 78 kPa

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For graphic presentation of a frequency distribution, the paper to be used is: (a) Carbon paper (b) Ordinary paper (c) Graph paper (d) Butter paper

The human cell nucleus A. Has a membrane which is permeable to nucleic acid. B. In somatic cells contains 44 chromosomes. C. Stores its genetic material in the nucleolus. D. Has a skeleton of fine filaments. E. Is essential for cell division.

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If the camber on the two front wheels is not equal, the vehicle will try to pull towards the side where.a. The camber is higher. b. The camber is lower c. Both (a) & (b) d. None of the above