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(2) magnetic permeability 

[MLT^–2A ^–2] = Magnetic permeability

Answer (4) [ML^-1T^-2]

Pressure of steam, p = 120 bar

Specific volume, v = 0.01721 m³/kg 

(i) Temperature : 

First it is necessary to decide whether the steam is wet, dry saturated or superheated. At 120 bar, v_g = 0.0143 m³/kg, which is less than the actual specific volume of 0.01721 m³/kg, and hence the steam is superheated. 

From the superheat tables at 120 bar, the specific volume is 0.01721 m³/kg at a temperature of 350°C.

(ii) Enthalpy :

From the steam tables the specific enthalpy at 120 bar, 350°C, 

h = 2847.7 kJ/kg.

(iii) Internal energy : 

To find internal energy, using the equation, 

u = h – pv

= 2847.7 – \(\cfrac{120\times10^5\times0.01721}{10^3}\) 

= 2641.18 kJ/kg.

Pressure of steam, p = 7 bar 

Enthalpy of steam, h = 2550 kJ 

From steam tables corresponding to 7 bar pressure : 

h_f = 697.1 kJ/kg, 

h_fg = 2064.9 kJ/kg, 

v_g = 0.273 m³/kg, 

u_f = 696 kJ/kg, 

u_g = 2573 kJ/kg

(i) Dryness fraction, x : 

At 7 bar, h_g = 2762 kJ/kg, hence since the actual enthalpy is given as 2550 kJ/kg, the steam must be in the wet vapour state. 

Now, using the equation, 

h = h_f + xh_fg 

∴ 2550 = 697.1 + x × 2064.9

x = \(\cfrac{2550-697.1}{2064.9}\) = 0.897

Hence, dryness fraction = 0.897.

(ii) Specific volume of wet steam, 

v = xv_g = 0.897 × 0.273 = 0.2449 m³/kg.

(iii) Specific internal energy of wet steam, 

u = (1 – x)u_f + xu_g 

= (1 – 0.897) × 696 + 0.897 × 2573 

= 2379.67 kJ/kg.

Pressure of steam, p = 18 bar 

Dryness fraction, x = 0.85 

From steam tables corresponding to 18 bar pressure : 

h_f = 884.6 kJ/kg, 

hfg = 1910.3 kJ/kg, 

vg = 0.110 m³/kg, 

u_f = 883 kJ/kg, 

ug = 2598 kJ/kg. 

(i) Specific volume of wet steam, 

v = xvg = 0.85 × 0.110 = 0.0935 m³/kg. 

(ii) Specific enthalpy of wet steam, 

h = h_f + xhfg = 884.6 + 0.85 × 1910.3 

= 2508.35 kJ/kg.

(iii) Specific internal energy of wet steam, 

u = (1 – x)uf + xu_g 

= (1 – 0.85) × 883 + 0.85 × 2598 

= 2340.75 kJ/kg.

(i) At 0.75 bar. From steam tables ;

At 100°C, h_sup = 2679.4 kJ/kg 

At 150°C, h_sup = 2778.2 kJ/kg 

∴ 2778.2 = 2679.4 + c_ps (150 – 100)

c_ps = \(\cfrac{2778.2-26794}{50}\) = 1.976.

(ii) At 0.5 bar. From steam tables ; 

At 300°C, h_sup = 3075.5 kJ/kg 

At 400°C, h_sup = 3278.9 kJ/kg 

∴ 3278.9 = 3075.5 + c_ps (400 – 300)

c_ps = \(\cfrac{3278.9-30755}{100}\) = 2.034.

Mass of steam in the cooker = 1.5 kg 

Pressure of steam, p = 5 bar 

Initial dryness fraction of steam, x₁ = 1 

Final dryness fraction of steam, x₂ = 0.6 

Heat to be rejected : 

Pressure and temperature of the steam at the new state : 

At 5 bar. From steam tables,

t_s = 151.8°C ; hf = 640.1 kJ/kg ; 

h_fg = 2107.4 kJ/kg ; v_g = 0.375 m³/kg 

Thus, the volume of pressure cooker

= 1.5 × 0.375 = 0.5625 m³ 

Internal energy of steam per kg at initial point 1,

u₁ = h₁ – p₁v₁ 

= (h_f + h_fg) – p₁v_g1 (\(\because\)v₁ = vg₁) 

= (640.1 + 2107.4) – 5 × 10⁵ × 0.375 × 10^–3 

= 2747.5 – 187.5 = 2560 kJ/kg

Also, V₁ = V₂ (V₂ = volume at final condition)

0.5625 = 1.5[(1 – x₂) v_f2 + x₂v_g2]

= 1.5 x₂v_g2 (\(\because\)vf₂ negligible) 

= 1.5 × 0.6 × v_g2

v_g2 = \(\cfrac{0.5625}{1.5\times0.6}\) = 0.625 m³/kg. 

From steam tables corresponding to 0.625 m³/kg,

p₂ ~ 2.9 bar, t_s = 132.4°C, h_f = 556.5 kJ/kg, h_fg = 2166.6 kJ/kg 

Internal energy of steam per kg, at final point 2, 

u₂ = h₂ – p₂v₂ 

= (h_f2 + x₂h_fg2) - p₂xv_g2   (\(\because\) v₂ = xv_g2)

= (556.5 + 0.6 × 2166.6) – 2.9 × 10⁵ × 0.6 × 0.625 × 10^–3 

= 1856.46 – 108.75 = 1747.71 kJ/kg. 

Heat transferred at constant volume per kg 

= u₂ – u₁ = 1747.71 – 2560 = – 812.29 kJ/kg 

Thus, total heat transferred 

= – 812.29 × 1.5 = – 1218.43 kJ.  

Negative sign indicates that heat has been rejected.

Mass of dry steam, m_s = 50 kg

Mass of water in suspension, m_w = 1.5 kg

∴ Dryness fraction,

x = \(\cfrac{Mass \,of \,dry \,steam}{Mass \,of \,dry \,steam \,mass \,of \,water \,in \,suspension}\) 

= \(\cfrac{m_s}{m_s+m_w}\) = \(\cfrac{50}{50+1.5}\) = 0.971.

Volume of the vessel, V = 0.6 m³

Mass of liquid water and water vapour, m = 3.0 kg

Pressure, p = 0.5 MPa = 5 bar

Thus, specific volume, v =  \(\cfrac Vm\) = \(\cfrac{0.6}{3.0}\) = 0.2 m³/kg

At 5 bar : From steam tables,

v_fg = v_g – v_f = 0.375 – 0.00109 = 0.3739 m³/kg

We know that, v = v_g – (1 – x) v_fg, where x = quality of the vapour.

0.2 = 0.375 – (1 – x) × 0.3739

(1 – x) = \(\cfrac{(0.375-0.2)}{0.3739}\) = 0.468

x = 0.532

(i) Mass and volume of liquid, m_liq. = ? V_liq. = ?

m_liq. = m(1 – x) = 3.0 × 0.468 = 1.404 kg.

V_liq. = m_liq. v_f = 1.404 × 0.00109 = 0.0015 m³.

(ii) Mass and volume of vapour, m_vap. = ? V_vap. = ?

m_vap. = m.x = 3.0 × 0.532 = 1.596 kg

Vv_ap. = m_vap. v_g = 1.596 × 0.375 = 0.5985 m³.

Mass of steam generated, m = 1000 kg/h

Pressure of steam, p = 16 bar 

Dryness fraction, x = 0.9 

Temperature of superheated steam, 

T_sup = 380 + 273 = 653 K

Temperature of feed water = 30°C 

Specific heat of superheated steam, c_ps = 2.2 kJ/kg K. 

At 16 bar. From steam tables, 

t_s = 201.4°C (T_s = 201.4 + 273 = 474.4 K) ; 

h_f = 858.6 kJ/kg ; h_fg = 1933.2 kJ/kg 

(i) Heat supplied to feed water per hour to produce wet steam is given by : 

H = m [(h_f + xh_fg) – 1 × 4.18 × (30 – 0)] 

= 1000 [(858.6 + 0.9 × 1933.2) – 4.18 × 30] 

= 1000(858.6 + 1739.88 – 125.4) 

= 2473.08 × 10³ kJ. 

(ii) Heat absorbed by superheater per hour 

= m[(1 – x) h_fg + c_ps (T_sup – T_s)] 

= 1000[(1 – 0.9) × 1933.2 + 2.2 (653 – 474.4)] 

= 1000(193.32 + 392.92) 

= 586.24 × 10³ kJ.

Volume of wet steam, v = 0.15 m³

Pressure of wet steam, p = 4 bar 

Dryness fraction, x = 0.8 

At 4 bar. From steam tables, 

v_g = 0.462 m³/kg, h_f = 604.7 kJ/kg, h_fg = 2133 kJ/kg

∴ Density = \(\cfrac{1}{xv_g}\) = \(\cfrac{1}{0.8\times0.462}\) = 2.7056 kg/m³

Mass of 0.15 m³ of steam 

= 0.15 × 2.7056 = 0.4058 kg. 

Total heat of 1 m3 of steam which has a mass of 2.7056 kg 

= 2.7056 h (where h is the total heat of 1 kg of steam) 

= 2.7056 (h_f + xh_fg) 

= 2.7056(604.7 + 0.8 × 2133) 

= 6252.9 kJ.

( b )

is this correct :)

δ = 30°, A = 60°, μ = √2 

Let us find the minimum value of angle of deviation δ that is δ_m

_{\(\boxed {µ = \frac {sin [\frac {A + δm}2]} {sin (\frac {A}{2})}} \)} This is the formula for minimum deviation of prism. Using this, we will find δ_m

√2 = [sin(60° + δ_m) / 2] / sin(60°/2) 

√2 = [sin(60° + δ_m) / 2] / 1/2

[sin(60° + δ_m) / 2] = √2 × 1/2 

[sin(60° + δ_m) / 2] = 1/√2 = sin45° 

(60° + δ_m) / 2 = 45° 

δ_m = 90 - 60 = 30°

As δ_m = δ, Hence we conclude that the prism is in minimum deviation position. So the ray inside the prism will be parallel to the base of the prism. Therefore, the angle made by the ray inside the prism with the base of prism is 0°. 

 U = -12

f = -10 

1/v - 1/u = 1/f

1/v + 1/12 = 1/10

1/v = 1/60 

\(-\frac{1}{v^2}\)dv or \(\frac{1}{u^2}\)du

du = \(\frac{v^2}{u^2}\)du

\(\frac{v}{u}\)= same path

△u₁ = △u₂ = 1mm

△u₁ = △v₂ 

△v₂ = v_f - v_i 

x₂ = △v₂ + 1mm

x₂ = 2mm

\(\frac{x_1}{x_2}=\frac{1}{2}\)

Given u = -10

f = 12

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}-(\frac{1}{-10})=\frac{1}{12}\)

\(\frac{1}{v}=\frac{1}{12}-\frac{1}{10}\)

\(\frac{1}{v}=\frac{10-12}{120}\)

\(\frac{1}{v}=\frac{-2}{120}\)

\(\frac{1}{v}=\frac{-1}{60}\)

\(-\frac{1}{v^2}\) dv + \(\frac{1}{u^2}\) du

du = \(\frac{v^2}{u^2}\) du

\(\frac{v}{u}\) = same path

△u₁ = △u₂ = 1mm

△u₁ = △v₂

△v₂ = v_f - v_i

x₂ = △v₂ + 1mm

x₂ = 2mm

\(\frac{x_1}{x_2}=\frac{1}{2}\)

(c) Optical cable

Such high frequency can travel in optical cables.

Correct option is (1) A-s, B-r, C-q, D-p 

we know that, the fringe width β = \(\frac{\lambda D }{d}\)

where λ = wavelength of light.

D = distance b/w slits and screen.

d = distance b/w slits.

β \(\prec\) \(\frac{1}{d}\)

it d become 3 folds.

then β become \((\frac{1}{3})\) times.

Correct option is (a) 2MR²/3

MOI of sphere about tangent 7/5 MR²

MOI of disc about diameter MR²/4

MOI of ring about diameter: MR²/2

MOI of hollow sphere a from axis passing through center 2MR²/3

(i) It is monocromatic cohrent light.

(ii) It travell large distance without deviation.

Electric dipole moment—Electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge of the electric dipole and the dipole length. Mathematically, the magnitude of dipole moment is

P = q × 2l

It is a vector quantity. The SI unit of electric dipole moment is coulomb metre (cm).

Two characteristics of LASER rays are as follows—

1. It is highly directional with parallel beam.

2. It is monochromatic and coherent.

Let Y₁(t) be position of dropped stone and Y₂(t) be position of projected stone.

Then:

\(​Y_1(t) = H - \frac12gt^2\)

\(Y_2(t) = v_ot - \frac12gt^2​\)

\(​Y_1(3) = Y_2(3)\implies H - \frac92g = 20 \frac ms\cdot 3s - \frac92g​\)

Therefore \(H = 60m\)

Answer is (4) – 5 m/sec

x = 2 - 5t + 6t². 

ν = dx/dt = 0 - 5 x 1 + 12t

= -5 + 12.0 = -5 m/s. 

Answer is (2) 200 cm/sec

ν = at = (F/m)t = (100 x 10)/5 = 200 cm/sec. 

Glycolysis is a linear metabolic pathway of enzyme-catalyzed reactions that convert glucose into two molecules of pyruvate in the presence of oxygen or into two molecules of lactate in the absence of oxygen. In the absence of oxygen, glycolysis is the only option that cells have for the production of ATP from glucose.

Atropine competitively blocks the effects of acetylcholine, including excess acetylcholine due to organophosphorus poisoning, at muscarinic cholinergic receptors on smooth muscle, cardiac muscle, secretory gland cells, and in peripheral autonomic ganglia and the central nervous system.