GATE ME 2018 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is complemented, complimented.

Key Points

• Let's understand the meaning of the marked words:
• Complement: add to (something) in a way that enhances or improves it; make perfect.
• ​Example: B**ecause Jim is a serious man who always thinks before he acts, he is the perfect complement for Lisa who lives by her impulses.
• Compliment: politely congratulate or praise (someone) for something.
• ​Example: I just wanted to compliment you for the wonderful speech you gave tonight.
• From the above-given explanation, we can say that 'complemented' will be used in the first blank and 'complimented' will be used in the second blank.

​Therefore, the correct sentence is: The dress complemented her so well that they all immediately complimented her on her appearance.

The correct answer is undiminished.

Key Points

• Let's understand the meaning of the marked word.
• Undiminished: not diminished, reduced, or lessened.
• Example:  Even in his eighties, the man’s eyesight was undiminished and he could see as well as his grandson.
• Thus, from the above-given explanation, we can say that 'undiminished' will be used in the blank as the sentence is trying to explain that the judge's standing in the legal community is still intact even after false allegations.

Therefore, the correct sentence is: The judge’s standing in the legal community, though shaken by false allegations of wrongdoing, remained undiminished.

Additional Information

• Let's explore other options:

According to the English alphabet series and its positional value:

The logic follows here is:

In Each term, one letter is increased, and in each term consecutive letters.

In First Term: 2 letters

In Second Term: 3 letters

In Third Term: 4 letters

So;

In Fourth Term: 5 letters

Hence, "TUVWX" is the correct answer.

Explanation:

The perimeter of Circle of radius r = 2πr

The perimeter of a square of side s = 4s

The perimeter of an equilateral triangle of side a = 3a

The perimeters of a circle, a square, and an equilateral triangle are equal.

2πr = 4s = 3a

• s = 3/4 a = 0.75 a
• r = 3/2π a = 0.477 a

Area of an equilateral triangle:

$A_{trg}=\frac{\sqrt 3}{4}a^2=0.433 a^2$

Area of square = s² = (0.75 a)² = 0.5625 a²

Area of Circle = 2πr2 = π(0.477 a)2 = 0.714 a2

A_circle > A_Square > A_trg

Concept:

Logarithmic Properties:

${\log _a}xy = {\log _a}x + {\log _a}y$

${\log _a}\frac{x}{y} = {\log _a}x - {\log _a}y$

${\log _a}{x^n} = n{\log _a}x$

${\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}$

${\log _a}b = \frac{1}{{{{\log }_b}a}}$

${\log _a}a = 1$

Calculation:

$y=\frac{1}{{1 + {{\log }_u}vw}} + \frac{1}{{1 + {{\log }_v}wu}} + \frac{1}{{1 + {{\log }_w}uv}}$

$y=\frac{1}{{\log_uu+ {{\log }_u}vw}} + \frac{1}{{\log_vv + {{\log }_v}wu}} + \frac{1}{{\log_ww + {{\log }_w}uv}}=\frac{1}{\log_uuvw}+\frac{1}{\log_vuvw}+\frac{1}{\log_wuvw}$

Using ${\log _a}b = \frac{1}{{{{\log }_b}a}}$

$y={\log _{uvw}}u + {\log _{uvw}}v + {\log _{uvw}}w={\log _{uvw}}uvw=1$

Each student watched either only one film or all three.

Let a = students who watched only A,

b = students who watched only B,

c = students who watched only C,

x = students who watched A, B and C i.e. all three

Given: P(A) = 13, P(B) = 16, P(C) = 19

a + x = 13

b + x = 16

c + x = 19

a + b + c + 3x = 13 + 16 + 19

a + b + c + 3x = 48

Total strength of class is 40,

a + b + c + x = 40

⇒ 2x = 8

x = 4

So 4 students watched all three films.

Concept:

Area of Square of side 'a' = a²

The perimeter of a square of side 'a' = 4s

Area of a circle of radius 'r' = πr²

Circumference of a circle of radius 'r' = 2πr

Calculation:

A wire would enclose an area of 1936 m2 if it is bent into a square. Side of square = ?

a² = 1936 m² ⇒ a = 44 m

As wire encloses a square so length of wire = Perimeter of a square

L = 4a = 176 m

The wire is cut into two pieces. The longer piece (k) is thrice as long as the shorter piece (s).

k = 3s and k + s = L = 176 m

3s + s = 176 m ⇒ s = 44 m ⇒ k = 3s = 132 m

So the longer piece is of 132 m length and the shorter piece is of 44 m length.

The long and the short pieces are bent into a square and a circle, respectively.

So the perimeter of a square = Length of Longer wire = 132 m

4a' = 132 m ⇒ a' = 33 m

Area of the square = a² = 33² = 1089 m²

The perimeter of a circle =  Length of shorter wire = 44 m

2πr = 44 ⇒ r = 7 m

Area of circle = πr² = 154 m²

Sum of the areas enclosed by the two pieces: A_square + A_circle = 1089 + 154 = 1243 m²

Concept:

Time and Work - Important Formula:

• If A can do a piece of work in n days, work done by A in 1 day = 1/n
• If A does 1/n work in a day, A can finish the work in n days
• If M₁ men can do W₁ work in D₁ days working H₁ hours per day and M₂ men can do W₂ work in D₂ days working H₂ hours per day (where all men work at the same rate), then M₁D₁H₁/W₁ = M₂D₂H₂/W₂
• If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq/(p+q) days
• If A is thrice as good as B in work, then
• The ratio of work done by A and B = 3: 1 and
• The ratio of time taken to finish a work by A and B = 1 : 3

Calculation:

125 robots operating for 7 hours a day are working for 39 days and completing five-seventh of the work

X robots operating for8 hours a day are working for 52 - 39 = 13  days and completing 1 - 5/7 = 2/7th of the work

$\frac{{{M_1}{D_1}{H_1}}}{{{W_1}}} = \frac{{{M_2}{D_2}{H_2}}}{{{W_2}}}$

$\frac{{125 \times 39 \times 7}}{{\frac{5}{7}}} = \frac{{X \times 13 \times 8}}{{\frac{2}{7}}} \Rightarrow X = 131.25$

So, the number of additional robots to be required

= 131.25 – 125 = 6.25 or 7 robots

• Condition 1: If the house number is a multiple of 3, then it is a number from 50 to 59.
• Possibilities: 51, 54, 57
• But if one of these is the right answer, then it should be a multiple of 4 and 6 which is not the case here.
• Condition 2: If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
• Possibilities: 61, 63, 65, 66, 67, 69
• But if one of these is the right answer, it should be multiple of six (i.e. 66) and not a multiple of 3 (but 66 is multiple of 3). So it is not a required sample space.
• Condition 3: If the house number is NOT a multiple of 6, then it is a number from 70 to 79.
• Possibilities: 70, 71, 73, 74, 75, 76, 77, 79
• But if one of these is the right answer, it should be multiple of 4 (i.e. 76) and not a multiple of 3 (i.e. 70, 71, 73, 74, 76, 77, 79).

So 76 is the correct answer.

Concept:

Out of six tosses, the result of four tosses is given as HHHT__ __.

For the last two tosses, the event of occurrence is: HH, HT, TH, TT

• The probability that two heads will occur, P(HH) = 1/4
• The probability that two tails will occur, P(TT) = 1/4
• The probability that one head and one tail will occur, P(HT/TH) = 2/4 = 1/2

P(HT/TH) > P(TT) or P(HH)

"One H and one T will occur" has the highest probability of being correct

Concept:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

Calculation:

 f(x) = cos2(x)

$f(x) = \frac{{1 + \cos 2x}}{2} = \frac{1}{2} + \frac{{\cos 2x}}{2}$

On comparing with: \(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

a₀ = 1

a₁ = 0

a₂ = 1/2

Concept:

The divergence of any vector field $\vec A$ is defined as:

$Div= \vec \nabla .\vec A$

The nabla operator is defined as:

$\vec \nabla ={\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}}}+\hat k\frac{\partial }{\partial z}$

Calculation:

Given:

vector $\vec u = {e^x}\left( {\cos y\hat i + \sin y\hat j} \right)$

Divergence of u will be

$Div = \vec \nabla .\vec u$

$= \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}}} \right).\left( {{e^x}.\cos y\hat i + {e^x}.{\rm{siny}}\hat j} \right)$

$= \frac{\partial }{{\partial x}}\left( {{e^x}.\cos y} \right) + \frac{\partial }{{\partial y}}\left( {{e^x}.\sin y} \right)$

$= {e^x}.\cos y + {e^x}.\cos y$

$\vec \nabla .\vec u = 2{e^x}\cos y$

Explanation:

3-D heat equation is given as below

$\left( {\frac{{{\partial ^2}T}}{{d{x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}}} \right) + \frac{{Q\left( {x,t} \right)}}{K} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}$

For 1 – D & without heat generation:

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}$

Where α ÷ thermal diffusivity.

Wave equation is given by:

$\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}.\frac{{{\partial ^2}u}}{{\partial {x^2}}}$      (1-D)

Laplace equation:

$\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0$     (3-D)

${\nabla ^2}u = 0$

Concept:

${y^3}\frac{{dy}}{{dx}} + {x^3} = 0$

Given y (0) = 1,      y(-1) = ?

${y^3}\frac{{dy}}{{dx}} + {x^3} = 0$

y³ dy = -x³dx

\(\mathop \smallint \nolimits{y^3}dy = - \mathop \smallint \nolimits {x^3}dx + c\)

$\frac{{{y^4}}}{4} = - \frac{{{x^4}}}{4} + C$

$\frac{{{x^4}}}{4} + \frac{{{y^4}}}{4} = C$      ----(1)

Now given y (0) = 1

$C = \left( {\frac{1}{4}} \right) + \frac{{{{\left( 0 \right)}^2}}}{4}$

$C = \frac{1}{4}$

$\frac{{{y^4}}}{4} + \frac{{{x^4}}}{4} = \frac{1}{4}$                       {By equation (1)}

Y (-1) =?

$\frac{{{y^4}}}{4} + \frac{{{{\left( { - 1} \right)}^4}}}{4} = \frac{1}{4}$

$\frac{{{y^4}}}{4} + \frac{1}{4} = \frac{1}{4}$

$\frac{{{y^4}}}{4} = 0 \Rightarrow y = 0$

y (-1) = 0

Concept:

Ends of the columns:

             

In this question:

${P_{min}} = \frac{{{\pi ^2}EI}}{{L{e^2}}}$

${P_{min}} = \frac{{{\pi ^2}EI}}{{{L^2}}}$

Concept:

Power transmitted through the gear remains constant 

$PowerP = \frac{{2\pi NT}}{{60}}$

For two mating gears A and B 

NATA = NBTB

where N is the angular speed in rpm and T is the torque

also, 

NAZA = NBZB 

Where Z is the number of teeth on the gear. 

Calculation:

Given, Z_A = 12 teeth, Z_B = 48 teeth

T_A = 100 N-m, Z_C = 12, Z_D = 60 teeth

T_D = ?

∵ Now N_AZ_A = N_BZ_B ⇒ N_A × 12 = N_B × 48

N_A = 4 N_B            ----(1)

$PowerP = \frac{{2\pi NT}}{{60}}$

P_A = P_B ⇒ N_AT_A = N_BT_B

4N_B T_A = N_B T_B                                                    [N_A = 4 N_B    By equation (1)]

T_B = 4 × 100 = 400 N-m

Now gear B & C are on the same shaft, therefore torque on gear B & C will be the same.

Thus T_C = 400 N-m

Now, N_CZ_C = N_DZ_D

N_C × 12 = N₀ × 60 ⇒ N_C = 5 N­_D        ----(2)

N_CT_C = N_DT_D

5 N_D× 400 = N_D T_D                                                [N_C = 5 N­_D By equation (2)]

T_D = 2000 N-m

Concept:

Transmissibility is given by:

$\varepsilon = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left[ {2\zeta \frac{\omega }{{{\omega _n}}}} \right]}^2}} }}$

$C = 2\zeta \sqrt {KM} \Rightarrow \zeta = \frac{C}{{2\sqrt {KM} }}$

Calculation:

After additional damper in parallel $\zeta ' = \frac{{C + C'}}{{2\sqrt {KM} }}$

Thus  $\zeta ' > \zeta$

Now,  ${\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}}$  

 $\zeta ' > \zeta$indicating that $\omega _d' < {\omega _d}$

$time\ period\left( {{T_d}} \right) = \frac{{2\pi }}{{{\omega _d}}}$

Thus, the time period of free oscillation will increase.

External tensile load applied to a no preloaded joint transmits entirely to the bolt. The stiffness of a bolted joint without a preload equals the bolt stiffness K_e = K_b. For a preloaded bolted joint, joint stiffness is the sum of bolt & clamped member stiffness k_e = K_b + K_c. Therefore, preloading increases the stiffness of the bolted joints, which also increases the resonant frequency of an assembly.

Note:

However, once a joint is preloaded, increasing the magnitude of the preload can not make a joint stiffer.

Concept:

From Wein's displacement law 

${\lambda _{max}}T = 2898;\mu m - k\left( {constant} \right)$

Thus, ${\lambda _{peak}}T = \lambda _{peak}'T'$

Calculation:

Given, 

Black body λpeak = 1.45 μm at 2000 K.

Now, $\lambda _{peak}' = 2.90;\mu m$

1.45 × 2000 = 2.90 × T'

T' = 1000 K

Explanation:

Entropy change for an ideal gas with constant properties undergoing a quasi-static process:

${s_2} - {s_1} = {C_p}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) + {C_v}\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)$

${s_2} - {s_1} = {C_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$

${s_2} - {s_1} = {C_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - R\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)$

• 50% reaction stage steam turbine also known as “Parsons turbine”.
• The velocity triangles for inlet & outlet conditions are symmetrical having.

 

\(\left{ {\begin{array}{{20}{c}} {{\alpha _1} = {\beta _2}}\ {{\alpha _2} = {\beta _1}} \end{array}} \right};;;;;;;& ;;;;\left{ {\begin{array}{{20}{c}} {{V_{{r_1}}} = {V_2}}\ {{V_{{r_2}}} = {V_1}} \end{array}} \right}\)

Explanation:

On cooling reaction will tend in forward direction and on heating cycle reaction will reverse in direction.

Explanation:

During solidification of pure molten metal, the grains in the casting near the mould wall are fine and randomly oriented because the rate of solidification is high at the surface of mould during solidification.

Explanation:

Centrifugal casting:

It is used to manufacture hollow products (such as pipes) having a rotationally symmetric hollow space & on-axis coinciding with the axis of rotation of the centrifugal casting machine.

Casting materials: Especially cast iron, cast steel, heavy & light metals.

Blow moulding:

• Can form intricate shapes, hollow shapes with thin walls.
• Ideal for mass production.
• Plastics used: HDPE, LDPE, PET, PP, PS, PVC.
• End-uses: Plastic bottles & containers of all sizes and shapes; e.g. soft drinks bottles & shampoo bottles.

Rolling: Is the process of reducing the thickness or changing the cross-section of a long workpiece by compressive forces applied through a set of rolls.

• Products: Structures, rails, threaded bolts by thread rolling.

Impact extrusion:

• It is similar to indirect extrusion, it is often included in the cold-extrusion category. The descends rapidly on the blank (slug), which is extruded backward.
• Performed at higher speeds & shorter strokes as compared to conventional extrusion.

Products:

• Made by this process included toothpaste tubes & battery cases.
• Can produce thin walled tubular sections, ones having thickness to diameter ratios as small as 0.005.

Explanation:

Feed rate in slab milling operation is given by, 

f_m = f_t × N × Z 

Where f_t is the feed per tooth.

N = Spindle rotational speed (in rpm)

Z = Number of teeth in cutter (teeth per rev)

Explanation:

The unconventional machining process and their characteristics and the application areas are discussed in the table below:

Explanation:

lathe chuck:

• A lathe chuck is a holding device which is used for holding the job firmly against the cutting tool.

There are two types of chuck:

Four Jaw Chuck:

• The four-jaw chuck is also called an independent chuck since each jaw can be adjusted independently.
• Four jaw chucks are used for a wide range of regular and irregular shapes.
• A work can be trued to within 0.02 mm accuracy, using this chuck.
• This type of chuck is much more heavily constructed than the self-centering chuck, and has much greater holding power.
• Each jaw is moved independently by a square thread screw.
• The jaws are reversible for holding large diameter jobs.
• The independent four-jaw chuck has four jaws each working independently of the others in its own slot in the chuck body and actuated by its own separate square threaded screw.
• By suitable adjustment of the jaws, a workpiece can be set to run either true or eccentric with the machine centre.
• Finished jobs when held in a four-jaw chuck can be trued with the help of a dial test indicator.

Three Jaw Chuck:

• It is also known as three jaws universal chuck, self-centring chuck and concentric chuck having three jaws which work at the same time.
• Three jaw chucks are used to hold only perfect round and regular jobs

Concept:

$EOQ = {Q} = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}}$

Where Ch is the holding cost

D is the annual demand

Q is the quantity ordered

Co is the setup or ordering cost

Calculation:

Given, D = 9600 tyres/year, C_h = Rs. 16 tyre/year, C_o = Rs. 75

$Economic;order\left( {EOQ} \right) = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}} = \sqrt {2 \times 600 \times 75}$ = 300 units

Concept:

Given $\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = A\pi i$ 

$\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = \frac{1}{5}\mathop \oint \limits_C^; \frac{1}{{z - \frac{4}{5}}}$

There is a pole at $z = \frac{4}{5 }$

Thus, \(\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = \frac{{2\pi i}}{5}\lim z \to \frac{4}{5}\left{ {\frac{{z - \frac{4}{5}}}{{z - \frac{4}{5}}}} \right}\)

Given,

$A\pi i = \frac{{2\pi i}}{5} \times 1$

$A = \frac{2}{5}$

Explanation:

x₁ & x₂: two independent exponentially distributed random variables with means 0.5 and 0.25.

A continuous random variable x is said to have an exponential (λ) distribution if it has probability density function.

\({f_x}\left( {x/\lambda } \right) = \left{ {\begin{array}{*{20}{c}} {\lambda {e^{ - \lambda x}}}&{for;x > 0}\ 0&{for;x \le 0} \end{array}} \right.\)

Where, λ > 0, is called the rate of distribution. The mean of the exponential (λ) distribution is calculated using integration by parts as:

\(E\left[ x \right] = \mathop \smallint \limits_0^\infty x\lambda .{e^{ - \lambda x}}dx\)

\(= \lambda \left{ {\left[ {\frac{{ - x.{e^{ - \lambda x}}}}{\lambda }} \right]_0^\infty + \frac{1}{\lambda }\mathop \smallint \limits_0^\infty {e^{ - \lambda x}}.dx} \right}\)

\(= \lambda \left{ {\lambda \left[ {0 + 0} \right] + \frac{1}{\lambda }\left[ {\frac{{ - {e^{ - \lambda x}}}}{\lambda }} \right]_0^\infty } \right}\)

\(= \lambda \left{ {0 + \frac{1}{\lambda }\left[ {0 + \frac{1}{\lambda }} \right]} \right}\)

$= \lambda .\frac{1}{{{\lambda ^2}}}$

$E\left[ x \right] = \frac{1}{\lambda }$

Let x₁, x₂ …..., x_n be independent random variables, with x_i having exponential (λ_i) distribution. Then the distribution of min (x₁, x₂ ……, x_n) is exponential (λ₁ + λ₂ + …. + λ_n)

Thus, mean of min $\left( {{x_1},{x_2}} \right) = \frac{1}{{{\lambda _1} + {\lambda _2}}}$

${\lambda _1} = \frac{1}{{E\left[ {{x_1}} \right]}} = \frac{1}{{0.5}} = 2$

${\lambda _2} = \frac{1}{{E\left[ {{x_2}} \right]}} = \frac{1}{{0.25}} = 4$

Mean of min $\left( {{x_1},{x_2}} \right) = \frac{1}{{2 + 4}}$

${\lambda _2} = \frac{1}{{E\left[ {{x_2}} \right]}} = \frac{1}{{0.25}} = 4$

Thus, y = min (x₁, x₂) is exponentially distributed with mean $\left( {\frac{1}{6}} \right)$.

Explanation:

Position vector $\vec r = x\hat i + y\hat j + zk$

$\left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}}$

$\phi = \ln \left| {\vec r} \right|$

\(Gradient;\nabla \phi = \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + k\frac{\partial }{{\partial z}}} \right).\left{ {\ln \sqrt {{x^2} + {y^2} + {z^2}} } \right}\)

\(= \frac{1}{2}\left{ {i\frac{\partial }{{\partial x}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + j.\frac{\partial }{{\partial y}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + k\frac{\partial }{{\partial z}}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right}\)

\(= \frac{1}{2}\left{ {i.\frac{{2x}}{{{x^2} + {y^2} + {z^2}}} + j\frac{{2y}}{{{x^2} + {y^2} + {z^2}}} + k\frac{{2z}}{{{x^2} + {y^2} + {z^2}}}} \right}\)

\(\nabla \phi = \left{ {\frac{{xi + yj + zk}}{{{x^2} + {y^2} + {z^2}}}} \right}\)

$Gradient;\nabla \phi = \frac{{\vec r}}{{\vec r.\vec r}}$

Explanation:

As the acceleration of point Q is zero, so assuming PQR a rigid body hinged at Q.

${\vec a_{RP}} = {\vec a_R} - {\vec a_p}$

Given,

Given: The point R is accelerating with respect to point P at 10∠180° m/s².

i.e. 10 m/s² acceleration at 180° of (P) so only radial acceleration exist i.e. ω = constant; α = 0

∴ a_RP = (RP) ω² = 10

⇒ ω² × 20 = 10

$\omega = \left( {\frac{1}{{\sqrt 2 }}} \right)*$

Now as α is zero so R w.r.t Q will also have a radial acceleration in the x-axis

${a_R} = \left( {RQ} \right){\omega ^2} = 16 \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = 8;m/{s^2}$

Now our reference is (PR) so the angle will be (180 + θ) *

$\sin \theta = \left( {\frac{{12}}{{20}}} \right) \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = 37^\circ$

180 + 37 = 217°

a_R = 8 ∠217°

Concept:

A translating body or a body in general plane motion is in pure rotation about its instantaneous centre of rotation.

and the velocity is given by 

V = ω × Distance of instantaneous centre of rotation

Calculation:

O is the instantaneous centre of the rigid rod, so rod will have pure rotation along the instantaneous centre.

Let rotational speed be = ω (about I.C.)

$OP = 1\sin 45^\circ = \frac{1}{{\sqrt 2 }}$

$OQ = 1\cos 45^\circ = \frac{1}{{\sqrt 2 }}$

V_P = rω = OP × ω

$5 = \frac{1}{{\sqrt 2 }}ω$     ----(1)

V_Q = OQ × ω

${V_Q} = \frac{1}{{\sqrt 2 }}ω$     ----(2)

By equation (1) & (2):

5 = V_Q

V_Q = 5 m/s

Concept:

The angle of twist on a shaft is given by 

$\theta=\frac{{{T} \times L}}{{G\times{I_P}}}$

Where T is the torque, L is the length of the shaft

G is the modulus of the rigidity and

I_P is the polar moment of inertia

Calculation:

Given T = 150 N.m

The torsional reactions at point P are TP  and at point Q it is TQ

TP + TQ = 150      ---- (1)

Since torque is being applied on the same shaft, thus twist of rod will be same.

Thus, $\frac{{{T_P} \times 100}}{{G{I_P}}} = \frac{{{T_Q} \times 200}}{{G{I_P}}}$

T_P = 2T_Q      ----  (2)

By equation (1) & (2):

T_Q = 50 N.m

T_P = 100 N.m

(T_P, T_Q) = (100, 50)

Concept:

Uniform Acceleration and Retardation

Under the given situation one can apply the laws of kinematics.

Uniform acceleration and uniform declaration

‘ω’ angular velocity, ‘δ’ angle turned by the cam

Total time of flight $\left( {{t_0}} \right) = \left( {\frac{\delta }{\omega }} \right)$

$S = ut + \frac{1}{2}a{t^2}$

Calculation:

Applying for the acceleration half

$\left( {S = \frac{h}{2}} \right)$ and (u = 0); $t = \left( {\frac{{{t_0}}}{2}} \right)$

$\Rightarrow \frac{h}{2} = \frac{1}{2}a \times {\left( {\frac{{{t_0}}}{2}} \right)^2};\frac{h}{2} = \frac{1}{2} \times a \times {\left( {\frac{\delta }{{2\omega }}} \right)^2}$

$\Rightarrow {a = \frac{{4h{\omega ^2}}}{{{\delta ^2}}}}$

Now maximum velocity will occur at the end of the acceleration half and after that declaration starts hence velocity will decrease.

[V² – u² = 2aS]

Using this equation again for acceleration half.

${v^2} = 2as = 2 \times \frac{{4h{\omega ^2}}}{{{\delta ^2}}} \times \frac{h}{2} \Rightarrow v = \frac{{2h\omega }}{\delta }$

${v_{maximum}} = \frac{{2h\omega }}{\delta }$

Concept:

The axial strain is given by 

$ϵ=\frac{{{P}}}{{{A}{E}}}$

Where P is the load acting in the axial direction, A is the cross-sectional area and E is the Modulus of elasticity

Calculation:

ES = 210 GPa

EAI = 70 GPa

Since the magnitude of axial strains in both bars same. Thus,

$\frac{{{P_s}}}{{{A_s}{E_s}}} = \frac{{{P_{Al}}}}{{{A_{Al}}{E_{Al}}}}$

Since, ${A_S} = {A_{Al}} = 1;{m^2}$

$\frac{{{P_S}}}{{210}} = \frac{{{P_{Al}}}}{{70}}$

P_S = 3 P_Al

P = P_S + P_AI      ----(1)

Now, ϵ = 10^-6

$\frac{{{P_S}}}{{{A_S}{E_S}}} = {10^{-6}}$

P_S = 10^-6 × 1 × 210 × 10⁹

P_S = 210 KN

P_AI = 70 KN

P = P_AI + P_S = 210 + 70 = 280 KN

Concept:

Bernoulli equation between section (1) & (2)

$\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {Z_2} + {h_L}$

h_L = 0 (Gradually reducing cross-section)

ρ = Constant (incompressible fluid)

Calculation:

ρair = 1.2 kg/m3

$\frac{{{P_2} - {P_1}}}{{\rho g}} = \frac{{V_1^2}}{{2g}} - \frac{{V_2^2}}{{2g}}$

${P_2} - {P_1} = \frac{\rho }{2}\left( {V_1^2 - V_2^2} \right)$

∴ Q = 1.5 m³/s

A₁V₁ = A₂V₂ = 1.5 m³/s

${A_1} = \frac{\pi }{4}d_1^2$

${A_1} = \frac{\pi }{4}{\left( {0.4} \right)^2} = \frac{{4\pi }}{{100}}{m^2}$

${A_2} = \frac{\pi }{4}{\left( {0.2} \right)^2} = \frac{\pi }{{100}}$

Thus, ${V_1} \times \frac{{4\pi }}{{100}} = 1.5$

${V_1} = \frac{{150}}{{4\pi }}$

${V_2} = \frac{{150}}{{\pi }}$

\({P_2} - {P_1} = \frac{{1.2}}{2}\left{ {{{\left( {\frac{{150}}{{4\pi }}} \right)}^2} - {{\left( {\frac{{150}}{\pi }} \right)}^2}} \right}\)

$= 0.6 \times {150^2}\left( {\frac{1}{{16{\pi ^2}}} - \frac{1}{{{\pi ^2}}}} \right)$

= -1282.346 N/m² (or pascal)

P₂ - P₁ = -1.28 KPa

Explanation:

Maximizing z = x₁ - x₂

Constraints x₁ + x₂ ≤ 10

x₁ ≥ 0

x₂ ≥ 0

x₂ ≤ 5

Corner points are: (0, 0) (5, 0), (5, 5), (0, 10).

Z (0, 0) = 0 - 0 = 0

Z (0, 5) = 0 - 5 = -5

Z (5, 5) = 5 - 5 = 0

Z (10, 0) = 10 - 0 = 10

Maximum Z = Z (10, 0) = 10

Thus, problem has one optimum solution.