I hope GATE 2014 was a good experience for you. I don’t remember values and figures but language of the questions can make you remind the question so, I will post here the manner in which the question was to be solved.This post contains the questions asked in set I(9:00AM to 12:00PM).
I will keep updating this post. I will give you the possible manner in which question was to be solved. Questions are not in exact serial number because it is difficult to remember.
Surveying:
Q.1 Match group I with Group II.
GroupI Group II
P. Alidade A. Chaining
Q. Arrow B. Plane Table
R.
Solution: P-B, Q-A,
Q.2 Following staff readings were taken using a level. Find the R.L. of station S.
Station B.S. F.S. R.L.(m)
P 100
Q.
R.
S.
Solution: In this question first you have to add the B.S. to the R.L. of bench mark(P), to get the Height of instrument and then minus the F.S. at Q, then add the B.S. at Q, then minus the F.S. at R, then add up the B.S. at R and minus the for sight at S and you will get the R.L. of the S.
Remember, the Fore sights were given with the negative signs at some stations, so automatically two negative signs will produce a positive sign.
Structural Engineering:
Q.1 The influence line diagram for the shear force at point P at a distance of L/4 from end A in a simply supported beam AB will be.
Solution: Remember, there were four diagrams. The diagram which has a negative linear variation of shear force on the left of the section from a value of 0 at A to 0.25 units just on left of P and then 0.75 when just on right of P and then varying linearly to a value of 0 at B, is the right diagram for the influence line of shear force at P.
Q.2 Central singly concentrated collapse load for the given propped cantilever beam of length L will be:
(A) 2Mp/L (B) 4Mp/L (C) 6Mp/L (D) 8Mp/L
Solution: (C) 6.Mp/L
Given beam is a propped cantilever so, there will be two hinges, one at the fixed end and one at the center, where the load is acting.
Equating the external work done to the internal work done,
Wp.L/2.O= Mp.O + 2.Mp.O
=> Wp = 6.Mp/L
Q.3 Find out the static indeterminacy:
(A) Two (B) One (C) Zero (D) Three.
Solution: (C) Zero
You were given a rigid frame with two pinned end supports and one internal hinge. Two pinned supports produce 4 external support reactions and also with one internal hinge our number of equations of static equilibrium gets increased by one so, total is four.
So, static indeterminacy = 4-4 = 0.
Q.4 A given reinforced concrete beam of M25, has a neutral axis at a distance of y from the top face. If the maximum stress developed in the top fiber is x, find out the curvature of the beam. (If you remember the exact values, please post in comment box).
Solution: Here you have to use the flexure formula,
f/y = E/R => 1/R = f/(E.y)
where, f= flexural stress in the top fiber, y = distance of the neutral axis from the top fiber
E is Thomas Young’s modulus of elasticity = 5000(fck)^(1/2) here, fck = 25 Mpa
Q.5 Find out the value of force P and its angle of application, in order to lift the given body of weight ‘W’ without any swing.
Solution: There were numbers of forces acting on the body with their respective angles mentioned. In order to lift up the body without any swing we had to use the following two equations:
Algebraic sum of all the horizontal forces = 0
Algebraic sum of all the vertical forces = W
We have two unknowns, P and its angle and also have two equations, so can be easily found.
Transportation Engineering:
Q.1 The design speed of a road is V km, and the co-efficient of longitudinal friction for the road and tyre is f, if the stopping sight distance of the vehicle is ‘s’ find out the total reaction time of the driver.
Solution: In this question we need to use the equation V.t + V^2/ (2gf) = s
here, all the values are given except t, which can be easily found.
Q.2 Equation of the velocity(Km/Hr) for a given traffic flow is u= 70+0.7k, where k(Veh/Km) is the density of the flow, the capacity of the flow is:
(A).. (B) .. (C) .. (D)..
Solution:
Hydraulics and water resource Engineering
Q.1 Two parallel pipes diverges from a single pipe and after same length again converge at the another end. If the diameter of the first pipe is four times the diameter of the second pipe, the velocity ratio in first first pipe to that in second pipe will be
(A) 1 (B) 2 (C) 4 (D) 3
Solution: (B)
As the head loss in two pipes will be same,
we can equate the head loss equations of the two. given d1=4.d2.
use Hl= f.l.v^2/(2gd).
Q.2 A1 and V1 are area of cross section and velocity at section 1, and A2 is the area of the section at section two of a section of a continuous flow. What is the velocity at section 2?
Solution: Continuity equation can be applied to get the velocity at section 2.
Note: Values were given in the question and it was a subjective question)
Q.3 Diameter of the Venturimeter and of a section are given, velocity head difference at the throat of Venturimeter and section is given also, velocity at the section is given. Find out the co-efficient of discharge for the Venturimeter.
Solution: Here first you have to calculate the velocity at the Venturimeter throat which, can be calculated from the given velocity head difference between the section and the throat.
The discharge at the section is calculated by the given data of the area and velocity,
finally equate the discharge to the formula of the discharge for the Venturimeter.
Final equation will give the value of co-efficient of discharge.
To be continued…
