GATE ME old paper Questions and Solutions Set 1

GATE ME old paper Questions and Solutions Set 1
Question:-
Option (A)
1.88
Option (B)
3.56
Option (C)
6.12
Option (D)
11.56
Correct Option:
D
Question Solution:
b = 80mm; D=250mm; m = 0.25; q = 270dgrees = 4.712 rad; Ft = 1000 N-m We know that Ft = (p1 - p2)R

(P1/p2) = emq, p1 = emq. P2 = e0.25x4.712. P2, p1 = 3.25p2

\ 2.25p2 = (Ft /R) = (1000 / 0.125) Þ p2 = 3.5KN

And P1=11.56 KN

Therefore Max tension in the belt = 11.56 KN.
Question:-
Option (A)
2/315
Option (B)
1/630
Option (C)
1/1260
Option (D)
1/2520
Correct Option:
C
Question Solution:
Given, (Washers /2) (Nuts /3)(Bolts /4) (Total /9)

The probability of drawing two washers from 9 objects is (1/9c2) = (1/36)

The probability of drawing 3 nuts out of remaining 7 objects is (1/ 7c3) = (1/35)

The probability of drawing 4 bolts out of remaining 4 bolts is 1.

Therefore required probability (1/ (36x35)) = (1/1260)
Question:-
Option (A)
4.4
Option (B)
8.8
Option (C)
17.6
Option (D)
35.2
Correct Option:
B
Question Solution:
Direct load=1000N on the bracket fixed at wall. There is no eccentricity therefore P at each rivet = (1000 /4) = 250N

\ Shear Stress (c) = (P/A) = (250 / (p/4xb2) = 8.8 Mpa
Question:-
Option (A)
0.08
Option (B)
5.0
Option (C)
7.0
Option (D)
7.8
Correct Option:
B
Question Solution:
d = 5mm; L=100mm; K = 400 w/mk; T0 = 130°C; T¥ = 30°c; H= 40W/m2k

Q = Ö(hpkA)(T0 - T¥)tanhml, A= (p/4)d2 = 1.96x10-5, p = pd = 0.0157

m = Ö(hp/KA) = 8.94997; ml = 0.895 \ Q = 5.0w
Question:-
Option (A)
177
Option (B)
354
Option (C)
500
Option (D)
707
Correct Option:
B
Question Solution:
We know that

Pu = (p /(H3 /2))

i.e. (1000 / 403/2) = (p /203/2 ® Power developed at two different heads, P=353.55KW
Question:-
Option (A)
37
Option (B)
74
Option (C)
118
Option (D)
237
Correct Option:
A
Question Solution:
Dj = 50mm; dB = 50.05mm; IB = 20mm, N = 1200 rpm; m = 0.03 pa-sec

Frictional factor (f) = shear stress x Area Þ (mu / h)xpdj|l

H = rb - rj Þ 0.00025m

U = (pDjN /60) = (px0.05x1200 / 60) = 3.141

(f) = (0.03x3.141 / 0.00025) xp(0.05)(0.02)

(Torque) T = fxrj

T = (0.03x3.141 / 0.00025)x p(0.05)(0.02)x0.025 = 0.2961

Power = Txw = (2pNT /60) = ((2px1200x0.2961) / 60) = 37.2w