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The longest leaf in a leaf spring is called centre leaf.
A
It is called middle leaf
B
It is called master leaf
C
Yes
D
None of the listed
Correct Answer:
It is called master leaf
It is called master leaf.
A Pickering governor is driving a gramophone. Each disc attached to the centre of a leaf spring has a mass of 20 g. The width of each spring is 5 mm and thickness of 0.125 mm. The effective length of each spring is 40 mm. The distance from the spindle axis to the centre of gravity of the mass when the governor is at rest, is 10 mm. Find the speed of the governor in rpm when the sleeve has risen to a height of 0.8 mm. (E = 210kN/mm2)
A
25.5
B
2.43
C
51.0
D
12.25
A Pickering governor is driving a gramophone. Each disc attached to the centre of a leaf spring has a mass of 20 g. The width of each spring is 5 mm and thickness of 0.125 mm. The effective length of each spring is 40 mm. The distance from the spindle axis to the centre of gravity of the mass when the governor is at rest, is 10 mm, find the speed of the turntable in rpm if ratio of the governor speed to the turntable speed is 10.5.
A
25.5
B
2.43
C
51.0
D
12.25
When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the deflection of the spring ($$\delta $$) is given by (where d = Diameter of spring wire, n = No. of turns of the spring and C = Modulus of rigidity for the spring material)
A
$$\frac{{{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
B
$$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
C
$$\frac{{4{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
D
$$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$
A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be
A
$$\frac{\tau }{2}$$
B
$$\tau $$
C
$$2\tau $$
D
$$4\tau $$
An open coiled helical compression spring 'A' of mean diameter 50 mm is subjected to an axial load ‘W’. Another spring 'B' of mean diameter 25 mm is similar to spring 'A' in all respects. The deflection of spring 'B' will be __________ as compared to spring 'A'.
A
One-eighth
B
One-fourth
C
One-half
D
Double
The part which permits variation in the distance between the spring eyes of a leaf spring as the spring flexes, is called
A
Shackle
B
U-bolt
C
Leaf
D
Bush
Given are the steps to draw a perpendicular to a line at a point within the line when the point is near the centre of line. Arrange the steps. Let AB be the line and P be the point in it i. Join F and P which is perpendicular to AB. ii. Now C as centre take the same radius and cut the arc at D and again D as centre with same radius cut the arc further at E. iii. With centre as P take any radius and draw an arc (more than a semicircle) cuts AB at C. iv. Now D, E as centre take radius (more than half of DE) draw arcs which cut at F.
A
i, iv, ii, iii
B
iii, ii, iv, i
C
iv, iii, i, ii
D
ii, i, iv, iii
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B'. The ratio of deflections in spring 'A' to spring 'B' is
A
$$\frac{1}{8}$$
B
$$\frac{1}{4}$$
C
$$\frac{1}{2}$$
D
2
When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)
A
$$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
B
$$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
C
$$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
D
$$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
When an open coiled helical compression spring is subjected to an axial load (W), the compression produced in the spring will be (where, n = No. of active turns of the spring and G = Modulus of rigidity for the spring material)
A
$$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{G}}{{\text{d}}^4}}}$$
B
$$\frac{{4{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{G}}{{\text{d}}^4}}}$$
C
$$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{G}}{{\text{d}}^4}}}$$
D
$$\frac{{16{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{G}}{{\text{d}}^4}}}$$