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This mock test of Test: Design Of Bearings (Level - 1) for Mechanical Engineering helps you for every Mechanical Engineering entrance exam.
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QUESTION: 1

In thrust bearing, the load acts

Solution:

QUESTION: 2

If the load on a ball bearing is reduced to half, the life of the ball bearing will be B

Solution:

Given, Initially, w_{1} = w

Life = L_{1}

After half of load, w_{2} =w/2

Life = L_{2}

Considering the relation,

Lw^{3 }= c

L_{2} = 8L_{1} times

QUESTION: 3

Which of the following is not the part of roller bearing?

Solution:

QUESTION: 4

The life of bearing is expressed in

Solution:

Life of bearings is generally expressed as millions of revolution.

QUESTION: 5

Spherical roller bearings are normally used

Solution:

QUESTION: 6

Antifriction bearings are termed as

Solution:

QUESTION: 7

The rolling element bearings are

Solution:

QUESTION: 8

In a journal bearing P = average bearing pressure, Z = absolute viscosity of the lubricant, N = rotational speed of the journal. The bearing characteristic number is given by

Solution:

QUESTION: 9

With a dynamic load capacity of 2.2 kN, a roller bearing can operate at 600 rpm for 2000 hours. Then its maximum radial load will be equal to

Solution:

Given: Dynamic load capacity (C) = 2.2 kN

Life of bearing (L_{f}) = 60 x N x time duration

= 60 x 600 x 2000

= 72 x 10^{6} revolutions

Using the following relation,

K = Constant = 3.3

On solving, we get, w = 609.8 (approx)

QUESTION: 10

A sliding bearing that can support steady loads without any relative motion between the journal and the bearing is called as

Solution:

QUESTION: 11

The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 rpm is 3000 hours. Then the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm will be

Solution:

Given: Load(w_{1 }9800 N

Speed(N_{1}) = 100 rpm

(Life) (L_{1})time duration = 3000 hours

Now, if load (w_{2}) = 4900 N

Speed(N_{2}) = 2000rpm

Then life(L_{2}) = ？

Considering the following,

QUESTION: 12

The dynamic load capacity of 6306 bearing is 22 kN. The maximum radial load it can sustain to operate at 600 rev/min, for 2000 hours will be equal to

Solution:

Given:dynamic load capacity = 22 x 10^{3} N

Speed(N) = 600 rav/min

Life = 2000 x 60 x 600

72 x 10^{6} rev

Using following relation,

w = 5288.24 N

w = 5.288 kN

QUESTION: 13

Which of the following assumptions regarding the lubricant film is made in Petroff’s equation?

Solution:

QUESTION: 14

In hydrodynamic bearings

Solution:

QUESTION: 15

The rated life of a bearing varies

Solution:

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