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Test: Torsion Level - 2 - Mechanical Engineering MCQ


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10 Questions MCQ Test Strength of Materials (SOM) - Test: Torsion Level - 2

Test: Torsion Level - 2 for Mechanical Engineering 2024 is part of Strength of Materials (SOM) preparation. The Test: Torsion Level - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Torsion Level - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Torsion Level - 2 below.
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Test: Torsion Level - 2 - Question 1

A shaft of 9.0m length is subjected to a torque of 300.0kN-m at a distance of 3.0m from one end. The maximum torque in the shaft is

Detailed Solution for Test: Torsion Level - 2 - Question 1

Short-cut method

TB = = 100kN­m

∴ Maximum torque = 200kN­m

Test: Torsion Level - 2 - Question 2

The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal to

Detailed Solution for Test: Torsion Level - 2 - Question 2
Given, θ = 3° = 3 ×

I = 6m; T = 14 kN − m; G = 83GN/m2

= 118mm

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Test: Torsion Level - 2 - Question 3

A shaft of J polar moment of inertia and C modulus of rigidity is fixed at one end and subjected to torque T at the free end and the same torque at mid length in opposite direction as shown in figure then the difference in the twist between the free end and the midpoint is equal to

Detailed Solution for Test: Torsion Level - 2 - Question 3

We can divide this shaft into two parts AB and BC

θC − θB is asked

θB = 0

Since there is no torque in AB portion

θC − θB =

Test: Torsion Level - 2 - Question 4

If the diameter of a shaft subjected to torque alone is doubled, then horse power P can be increase to

Detailed Solution for Test: Torsion Level - 2 - Question 4
Given: P =

P ∝ T But, T ∝ d3

P ∝ d3

P2 = 8P1

Test: Torsion Level - 2 - Question 5

The ratio of torque carrying capacity of a solid shaft to that of a hollow shaft is given Where K = = Inside diameter of hollow shaft and Do = Outside diameter of hollow shaft. Shaft material is the same.

Detailed Solution for Test: Torsion Level - 2 - Question 5

ZP = Polar section modulus Since the material is same, τper will be the same for hollow and solid shaft.

Test: Torsion Level - 2 - Question 6

Shafts of the same material and same lengths are subjected to the same torque. If the first shaft is a solid circular section (D), and the second shaft is of hollow circular section whose internal diameter is 2⁄3 of the outside diameter (Do ), and the max. shear stress developed in each is same D ⁄ D0 = ?

Detailed Solution for Test: Torsion Level - 2 - Question 6
Maximum shear stress is same τsolid = τhollow

Zhollow = Zsolid

Test: Torsion Level - 2 - Question 7

The ratio of the strength of a solid shaft and hollow shaft of the same external diameter and internal diameter half of the external diameter is

Detailed Solution for Test: Torsion Level - 2 - Question 7

= 1.07

=

= 1.07

Test: Torsion Level - 2 - Question 8

Two shafts A and B are of same length and subjected to same torque T. If the diameter of shaft B is twice that of shaft A, the ratio of shear stresses developed in shafts A and B is

Detailed Solution for Test: Torsion Level - 2 - Question 8
Given: We know,

τ =

= 23 = 8

Test: Torsion Level - 2 - Question 9

Two shafts A and B made of the same material transmit 100kW each. Shaft A turns at 250rpm while B at 300rpm. Which one has greater diameter?

Detailed Solution for Test: Torsion Level - 2 - Question 9
P = 2πNT

↑ dia: ↓ N = ′A′

Test: Torsion Level - 2 - Question 10

Find the maximum shear stress induced in a solid circular shaft of diameter 200mm when the shaft transmits 190 kW power at 200 rpm.

Detailed Solution for Test: Torsion Level - 2 - Question 10
Power, P =

190 × 103 =

τ = 9.071kN/m

τ =

τ = 5.78N/mm2

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