Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
246 views
in General by (95.4k points)
closed by
A solid shaft of 100 mm diameter transmits 160 HP at 200 rpm. The modulus of rigidity G = 8 × 105 kg/cm2. Then the maximum angle of twist for a length of 6 meter is
1. 5° 
2. 2.5°
3. 3.2° 
4. 2° 

1 Answer

0 votes
by (95.2k points)
selected by
 
Best answer
Correct Answer - Option 2 : 2.5°

Concept:

We know the power transmitted by the solid shaft is given by,

P = \(\frac{{2\pi NT}}{{60}}\) [where N = angular velocity of the shaft in rpm, T = transmitted torque]

Now, according to the torsion equation,

\(\frac{T}{J}\; =\frac{\tau_{max}}{r}= \;\frac{{G\theta }}{l}\)

where, T = transmitted torque, J = polar moment of inertia, G = modulus of rigidity, θ = angle of twist, l = length of the shaft

Calculation:

Given:

The power transmitted by the solid shaft = 160 HP = 160 × 746 W = 119360 W = 119.360 kW, d = 100 mm, G = 8 × 105 kg/cm2,  N = 200 rpm, l = 6 m

So, 119.36 = \(\frac{{2 \times 3.14 \times 200 \times T}}{{60}}\)

⇒  T = 5.70 kN/m

According to the torsion equation,

\(\frac{{5.70 \times 1000}}{{\frac{\pi }{{32}}{1^4}}} = \frac{{8 \times {{10}^5} \times {{10}^4} \times \theta }}{{6 \times 1000}}\)

\(\frac{{32 \times 5.70 \times 1000}}{{\pi \times {1^4}}} = \frac{{8 \times {{10}^9} \times \theta }}{{6 \times 1000}}\)

θ ≈ 0.044 rad = \(\frac{{0.044 \times 180}}{\pi } \approx 2.5^\circ \)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...