LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
94 views
in General by (24.2k points)
closed by
A belt drives a pulley of 200 mm diameter such that the ratio of tensions in the tight side and the slack side is 1.2, the maximum tension in the belt is not to exceed 240 kN. The speed of pulley is 60 rpm. Find the safe power transmitted by the pulley. 
1. 50 kW
2. 20 kW
3. 35 kW
4. 25 kW

1 Answer

0 votes
by (24.2k points)
selected by
 
Best answer
Correct Answer - Option 4 : 25 kW

Concept:

Power transmitted by a belt:

P = (T1 – T2)V

T1 = Tension on the tight side, T2 = Tension on the slack side, V = velocity of the belt

Calculation:

Given:

D = 200 mm = 0.2 m, \(\frac{{T_1}}{{T_2}}=1.2\), Maximum tension = 240 kN, N = 60 rpm

Since, Between T1 and T2, T1 is always greater than T2.

Therefore, T1 = 240 kN, \(\frac{{T_1}}{{T_2}}=1.2 = T_2=\frac{{240}}{{1.2}}=200~ kN\)

\(V=\frac{{\pi DN}}{{60}}=\frac{{\pi ~×~0.2~×~60}}{{60}}=0.628 ~m/s\)

Power transmitted by the pulley, P = (T1 – T2)V = (240 - 200) × 0.628 = 25.12 kW

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

...