- 32
- 31
- 30.55
- 3.05

Option 3 : 30.55

Free

SSC JE ME Full Test 4

5500

200 Questions
200 Marks
120 Mins

Power transmitted by a belt:

P = (T_{1} – T_{2}).v

For maximum power:

T_{1} = T/3 where T_{1} = Tension in the tight side, T = Maximum tension to which the belt is subjected Velocity of the belt for maximum power:

\(v = \sqrt {\frac{T}{{3m}}} \) where m is mass of belt per unit length.

m = Area × length × density = b.t.l.ρ = 0.1 × 0.005 × 1000 = 0.5 kg/m

\(v = \sqrt {\frac{T}{{3m}}} = \sqrt {\frac{{1400}}{{3 \times 0.5}}} = 30.55\;m/s\)

India’s **#1 Learning** Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Trusted by 2,16,89,624+ Students

Start your FREE coaching now >>

Testbook Edu Solutions Pvt. Ltd.

1st & 2nd Floor, Zion Building,

Plot No. 273, Sector 10, Kharghar,

Navi Mumbai - 410210

[email protected]
Plot No. 273, Sector 10, Kharghar,

Navi Mumbai - 410210

Toll Free:1800 833 0800

Office Hours: 10 AM to 7 PM (all 7 days)