A uniform chain of length 2 m is kept on a table such that a length of  60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg . What is the work done in pulling the entire chain on the table ?

Option 1)

7.2 J

Option 2)

3.6 J

Option 3)

120 J

Option 4)

1200 J.

Answers (2)

As we learnt in

Definition of work done by variable force -

W=\int \vec{F}\cdot \vec{ds}

- wherein

\vec{F} is variable force and \vec{ds} is small displacement

 

 

Consider a small part dx at a depth x from table. 

Work done in lifting this small portion is dw = dm gx

Total work done =\int dw=\int_{0}^{h}(\frac{m}{l}dx)gx

    =\frac{mg}{l}\int_{0}^{h}x dx=\frac{4\times 10}{2}\times \frac{(0.6)^{2}}{2}

=\frac{mg}{l}\int_{0}^{h}x dx=\frac{4\times 10}{2}\times \frac{(0.6)^{2}}{2}=3.6J

Correct answer = 2


Option 1)

7.2 J

Option 2)

3.6 J

Option 3)

120 J

Option 4)

1200 J.

N neha

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions