R P Goyal and S P Tripathi Solutions for Chapter: Force, Exercise 2: EXERCISE-1(B)

Write two factors on which moment of force about a point depends.

A uniform metre rule is pivoted at its mid-point. A weight of $50 \mathrm{gf}$ is suspended at one end of it. Where should a weight of $100 \mathrm{gf}$ be suspended to keep the rule horizontal?

A uniform metre rule balances horizontally on a knife-edge placed at the $58 \mathrm{cm}$ mark when a weight of $20 \mathrm{gf}$ is suspended from one end.
(i) Draw a diagram of the arrangement. (ii) What is the weight of the rule?

The diagram below shows a uniform bar supported at the middle point $O$. A weight of $40 \mathrm{gf}$ is placed at a distance $40 \mathrm{cm}$ to the left of point $O$. How can you balance the bar with a weight of $80 \mathrm{gf}$?

When a boy weighing $20 \mathrm{kgf}$ sits at one end of a $4 \mathrm{m}$ long see-saw, it gets depressed at this end. How can it be brought to the horizontal position by a man weighing $40 \mathrm{kgf}$.

A uniform metre scale can be balanced at the $70.0 \mathrm{cm}$ mark when a mass $0.05 \mathrm{kg}$ is hung from the $94.0 \mathrm{cm}$ mark. Draw a diagram of the arrangement.

A uniform metre scale can be balanced at the $70.0 \mathrm{cm}$ mark when a mass $0.05 \mathrm{kg}$ is hung from the $94.0 \mathrm{cm}$ mark. Find the mass of the metre scale.

A uniform metre rule of mass $100 \mathrm{g}$ is balanced on a fulcrum at mark $40 \mathrm{cm}$ by suspending an unknown mass $m$ at the mark $20 \mathrm{cm}$.

(i) Find the value of $m$.
(ii) To which side the rule will tilt if the mass $m$ is moved to the mark $10 \mathrm{cm}$? 
(iii) What is the resultant moment now?
(iv) How can it be balanced by another mass of $50 \mathrm{g}$?