Question

A uniform metre scale can be balanced at the 70.0 cm mark when a mass of 0.05 kg is hung from the 94.0 cm
mark
a. Draw a diagram of the arrangement.
b. Find the mass of the metre scale.
(Ans. b. 60 g)
plzzz draw the diagram​

Answer 1

Gɪᴠᴇɴ :-

• A uniform metre scale can be balanced at the 70 cm mark , when a mass of 0.05 kg is hung from 94 cm mark .

Tᴏ FɪɴD :-

• (a) Draw a diagram of the arrangement.
• (b) Mass of the metre scale .

AɴsᴡᴇR :-

$\sf ( For \ diagram \ refer \ to \ attachment :- )$

$\underline{\red{\sf \mathscr{A}ccording \ \mathscr{T}o\ \mathscr{Q} uestion :- }}$

$\sf\qquad\blue{\bullet} \: Length \ of \ scale \ = \ 1m \ = 100cm . \\\\\sf\qquad\blue{\bullet}\: Scale \ is\ of\ uniform \ mass .\\\\\sf\qquad\blue{\bullet}\:Mass\ of \ 0.05kg \ is \ hanged\ from \ 94cm \ mark.$

Now , here the lenght of the scale is 100 cm . Since the scale is of uniform mass , its centre of gravity will lie at the middlemost position. That is 100cm ÷ 2 = 50 cm .

$\underline{\purple{\sf Diagram\ of \ arrangement :- }}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\put(7.5,3.5){\vector(0,-1){1.5}}\put(5,1.23){\line(1,0){2}}\put(5,1.23){\line(1,2){1}}\put(6,3.23){\line( 1, - 2){1}}\put(4.2,3.5){\vector(0,-1){1.5}}\put(4,4){ \bf 50cm } \put(4,4.5){ \bf 1m } \put(4.59,4.5){\vector(1,0){3.1}}\put(3.9,4.5){\vector(-1,0){3.9}}\linethickness{0.5cm}\put(0,3.5){\line(1,0){8}}\put(7.5,4){ \bf 94cm } \put(5.7,4){ \bf 70cm } \put(7,1.75){\line(1,0){1}}\put(3.5,1.5){ \bf C.O.G }\end{picture}$

$\rule{200}2$

$\Large\underline{\boxed{\orange{\tt Law \ of \ Moment :-} }}$

According to this law , when a number of parallel forces act on a rigid body and the body is in equilibrium then the algebraic sum of anticlockwise moments of force is equal to the clockwise moments about any point .

$\rule{200}2$

$\underline{\sf{\pink{Hence \ here \ According \ to \ this \ law :- }}}$

$\tt:\implies m_{scale} \times AO = m_{hanged} \times OC \\\\\tt:\implies m_{scale} \times (70cm - 50cm ) = (94cm - 70cm )\times 0.05kg \\\\\tt:\implies m_{scale}\times 20 cm = 0.05 kg \times 24 cm \\\\\tt:\implies m_{scale}=\bigg\lgroup \dfrac{0.05kg \times 24cm }{20cm }\bigg\rgroup\\\\\tt:\implies m_{scale}= \bigg\lgroup\dfrac{5\times 24 cm }{100\times 20 }\bigg\rgroup \\\\\tt:\implies m_{scale}=\dfrac{6}{100}kg \\\\\underline{\boxed{\red{\tt\longmapsto Mass_{scale}= 0.06 kg = 60g }}}$

$\boxed{\green{\bf \pink{\dag}\:\:Hence\ the \ mass \ of \ the \ scale \ is \ 60g .}}$

Answer 2

$\large\bf{\underline\red{Good \: Afternoon♡}}$

Question :-

A uniform metre scale can be balanced at the 70.0 cm mark when a mass of 0.05 kg is hung from the 94.0 cm

a. Draw a diagram of the arrangement.

b. Find the mass of the metre scale.

Answer :-

a. Diagram of the given arrangement is shown above. ↑

b. As the given meter scale is a uniform scale. So its centre of gravity lies at 50 cm.

Let mass of meter scale be W1 kg.

By principle of moments,

m1x1 = m2x2

m1 x (70 - 50) = 0.05 x (94 - 70)

m1 = 0.05 x 24/20

= 0.06 kg.

= 60 gm.