Question

give correct answer. ​

Answer 1

$\huge\bf{\underline{\color{red}GIVEN}}$

• ☞︎︎︎ First of all, here we elaborate the diagram of this problem i.e. attached in part-1 (See the attachment diagram.) .

• ☯︎ It is given that a metre scale PQ of uniform thickness and weighing $\bf{100~g_{wt}}$ is suspended by a string. And the point at which the scale is attached that point be ‘R’. At point ‘P’ a load of $\bf{85~g_{wt}}$ is attached.

And consider that at point ‘Q’ a load of $\bf{m_o~g_{wt}}$ is attached .

Now it is given that,

$\pink\longmapsto~\bf\gray{PR~:~QR~=~2~:~3~}$

So,

Consider that $\bf{\red{PR~=~2L~\&~QR~=~3L}}.$

$\huge\bf{\underline{\color{orange} SOLUTION}}$

$:\implies~\bf{\dfrac{PR}{QR}~=~\dfrac{2}{3}}$

$:\implies~\bf{PR~=~\dfrac{2}{3}~QR}$

$:\implies~\bf{QR~=~\dfrac{3}{2}~PR}--(i)$

☯︎ PQ is a metre scale. So the length of the scale is 100cm.

• ➝ PQ = 100 cm

➝ PR + QR = 100 cm

• ➝ PR = 100 - QR

• ➝ PR = 100 - $\rm{\dfrac{3}{2}\:PR~}$ (from i)

• ➝ PR + $\rm{\dfrac{3}{2}\:PR~}$ = 100

• ➝ $\rm{\dfrac{5}{2}\:PR~}$ = 100

• ➝ PR = $\sf{100\times{\dfrac{2}{5}}~}$

• ➝ PR = 40 cm

• ➝ QR = 60 cm

☯︎ If we put “PQ = $\bf{100~g_{wt}}$”, then

➝ PR = $\bf{40~g_{wt}}$

➝ QR = $\bf{60~g_{wt}}$

✈︎ Now, for the scale to be equilibrium

$:\implies~\bf{\Big(40\times{\dfrac{2L}{2}}\Big)~+~85\times{2L}~=~\Big(60\times{\dfrac{3L}{2}}\Big)~+~m_o\times{3L}~}$

$:\implies~\bf{40L~+~170L~=~30\times{3L}~+~3m_o~L~}$

$:\implies~\bf{210L~=~90L~+~3m_o~L~}$

$:\implies~\bf{3m_o~L~=~210L~-~90L}$

$:\implies~\bf{3m_o~L~=~120L~}$

$:\implies~\bf{m_o~=~\dfrac{120L}{3L}~}$

$:\implies~\bf\green{m_o~=~40~g_{wt}~}$

$\huge\red\therefore$ The load attached at Q is weighing $\bf{\pink{40~g_{wt}}}.$

☯︎ Now, it is given that S is a point on the scale such that $\bf\gray{PR~:~SR~=~2~:~1~}$ and the load at Q is shifted to S. (See the attachment diagram part-2.)

So,

Consider that $\bf{\red{PR~=~2L~\&~SR~=~L}}.$

$:\implies~\bf{\dfrac{PR}{SR}~=~\dfrac{2}{1}}$

$:\implies~\bf{SR~=~\dfrac{1}{2}\times{PR}}$

➪ If PR = 40 cm,

$:\implies~\bf{SR~=~\dfrac{1}{2}\times{40}}$

$:\implies~\bf{SR~=~20cm~}$

➪ If PR = $\bf{40~g_{wt}}$

$:\implies~\bf{SR~=~20~g_{wt}~}$

☯︎ So it is sure that the load at Q is shifted towards the point R.

✈︎ If $\bf{m'_o}$ is the mass of load attached at S, then for the scale to be equilibrium

$:\implies~\bf{85\times{2L}~-~\Big(40\times{\dfrac{2L}{2}}\Big)~=~m'_o\times{L}~+~\Big(20\times{\dfrac{L}{2}}\Big)~}$

$:\implies~\bf{170L~-~40L~=~m'_o\times{L}~+~10L~}$

$:\implies~\bf{130L~=~m'_o\times{L}~+~10L~}$

$:\implies~\bf{130L~-~10L~=~m'_o\times{L}~}$

$:\implies~\bf{120L~=~m'_o\times{L}~}$

$:\implies~\bf\green{m'_o~=~120~g_{wt}~}$

Now,

$\longmapsto~\bf{Change~in~\%~=~\dfrac{120~-~40}{40}\times{100}~}$

$\longmapsto~\bf{Change~in~\%~=~\dfrac{80}{40}\times{100}~}$

$\longmapsto~\bf{Change~in~\%~=~2\times{100}~}$

$\longmapsto~\bf\purple{Change~in~\%~=~200\%~}$

$\huge\red\therefore$ The load at point S, which is shifted from Q is weighing $\bf{\pink{120~g_{wt}}}.$ And the change in percentage is 200% .

Answer 2

Answer:

☞︎︎︎ First of all, here we elaborate the diagram of this problem i.e. attached in part-1 (See the attachment diagram.) .

☯︎ It is given that a metre scale PQ of uniform thickness and weighing \bf{100~g_{wt}}100 g

wt

is suspended by a string. And the point at which the scale is attached that point be ‘R’. At point ‘P’ a load of \bf{85~g_{wt}}85 g

wt

is attached.

And consider that at point ‘Q’ a load of \bf{m_o~g_{wt}}m

o

g

wt

is attached .

Step-by-step explanation:

$thx$