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Question Number 163432 by Tawa11 last updated on 06/Jan/22
Answered by mr W last updated on 07/Jan/22
(5) W_(air) =Vρ W_(water) =V(ρ−ρ_(water) ) W_(liquid) =V(ρ−ρ_(liquid) ) W_(air) −W_(water) =Vρ_(water) W_(air) −W_(liquid) =Vρ_(liquid) (ρ_(liquid) /ρ_(water) )=((W_(air) −W_(liquid) )/(W_(air) −W_(water) ))=((0.30−0.27)/(0.30−0.25))=0.6 ⇒answer B
$$\left(\mathrm{5}\right) \\ $$$${W}_{{air}} ={V}\rho \\ $$$${W}_{{water}} ={V}\left(\rho−\rho_{{water}} \right) \\ $$$${W}_{{liquid}} ={V}\left(\rho−\rho_{{liquid}} \right) \\ $$$${W}_{{air}} −{W}_{{water}} ={V}\rho_{{water}} \\ $$$${W}_{{air}} −{W}_{{liquid}} ={V}\rho_{{liquid}} \\ $$$$\frac{\rho_{{liquid}} }{\rho_{{water}} }=\frac{{W}_{{air}} −{W}_{{liquid}} }{{W}_{{air}} −{W}_{{water}} }=\frac{\mathrm{0}.\mathrm{30}−\mathrm{0}.\mathrm{27}}{\mathrm{0}.\mathrm{30}−\mathrm{0}.\mathrm{25}}=\mathrm{0}.\mathrm{6} \\ $$$$\Rightarrow{answer}\:{B} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
Ohh, I have seen my mistake sir. I have used your workings to solve number 7 and 8. I don′t get number 6. It remaining only number 6 sir. I really appreciate sir. I got (7) C (8) C
$$\mathrm{Ohh},\:\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{my}\:\mathrm{mistake}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{used}\:\mathrm{your}\:\mathrm{workings}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{number}\:\:\mathrm{7}\:\:\mathrm{and}\:\:\mathrm{8}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{number}\:\:\mathrm{6}. \\ $$$$\mathrm{It}\:\mathrm{remaining}\:\mathrm{only}\:\mathrm{number}\:\:\mathrm{6}\:\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{got}\:\:\:\:\left(\mathrm{7}\right)\:\mathrm{C}\:\:\:\:\:\:\left(\mathrm{8}\right)\:\mathrm{C} \\ $$
Commented by mr W last updated on 07/Jan/22
if you really have understood the things, not just only blindly use my workings, then you should be able to solve all these questions. if you get a different result as the question gives, it can also mean that the question′s answer is wrong, not yours. what′s special in Q.6 to you such that you can′t solve?
$${if}\:{you}\:{really}\:{have}\:{understood}\:{the}\: \\ $$$${things},\:{not}\:{just}\:{only}\:{blindly}\:{use}\:{my}\: \\ $$$${workings},\:{then}\:{you}\:{should}\:{be}\:{able}\:{to}\: \\ $$$${solve}\:{all}\:{these}\:{questions}. \\ $$$${if}\:{you}\:{get}\:{a}\:{different}\:{result}\:{as}\:{the} \\ $$$${question}\:{gives},\:{it}\:{can}\:{also}\:{mean}\:{that} \\ $$$${the}\:{question}'{s}\:{answer}\:{is}\:{wrong},\:{not} \\ $$$${yours}. \\ $$$${what}'{s}\:{special}\:{in}\:{Q}.\mathrm{6}\:{to}\:{you}\:{such}\:{that} \\ $$$${you}\:{can}'{t}\:{solve}? \\ $$
Commented by Tawa11 last updated on 07/Jan/22
I got 5.33 I did not say: 60 − 5.33
$$\mathrm{I}\:\mathrm{got}\:\:\:\mathrm{5}.\mathrm{33} \\ $$$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{say}:\:\:\:\:\:\mathrm{60}\:\:\:−\:\:\:\mathrm{5}.\mathrm{33} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 07/Jan/22
you think 5.33 is wrong?
$${you}\:{think}\:\mathrm{5}.\mathrm{33}\:{is}\:{wrong}? \\ $$
Commented by Tawa11 last updated on 07/Jan/22
Yes sir
$$\mathrm{Yes}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
I got 5.33 So, I say new weight = 60 − 5.33
$$\mathrm{I}\:\mathrm{got}\:\:\:\mathrm{5}.\mathrm{33} \\ $$$$\mathrm{So},\:\mathrm{I}\:\mathrm{say}\:\:\:\mathrm{new}\:\mathrm{weight}\:\:\:\:=\:\:\:\mathrm{60}\:\:\:−\:\:\:\mathrm{5}.\mathrm{33} \\ $$
Commented by mr W last updated on 07/Jan/22
i can′t bring you to understand. i just show how it should be done.
$${i}\:{can}'{t}\:{bring}\:{you}\:{to}\:{understand}.\: \\ $$$${i}\:{just}\:{show}\:{how}\:{it}\:{should}\:{be}\:{done}. \\ $$
Commented by mr W last updated on 07/Jan/22
(6) W_(air) =Vρ ⇒V=(W_(air) /ρ) W_(wax) =V(ρ−ρ_(wax) ) =((W_(air) (ρ−ρ_(wax) ))/ρ)=(1−(ρ_(wax) /ρ))W_(air) =(1−((800)/(900)))×60≈6.7 N
$$\left(\mathrm{6}\right) \\ $$$${W}_{{air}} ={V}\rho\:\Rightarrow{V}=\frac{{W}_{{air}} }{\rho} \\ $$$${W}_{{wax}} ={V}\left(\rho−\rho_{{wax}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{W}_{{air}} \left(\rho−\rho_{{wax}} \right)}{\rho}=\left(\mathrm{1}−\frac{\rho_{{wax}} }{\rho}\right){W}_{{air}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}−\frac{\mathrm{800}}{\mathrm{900}}\right)×\mathrm{60}\approx\mathrm{6}.\mathrm{7}\:{N} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
That means the textbook is wrong sir. They choose (b). But your solution shows that all the answers is wrong
$$\mathrm{That}\:\mathrm{means}\:\mathrm{the}\:\mathrm{textbook}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{sir}.\:\mathrm{They}\:\mathrm{choose}\:\:\left(\mathrm{b}\right). \\ $$$$\mathrm{But}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{all}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by mr W last updated on 07/Jan/22
the textbooks says a metal has a density 900kgm^(−3) . it′s even more lightweight than water. does such a metal exist?
$${the}\:{textbooks}\:{says}\:{a}\:{metal}\:{has}\:{a} \\ $$$${density}\:\mathrm{900}{kgm}^{−\mathrm{3}} .\:{it}'{s}\:{even}\:{more} \\ $$$${lightweight}\:{than}\:{water}.\:{does}\:{such} \\ $$$${a}\:{metal}\:{exist}? \\ $$
Commented by Tawa11 last updated on 07/Jan/22
I understand you now sir. Thanks sir. I appreciate.
$$\mathrm{I}\:\mathrm{understand}\:\mathrm{you}\:\mathrm{now}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Tawa11 last updated on 07/Jan/22
Their mistake is that they wrote 900 instead of 9000 So that (1 − ((800)/(9000))) × 60 = 54.67
$$\mathrm{Their}\:\mathrm{mistake}\:\mathrm{is}\:\mathrm{that}\:\mathrm{they}\:\mathrm{wrote}\:\:\mathrm{900}\:\:\:\mathrm{instead}\:\mathrm{of}\:\:\:\mathrm{9000} \\ $$$$\mathrm{So}\:\mathrm{that}\:\:\:\:\left(\mathrm{1}\:\:−\:\:\frac{\mathrm{800}}{\mathrm{9000}}\right)\:×\:\mathrm{60}\:\:\:\:\:=\:\:\:\:\mathrm{54}.\mathrm{67} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
I appreciate sir.
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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