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# in the figure, a man of true mass m is standing on a weighing machine placed in a cabin. the cabin is joined by a string with a body of mass m. assuming no friction, and negligible mass of cabin and weighing machine,

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get in the figure, a man of true mass m is standing on a weighing machine placed in a cabin. the cabin is joined by a string with a body of mass m. assuming no friction, and negligible mass of cabin and weighing machine, from screen.

## In the figure, a man of true mass m is standing on a weighing m

Solution for the question - in the figure, a man of true mass m is standing on a weighing machineplaced in a cabin. the cabin is joined by a string with a body

Physics- GENERAL Easy Question

## In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string with a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass)

measured mass of man is

acceleration of man is

acceleration of man is

measured mass of man is M.

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## In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string with a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is:

Click here👆to get an answer to your question ✍️ In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string with a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is: Question

## In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string with a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is: A

M+m Mm ​

B

M−m Mm ​

C

D

## depends on g

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

weight measure by weighting machine =M(g−a)

from F.B.D of lift T=m(g−a).....(1) from F.B.D of pully T=ma......(2) ma=m(g−a) mg=ma+ma mg=a(m+m) a= m+m M ​ g weight=m(g−a) =m(g− m+m mg ​ ) = m+m m(m+m−m)g ​ w= m+m mmg ​ mass= m+m mm ​ ​ Solve any question of Systems of Particles and Rotational Motion with:-

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## In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass)

Mg-T=Ma ..(i) T=ma ..(ii) Solving Eqs. (i) and (ii) , a=(Mg)/(M+m) FBD of man, Mg-N = Ma implies N=(Mmg)/((M+m)). Home > English > Class 11 > Physics > Chapter >

Newton's Laws Of Motion 1

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In fig., a man true mass M is ...

In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass) Updated On: 27-06-2022

00 : 30

Text Solution Open Answer in App A

The measured mass of man is

Mm (M+m) Mm(M+m) B

The acceleration of man is

mg (M+m) mg(M+m) . C

The acceleration of man is

Mg (M+m) Mg(M+m) . D

The measured mass of man is

Solution Mg−T=Ma Mg-T=Ma ..(i) T=ma T=ma ..(ii)

Solving Eqs. (i) and (ii) ,

a= Mg M+m a=MgM+m FBD of man, Mg−N=Ma⇒N= Mmg (M+m) Mg-N=Ma⇒N=Mmg(M+m) .  Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass)

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In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass)

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In fig., the mass of the man is M. Calculate the mass of the man as registered by weighing machine. Assume weighing machine, man, and wedge all are stationary.

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shows a man of mass 60 kg standing on a light weighing machine kept in a cabin of mass 40 kg The cabin is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself if the man manages to keep the cabin at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine?

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is standing on a weighing machine kept in a box of mass

as shown in the diagram If the man man ages to keep the box stationary, find the read ing of the weighing machine

30kg 30kg

.

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Mohammed 2 month ago

Guys, does anyone know the answer?