A block of aluminum of mass 1Kg and volume 3.6 X 10^-4m^3 is suspended

1.	A block of aluminum of mass 1Kg and volume 3.6 X 10^-4m^3 is suspended from a string and then completelyimmersed in a container of water. The decrease in tension in the string after immersion is(Use g =10 m/s^2)
Answer» Answer: 3.6 Explanation: when block is in air,tension=weight of block as accelaration is zero. so tension when block is in air=mg=10N when immersed in WATER, tension=weight of block-buoyant FORCE buoyant force = volume of substancedensity of liquidacceleration DUE to gravity B.F.=3.6/10000100010=3.6N tension=weight-3.6 tension=10-3.6=6.4N decrease in tension=10-6.4=3.6N hope it helps you.

A block of aluminum of mass 1Kg and volume 3.6 X 10^-4m^3 is suspended from a string and then completelyimmersed in a container of water. The decrease in tension in the string after immersion is(Use g =10 m/s^2)

Answer»

Answer:

3.6

Explanation:

when block is in air,tension=weight of block as accelaration is zero.

so tension when block is in air=mg=10N

when immersed in WATER,

tension=weight of block-buoyant FORCE

buoyant force = volume of substance*density of liquid*acceleration DUE to gravity

B.F.=3.6/10000*1000*10=3.6N

tension=weight-3.6

tension=10-3.6=6.4N

decrease in tension=10-6.4=3.6N

hope it helps you.

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A block of aluminum of mass 1Kg and volume 3.6 X 10^-4m^3 is suspended from a string and then completelyimmersed in a container of water. The decrease in tension in the string after immersion is(Use g =10 m/s^2)​

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A block of aluminum of mass 1Kg and volume 3.6 X 10^-4m^3 is suspended from a string and then completelyimmersed in a container of water. The decrease in tension in the string after immersion is(Use g =10 m/s^2)