# Answer on Mechanics | Relativity Question for jojo

Question #72345

A solid iron rectangular block of dimensions 127x52x15 mm and a density of 7860 kg/ m3 is suspended in

brine solution. When the iron block is completely immersed in the solution.Determine

i) the upthrust on the Iron block

ii) tension in the supporting cable

brine solution. When the iron block is completely immersed in the solution.Determine

i) the upthrust on the Iron block

ii) tension in the supporting cable

Expert's answer

The volume of the block is 127 X 52 X 15 mm^3

=0.127 × 0.052 × 0.015 m^3

=9.906 × 10^-5 m^3

Again density of the block was = 1786 kg/m^3

Therefore the mass of the block will be = 9.906 ×10^-5 ×1786 kg

=0.1769 kg

Again , the density of saturated brine solution at room temperature = 1202 kg / m^3

i) The upthrust will be = volume of displaced liquid × density of the liquid × g

= 9.906 × 10^-5 × 1202 x 9.8 N

=1.167 N

ii) tension of the supporting thread is

= Weight of the block - upthrust

= 0.1769 × 9.8 - 1.167 N

= 1.734 - 1.167 N

= 0.033 N

=0.127 × 0.052 × 0.015 m^3

=9.906 × 10^-5 m^3

Again density of the block was = 1786 kg/m^3

Therefore the mass of the block will be = 9.906 ×10^-5 ×1786 kg

=0.1769 kg

Again , the density of saturated brine solution at room temperature = 1202 kg / m^3

i) The upthrust will be = volume of displaced liquid × density of the liquid × g

= 9.906 × 10^-5 × 1202 x 9.8 N

=1.167 N

ii) tension of the supporting thread is

= Weight of the block - upthrust

= 0.1769 × 9.8 - 1.167 N

= 1.734 - 1.167 N

= 0.033 N

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